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maximize-total-cost-of-alternating-subarrays | C++ solution with explanation | c-solution-with-explanation-by-roy258-h7g0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | roy258 | NORMAL | 2024-07-20T07:45:39.589326+00:00 | 2024-07-20T07:45:39.589349+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n#define ll long long \nclass Solution {\npublic:\n // for every number we have two choose either add or subtract it\n // if we add it will add to max of both cases addResult and subtractResult\n // if we subtract it we want to subtract from addResult as we are trying to maximise addResult\n long long maximumTotalCost(vector<int>& nums) {\n ll addResult = nums[0];\n ll subtractResult = nums[0];\n for(int i=1;i<nums.size();i++){\n ll a = max(addResult,subtractResult) + nums[i];\n ll b = addResult - nums[i];\n \n addResult = a;\n subtractResult = b;\n }\n return max(addResult,subtractResult); \n }\n};\n``` | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Greedy Solution with Explanation | greedy-solution-with-explanation-by-umes-2by3 | Intuition\n Describe your first thoughts on how to solve this problem. \nwhenever a new_element is added in front of a subarray the sum of subarray(prev) become | umesh_346 | NORMAL | 2024-07-20T03:55:06.125562+00:00 | 2024-07-20T03:55:06.125587+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nwhenever a **new_element** is added in front of a subarray the sum of subarray(**prev**) becomes (**new_element**) - (**prev**) lets call it **temp**. \nif the **prev** is positive, solution maximizes if we split, \n**temp = new_element + prev**.\notherwise the new_element is added in subarray and \n**temp = new_element - prev**.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIterating Backwards, if **prev >= 0**, we split, \n**ans += prev** and update the prev with **prev = nums[i]**,\nelse only update the prev, **prev = nums[i]-prev**.\nAt the end return **ans + prev**.\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long prev = 0, ans = 0;\n for(int i = nums.size()-1; i>=0; i--){\n if(prev >= 0){\n ans += prev;\n prev = nums[i];\n }\n else{\n prev = nums[i]-prev;\n }\n }\n return ans + prev;\n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP | dp-by-112115046-gx03 | \n\n# Code\n\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n @cache\n def dp(i, ok):\n | 112115046 | NORMAL | 2024-07-18T07:27:00.619582+00:00 | 2024-07-18T07:27:00.619613+00:00 | 2 | false | \n\n# Code\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n @cache\n def dp(i, ok):\n if i == n: return 0\n if ok == 1:\n ans = nums[i] + dp(i+1, 0)\n else:\n ans = max(-nums[i] + dp(i+1, 1), dp(i,1))\n return ans\n return dp(0, 1)\n``` | 0 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | JAVA MEMO EASY SOLUTION USER FRIENDLY>>> | java-memo-easy-solution-user-friendly-by-4txp | Intuition\n Describe your first thoughts on how to solve this problem. \nIt is the simple DP Memoization Approach, Here in this case we can take or not-take the | ujjalmodak2000 | NORMAL | 2024-07-17T20:32:26.987364+00:00 | 2024-07-17T20:32:26.987385+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is the simple DP Memoization Approach, Here in this case we can take or not-take the next element of the array,If we don\'t take the next element,means it is not a subarray, so start from the next element to find a subarray. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIf we take the element of this Array, We take a multi variable where the initial **multi=-1**, \n We mulitiply the multi to **-1**, And also add the **nums[i]*(-multi)** which give me the negative next element and adding to the existing subarray, making the subarray sum. \n Now Atlast we do the **max(take,ntake)** to get the maximum total cost. \n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n Here the Time Complexity is O(n^2)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n Here the Time Complexity is O(n^2), Using The 2D Dp memoization.\n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n int n=nums.length;\n long dp[][]=new long[n][3];\n for(int i=0;i<n;i++){\n for(int j=0;j<3;j++){\n dp[i][j]=-1;\n }\n }\n return maximum(nums,0,-1,n,dp);\n }\n public long maximum(int nums[],int i,int multi,int n,long dp[][]){\n if(i>=n) return 0;\n if(dp[i][multi+1]!=-1) return dp[i][multi+1];\n long ntake=maximum(nums,i+1,-1,n,dp)+(long)nums[i];\n long take=maximum(nums,i+1,-1*multi,n,dp)+(long)nums[i]*(-multi);\n return dp[i][multi+1]=(long)Math.max(ntake,take);\n }\n}\n``` | 0 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Most Simple way to Understated | most-simple-way-to-understated-by-mentoe-taeg | \n\n# Approach\nrecursion : Take Not-take\n\n# Complexity\n- Time complexity : O(N)2 = O(N);\n\n- Space complexity:\nO(n2)\n\n# Code\n```\nclass Solution {\npub | Mentoes22 | NORMAL | 2024-07-17T12:41:52.817688+00:00 | 2024-07-17T12:41:52.817711+00:00 | 0 | false | \n\n# Approach\nrecursion : Take Not-take\n\n# Complexity\n- Time complexity : O(N)*2 = O(N);\n\n- Space complexity:\nO(n*2)\n\n# Code\n```\nclass Solution {\npublic:\n long long int helper(vector<int>& nums, int i, int state, vector<vector<long long int>>& dp) {\n if (i >= nums.size())\n return 0;\n\n if (dp[i][state] != -1)\n return dp[i][state];\n\n long long int includ = -1e16;\n long long int exclud = -1e16;\n\n if (state == 0) {\n includ = -1 * (nums[i]) + helper(nums, i + 1, 1, dp);\n } else if (state == 1) {\n includ = (nums[i]) + helper(nums, i + 1, 0, dp);\n }\n\n exclud = nums[i] + helper(nums, i + 1, 0, dp);\n\n return dp[i][state] = max(includ, exclud);\n }\n\n long long int maximumTotalCost(vector<int>& nums) {\n if (nums.size() <= 1)\n return nums[0];\n\n vector<vector<long long int>> dp(nums.size() + 1, vector<long long int>(2, -1));\n\n long long int temp1 = 0;\n temp1 = helper(nums, 0, 1, dp);\n\n return temp1;\n }\n};\n | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 5 Lines of code: simple DP with constant space. 0ms, beats 100% | 5-lines-of-code-simple-dp-with-constant-q90fk | Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(1)\n\n# Code\n\nimpl Solution {\n pub fn maximum_total_cost(nums: Vec<i32>) -> i64 { \n | germanov_dev | NORMAL | 2024-07-14T17:59:05.261536+00:00 | 2024-07-14T18:07:38.607491+00:00 | 1 | false | # Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nimpl Solution {\n pub fn maximum_total_cost(nums: Vec<i32>) -> i64 { \n let (mut dp0, mut dp1) = (nums[0] as i64,nums[0] as i64);\n for idx in 1..nums.len() {\n (dp0, dp1) = (dp0.max(dp1) + nums[idx] as i64,dp0 - nums[idx] as i64); \n }\n dp0.max(dp1) \n }\n}\n``` | 0 | 0 | ['Rust'] | 0 |
maximize-total-cost-of-alternating-subarrays | Simple 2D dp | simple-2d-dp-by-tayal-np1h | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | tayal | NORMAL | 2024-07-13T08:22:05.471103+00:00 | 2024-07-13T08:22:05.471119+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n\n long long compute(int ind, int sign, vector<vector<long long>>& dp, int& N, vector<int>& nums){\n if(ind == N){\n return 0;\n }\n\n if(dp[ind][sign] != -1)\n return dp[ind][sign];\n\n // take this element with prev element\n long long ans; \n\n ans = (sign == 1 ? nums[ind] : -1 * nums[ind]) + compute(ind + 1, 1 - sign, dp, N, nums);\n\n // start a-new\n ans = max(ans, nums[ind] + compute(ind+1, 0, dp, N, nums));\n\n return dp[ind][sign] = ans;\n }\n\n long long maximumTotalCost(vector<int>& nums) {\n int N = nums.size();\n vector<vector<long long>> dp(N + 10, vector<long long>(2, -1L));\n return compute(0, 1, dp, N, nums); \n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | JAVA - 100% Faster - (2D) DP - (Positive - Negative) Approach | java-100-faster-2d-dp-positive-negative-vrg2i | \n# Code\n\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n \n int n = nums.length;\n long add =0, sub=0;\n if(n= | AbhirMhjn | NORMAL | 2024-07-13T03:41:33.832133+00:00 | 2024-07-13T03:41:33.832156+00:00 | 1 | false | \n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n \n int n = nums.length;\n long add =0, sub=0;\n if(n==1){\n return nums[0];\n }\n // long[][] dp = new long[n][2];\n \n // dp[0][0] = nums[0];\n // dp[0][1] = nums[0];\n add = nums[0];\n sub = nums[0];\n\n for(int i=1;i<n;i++){\n long temp = Math.max(add, sub) +nums[i];\n sub = add - nums[i];\n add = temp;\n }\n\n return Math.max(add,sub);\n }\n}\n``` | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Clean solution | clean-solution-by-hawtinzeng-e2c1 | Intuition\n Describe your first thoughts on how to solve this problem. \ndon\'t rely on the test set, maybe the test set will give you some misunderstanding.\n\ | HawtinZeng | NORMAL | 2024-07-09T01:12:46.047864+00:00 | 2024-07-09T01:12:46.047882+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ndon\'t rely on the test set, maybe the test set will give you some misunderstanding.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/*\n addSum\n subSum\n*/\n\nfunction maximumTotalCost(nums: number[]): number {\n let addSum = nums[0], subSum = nums[0];\n\n nums.forEach((n, i) => {\n if (i === 0) return;\n const tempAdd = Math.max(addSum + n, subSum + n);\n subSum = addSum - n;\n addSum = tempAdd;\n\n })\n\n return Math.max(addSum, subSum);\n};\n\n``` | 0 | 0 | ['TypeScript'] | 0 |
maximize-total-cost-of-alternating-subarrays | Python || DP || Binary length Subarray | python-dp-binary-length-subarray-by-in_s-peku | Consider Subarray of length 1 or 2 to maximazie your answer.\nTC : O(n)\n\nCode:\n\n\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n | iN_siDious | NORMAL | 2024-07-04T18:12:18.002937+00:00 | 2024-07-04T18:12:18.002974+00:00 | 1 | false | Consider Subarray of length 1 or 2 to maximazie your answer.\nTC : O(n)\n\nCode:\n\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n=len(nums)\n @cache\n def dp(idx):\n if idx>=n: return 0\n #take one length subarray\n ans=dp(idx+1)+nums[idx]\n #check and take two length subarray\n if idx+1<n: ans=max(ans,dp(idx+2)+nums[idx]-nums[idx+1])\n return ans\n return dp(0)\n```\n | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | Clean java solution | DP easy to understand | clean-java-solution-dp-easy-to-understan-ev0j | Code\n\nclass Solution {\n Map<String, Long> memo;\n public long maximumTotalCost(int[] nums) {\n memo = new HashMap<>();\n return dfs(nums, | SG-C | NORMAL | 2024-07-04T12:22:43.379979+00:00 | 2024-07-04T12:22:43.380009+00:00 | 5 | false | # Code\n```\nclass Solution {\n Map<String, Long> memo;\n public long maximumTotalCost(int[] nums) {\n memo = new HashMap<>();\n return dfs(nums, 0, true);\n }\n private long dfs(int[] nums, int i, boolean sign){\n if(i == nums.length)\n return 0;\n\n String state = i + "," + sign;\n if(memo.containsKey(state))\n return memo.get(state);\n\n long res = nums[i] * (sign ? 1: -1);\n res += Math.max(dfs(nums, i + 1, true), dfs(nums, i + 1, !sign));\n memo.put(state, res);\n\n return res;\n }\n}\n``` | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Recursion + Memoization | recursion-memoization-by-surendrapokala1-ocnd | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | surendrapokala111 | NORMAL | 2024-07-04T09:38:25.439949+00:00 | 2024-07-04T09:38:25.439984+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n long long maxi = 0;\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long n = nums.size();\n vector<vector<long long>> dp(n + 1, vector<long long>(3, -1));\n return solve(0, nums, 1, dp);\n }\n long long solve(int ind, vector<int>& nums, int flag, vector<vector<long long>>& dp) {\n if (ind >= nums.size()) return 0;\n if (dp[ind][flag+1] != -1) return dp[ind][flag+1];\n long long ans = nums[ind] * flag;\n long long res = solve(ind + 1, nums, flag * -1, dp);\n long long res1 = solve(ind+1, nums, 1, dp);\n return dp[ind][flag+1] = max(res, res1) + ans;\n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Elegant DP Solution with Clear Mathematical & Constructive Proof | elegant-dp-solution-with-clear-mathemati-ulel | Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition | Sambosa123 | NORMAL | 2024-07-04T04:03:26.068081+00:00 | 2024-09-26T23:19:35.056721+00:00 | 7 | false | # Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition-subtraction manner. This can be approached using dynamic programming by considering only subarrays of length 1 or 2, simplifying the calculation of costs.\n\nthat was my notepad notes, it might be helpful for someone.\n```\ncost(l, r) = a[l] - a[l + 1] .... a[r] * -1 ^ ((r % l) % 2)\n\ncost(l, r) = (a[l] - a[l + 1]) + (a[l + 2] - a[l + 3]) .....\n\ncost(l, r) = cost(l, l + 1) + cost(l + 2, l + 3) + .... \n\nso to construct cost(l, r), we can only assume \nthat we will only take subarrays of length 1 or 2.\n\nstate : dp[i] represents maximum sum to make from prefix[0...i]\n\ndp[i] = max(dp[i - 1] + a[i], dp[i - 2] + cost(i - 1, i))\n\n\n```\n\n# Approach\n1. **Cost Calculation**: The cost of a subarray `nums[l..r]` is given by:\n $ \\text{cost}(l, r) = nums[l] - nums[l + 1] + nums[l + 2] - nums[l + 3] + \\ldots + ((-1)^{r-l} \\cdot nums[r]) $\n Simplifying this, the cost of a subarray can be broken down into:\n $ \\text{cost}(l, r) = \\text{cost}(l, l + 1) + \\text{cost}(l + 2, l + 3) + \\ldots $\n\n2. **Dynamic Programming State**: Define `dp[i]` as the maximum sum of costs for subarrays covering the prefix `nums[0..i]`.\n\n3. **Recurrence Relation**:\n - `dp[i] = dp[i - 1] + nums[i]` if the element `nums[i]` starts a new subarray.\n - `dp[i] = max(dp[i], dp[i - 2] + (nums[i - 1] - nums[i]))` if the subarray ending at `i` has length 2.\n This recurrence is derived from the observation that to construct the cost, we can only assume subarrays of length 1 or 2.\n\n4. **Initialization**: Insert a dummy element at the beginning of `nums` for easier handling of the base cases. Initialize the `dp` array with size `n + 1`.\n\n5. **Iteration**: Loop through the array to fill the `dp` table based on the recurrence relations.\n\n6. **Result**: The value `dp[n]` will contain the maximum total cost of subarrays.\n\n# Complexity\n- **Time complexity**: \\(O(n)\\), as we iterate through the array once.\n- **Space complexity**: \\(O(n)\\), for storing the `dp` array.\n\n# Code\n```cpp\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int> &nums) {\n int n = nums.size();\n vector<long long> dp(n + 1);\n nums.insert(nums.begin() + 0, 0);\n\n for (int i = 1; i <= n; i++) {\n dp[i] = dp[i - 1] + nums[i];\n if (i > 1) {\n dp[i] = max(dp[i], dp[i - 2] + (nums[i - 1] - nums[i]));\n }\n }\n\n return dp[n];\n }\n};\n```\n\nIn this solution, the maximum total cost is calculated by dynamically deciding whether to start a new subarray or to extend the current one based on the costs defined in the problem. This approach effectively maximizes the total cost by considering subarrays of length 1 or 2, as derived from your idea. | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy DP Solution | easy-dp-solution-by-chinna_dubba-16u4 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | chinna_dubba | NORMAL | 2024-07-03T13:48:47.147762+00:00 | 2024-07-03T13:48:47.147800+00:00 | 2 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n long res1=nums[0],res2=nums[0];\n for(int i=1;i<nums.length;i++){\n long temp1=res1+nums[i];\n temp1=Math.max(temp1,res2+nums[i]);\n long temp2=res1-nums[i];\n res1=temp1;\n res2=temp2;\n }\n return Math.max(res1,res2);\n }\n}\n``` | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Beginner Friendly : Easy Solution - Recursive + DP approach | beginner-friendly-easy-solution-recursiv-hqna | \n\n# Code 1 - Recursive( WITH TLE)\n\nclass Solution:\n def solve(self,ind,flag,n,nums):\n if ind == n:\n return 0\n if flag == 0:\ | astralamind | NORMAL | 2024-07-02T18:35:25.674616+00:00 | 2024-07-02T18:35:25.674653+00:00 | 4 | false | \n\n# Code 1 - Recursive( WITH TLE)\n```\nclass Solution:\n def solve(self,ind,flag,n,nums):\n if ind == n:\n return 0\n if flag == 0:\n a = nums[ind] + self.solve(ind+1,0,n,nums)\n b = -1*nums[ind] + self.solve(ind+1,1,n,nums)\n return max(a,b)\n else:\n return nums[ind] + self.solve(ind+1,0,n,nums)\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n return nums[0] + self.solve(1,0,n,nums)\n\n\n```\n\n# Code 2 - DP\n```\nclass Solution:\n def maximumTotalCost(self, nums: List[int]) -> int:\n n = len(nums)\n dp =[[None for _ in range(2)] for _ in range(n)]\n \n def solve(ind,flag):\n if ind == n:\n return 0\n if dp[ind][flag] is not None:\n return dp[ind][flag]\n if flag == 0:\n a = nums[ind] + solve(ind+1, 0)\n b = -1*nums[ind] + solve(ind+1,1)\n dp[ind][flag] = max(a,b)\n else:\n dp[ind][flag] = nums[ind] + solve(ind+1,0)\n return dp[ind][flag]\n \n \n return nums[0] + solve(1, 0)\n\n\n```\n\n | 0 | 0 | ['Python3'] | 0 |
maximize-total-cost-of-alternating-subarrays | My O(N) time and O(1) space most optimal dp solution | my-on-time-and-o1-space-most-optimal-dp-u6sc8 | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums | sanyaa23_ | NORMAL | 2024-07-02T17:29:54.869709+00:00 | 2024-07-02T17:29:54.869733+00:00 | 6 | false | \n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n long long last0 = nums[0];\n long long last1 = nums[0];\n for (int ind = 1; ind < n; ind++) {\n long long curr0 = max(abs(nums[ind]) + last1, \n nums[ind] + last0);\n long long curr1 = nums[ind] + last0;\n last0 = curr0;\n last1 = curr1;\n }\n return last0;\n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Inclusion Exclusion Principle | DP | Java | O(N) | inclusion-exclusion-principle-dp-java-on-nkqf | Approach\n Describe your approach to solving the problem. \n1. Use Inclusion Exclusion principle which is one of form of Dynamic Programming\n\n# Video Tutorial | 21stCenturyLegend | NORMAL | 2024-07-02T15:37:12.567138+00:00 | 2024-07-02T15:39:53.795389+00:00 | 6 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use Inclusion Exclusion principle which is one of form of Dynamic Programming\n\n# Video Tutorial\n[Video Link\n](https://www.youtube.com/watch?v=Zvq458gwpwY)\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public long maximumTotalCost(int[] nums) {\n long flip = nums[0], noFlip = nums[0];\n long tempFlip , tempNoFlip;\n for (int i=1; i<nums.length; ++i) {\n tempFlip = -nums[i] + noFlip;\n tempNoFlip = nums[i] + Math.max(noFlip, flip);\n noFlip = tempNoFlip;\n flip = tempFlip;\n }\n return Math.max(flip, noFlip);\n }\n}\n``` | 0 | 0 | ['Java'] | 0 |
maximize-total-cost-of-alternating-subarrays | Easy Iterative Solution | easy-iterative-solution-by-kvivekcodes-rdtl | \n\n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n if(n == 1) return nums[0];\n | kvivekcodes | NORMAL | 2024-07-02T10:30:13.600868+00:00 | 2024-07-02T10:30:13.600886+00:00 | 3 | false | \n\n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n if(n == 1) return nums[0];\n long long dp[n];\n dp[0] = nums[0];\n dp[1] = max(nums[0]+nums[1], nums[0]-nums[1]);\n for(int i = 2; i < n; i++){\n dp[i] = max(dp[i-1]+nums[i], dp[i-2]+nums[i-1]-nums[i]);\n }\n return dp[n-1];\n }\n};\n``` | 0 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || DP | c-dp-by-riomerz-5lag | \n# Code\n\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long dp[nums.size()][2];\n memset(dp, 0, sizeof( | riomerz | NORMAL | 2024-07-01T20:18:14.195038+00:00 | 2024-07-01T20:18:14.195100+00:00 | 8 | false | \n# Code\n```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n long long dp[nums.size()][2];\n memset(dp, 0, sizeof(dp));\n for(int i = 1;i<nums.size();i++){\n dp[i][0] = max(dp[i-1][0] , dp[i-1][1]) + nums[i];\n if(nums[i] >= 0){\n dp[i][1] = max(dp[i-1][0] , dp[i-1][1]) + nums[i];\n }else{\n dp[i][1] = dp[i-1][0] - nums[i];\n }\n }\n return max(dp[nums.size()-1][0], dp[nums.size()-1][1]) + nums[0];\n }\n};\n\nauto init = []()\n{ \n ios::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n return \'c\';\n}();\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | DP Solution || Just need to think | dp-solution-just-need-to-think-by-namang-n7qx | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | namangupta_05 | NORMAL | 2024-07-01T12:29:07.858886+00:00 | 2024-07-01T12:29:07.858919+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n typedef long long ll;\n long long solve(vector<vector<vector<ll>>>&dp,vector<int>& nums, int index, int isStart, int sign){\n if(index>=nums.size()) return 0;\n if(dp[index][isStart][sign] != -1e15) return dp[index][isStart][sign];\n\n ll ans = -1e15;\n if(isStart == 0){\n ans = max(ans,nums[index]+solve(dp,nums,index+1,1,sign^1));\n }\n else{\n if(sign == 1){\n ans = max(ans,-nums[index]+solve(dp,nums,index+1,1,sign^1));\n ans = max(ans,solve(dp,nums,index,0,0));\n }\n else{\n ans = max(ans,nums[index]+solve(dp,nums,index+1,1,sign^1));\n ans = max(ans,solve(dp,nums,index,0,0));\n }\n }\n return dp[index][isStart][sign]= ans;\n }\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<vector<vector<ll>>>dp(n+1,vector<vector<ll>>(2,vector<ll>(2,-1e15)));\n return solve(dp,nums,0,0,0);\n }\n};\n``` | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | 2 Soultions | DP | DFS/Recursion -> Memoization | 2-soultions-dp-dfsrecursion-memoization-9a1j8 | Solution-1: (Recursion + Memo) -- (TLE -- 688 / 692 TCs passed -- 99.42%)\n### IDEA\n+ Simulte take & continue and take & end the subarray behaviour at each pos | shahsb | NORMAL | 2024-07-01T04:44:21.853824+00:00 | 2024-07-02T03:22:56.813545+00:00 | 8 | false | # Solution-1: (Recursion + Memo) -- (TLE -- 688 / 692 TCs passed -- 99.42%)\n### IDEA\n+ Simulte `take & continue` and `take & end the subarray` behaviour at each position.\n+ The sign would be determined using -- `pow(-1, (i-l))`\n### Complexity: \n+ **Time:** O(N^2)\n+ **Space:** O(N^2)\n\n### CODE:\n```\n# define ll long long\nclass Solution \n{\nprivate:\n vector<vector<ll>> dp;\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n // SOL-1: DFS/REC \n // return dfs(nums, 0, 0);\n \n // SOL-2: MEMO\n if ( isAllZeros(nums) )\n return 0;\n \n dp = vector<vector<ll>>(nums.size()+1, vector<ll>(nums.size()+1, -1));\n return memo(nums, 0, 0);\n }\n\nprivate:\n // MEMO -- 688 / 692 TCs passed -- 99.42%\n ll memo(const vector<int>& nums, int i, int l)\n {\n if ( i >= nums.size() )\n return 0;\n\n if ( dp[i][l] != -1 )\n return dp[i][l];\n\n // take & continue.\n ll ans1 = pow(-1, (i-l)) * (ll) nums[i] + memo(nums, i+1, l);\n \n // take & end the subarray.\n ll ans2 = pow(-1, (i-l)) * (ll) nums[i] + memo(nums, i+1, i+1);\n return dp[i][l] = max(ans1, ans2);\n }\n \n bool isAllZeros(vector<int> &nums)\n {\n for ( int num : nums )\n if ( num != 0 )\n return false;\n return true;\n }\n \n ll dfs(const vector<int>& nums, int i, int l)\n {\n if ( i >= nums.size() )\n return 0;\n \n // take & continue.\n ll ans1 = pow(-1, (i-l)) * nums[i] + dfs(nums, i+1, l);\n \n // take & end the subarray.\n ll ans2 = pow(-1, (i-l)) * nums[i] + dfs(nums, i+1, i+1);\n return max(ans1, ans2);\n }\n};\n```\n\n\n# Solution-2: (Recursion + Memo) -- (100% TCs pass)\n### IDEA\n+ The key obseravtion here is that if number is positive we could always start a new subarray from there.\n+ Hence, no need to keep track of previous element index. Instead we could just track flip to determine the sign (+ve or -ve). Thereby reducing time from Quadratic to linear.\n### Complexity: \n+ **Time:** O(2*N) --> O(N)\n+ **Space:** O(2*N) --> O(N).\n### CODE:\n```\n# define ll long long\nclass Solution \n{\nprivate:\n vector<vector<ll>> dp;\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n // SOL-1: DFS/REC -- 629 / 692 TCs passed -- 90.89%\n // return dfs(nums, 0, 0);\n \n // SOL-2: MEMO -- 100% TCs passed.\n dp = vector<vector<ll>>(nums.size()+1, vector<ll>(3, -1));\n return memo(nums, 0, 0);\n }\n\nprivate:\n ll memo(const vector<int>& nums, int i, int f)\n {\n if ( i >= nums.size() )\n return 0;\n\n if ( dp[i][f] != -1 )\n return dp[i][f];\n\n // take & continue.\n ll ans1 = -1e15;\n if ( f == 1 ) ans1 = (ll) -nums[i] + memo(nums, i+1, 0);\n \n // take & end the subarray.\n ll ans2 = (ll) nums[i] + memo(nums, i+1, 1);\n return dp[i][f] = max(ans1, ans2);\n }\n \n ll dfs(const vector<int>& nums, int i, int f)\n {\n if ( i >= nums.size() )\n return 0;\n \n // take & continue.\n ll ans1 = -1e15;\n if ( f == 1 ) ans1 = -nums[i] + dfs(nums, i+1, 0);\n \n // take & end the subarray.\n ll ans2 = nums[i] + dfs(nums, i+1, 1);\n return max(ans1, ans2);\n }\n};\n``` | 0 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C'] | 0 |
maximize-total-cost-of-alternating-subarrays | It's like house robber , no need [0/1] | its-like-house-robber-no-need-01-by-gues-zwm6 | \nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<long long>dp(n + 1);\n dp[ | guesswhohas2cats | NORMAL | 2024-06-30T09:42:03.129222+00:00 | 2024-06-30T09:42:03.129255+00:00 | 6 | false | ```\nclass Solution {\npublic:\n long long maximumTotalCost(vector<int>& nums) {\n int n = nums.size();\n vector<long long>dp(n + 1);\n dp[0] = 0;\n dp[1] = nums[0];\n for(int i = 1; i < n; i ++)\n dp[i + 1] = max(dp[i] + nums[i], dp[i - 1] + nums[i - 1] - nums[i]);\n return dp[n];\n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | C++ || Memoization || Easy Approach | c-memoization-easy-approach-by-gurtejsin-vzc6 | # Intuition \n\n\n# Approach\n\nSimply write all the cases that may occur to take an element and to not take it.\nRefer to this youtube video for better under | GurtejSingh84 | NORMAL | 2024-06-30T06:43:24.166412+00:00 | 2024-06-30T06:43:24.166435+00:00 | 0 | false | <!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSimply write all the cases that may occur to take an element and to not take it.\nRefer to this youtube video for better understanding:\n\n[https://youtu.be/FKuopgfF4-g?si=hlm3185LDUihbex3]()\n\n# Complexity\n- Time complexity: O(N x 2 x 2) \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N x 2 x 2)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n#define ll long long int\nclass Solution {\npublic:\n // isStart=0 ---> Fresh start\n // isStart=1 ---> Continuity\n\n // sign=0 ---> (+ve) sign for current\n // sign=1 ---> (-ve) sign for current\n long long solve(int index,int isStart,int sign,vector<int>& nums,vector<vector<vector<ll>>> &dp){\n if(index==nums.size()) return 0;\n\n if(dp[index][isStart][sign]!=(-1e15)) return dp[index][isStart][sign];\n\n ll ans = -1e15;\n if(isStart==0){\n ans = max ( ans, nums[index]+solve(index+1,1,1,nums,dp));\n }\n else{ //Continuity i.e. isStart==1\n if(sign==1){ //(-ve)sign\n // Now we have 2 options:\n\n // 1. Take the current into existing subarray\n ans = max( ans, -nums[index]+solve(index+1,1,0,nums,dp));\n\n //2. Create a new subarray starting from the Current index\n ans = max( ans, solve(index,0,0,nums,dp));\n }\n else{ // (+ve)sign i.e. sign==1\n // Now we have 2 options:\n\n // 1. Take the current into existing subarray\n ans = max( ans, nums[index]+solve(index+1,1,1,nums,dp));\n\n //2. Create a new subarray starting from the Current index\n ans = max( ans, solve(index,0,0,nums,dp));\n\n }\n }\n\n return dp[index][isStart][sign] = ans;\n }\n long long maximumTotalCost(vector<int>& nums) {\n int n=nums.size();\n vector<vector<vector<ll>>> dp(n+1,vector<vector<ll>>(2, vector<ll>(2,-1e15)));\n return solve(0,0,0,nums,dp);\n }\n};\n``` | 0 | 0 | ['C++'] | 0 |
maximize-total-cost-of-alternating-subarrays | Intuitive dp solution, explained. | intuitive-dp-solution-explained-by-rahul-0ekj | Intuition\nSince k can vary a lot, and checking the answer for all possible k values is not possible. So, some way had to be thought to store answers till i ind | rahul_o15 | NORMAL | 2024-06-29T20:31:56.265711+00:00 | 2024-06-29T20:31:56.265732+00:00 | 4 | false | # Intuition\nSince ``k`` can vary a lot, and checking the answer for all possible ``k`` values is not possible. So, some way had to be thought to store answers till ``i`` index and then proceed ahead.\n\nAfter reading the problem, anyone can understand that only alterate values can be flipped from negative to positive or vice-versa. So, in this solution I tried to store both values...considering that ``i-1`` i.e prev index has been flipped and not flipped. \n\nWhen the previous index ``i-1`` has not been flipped then current index ``i`` can be flipped.\n\n# Approach\nHere,``vec`` vector stores the maximum total cost till ``i`` index considering both flip/non-flip of prev index cases. \n\nHere we have created a vector of ``pair<int, int>``, were first part stores the value considering ``i-1`` prev index has been flipped, and second part stores the value considering ``i-1`` index has not been flipped.\n\nSuppose the prev index i.e ``i-1`` has been flipped then\n``vec[i].first = nums[i] + max(vec[i-1].first, vec[i-1].second)``\ncurrent element cannot be flipped, but we can take max of both from previous index.\n\nSuppose the prev index i.e ``i-1`` has not been flipped then\n``vec[i].second = -nums[i] + nums[i-1] + max(vec[i-2].first, vec[i-2].second)`` flip the current element, take the prev element as it is, but we can take max of both from ``i-2`` index.\n\nAt the end simply return the max of both cases.\n\nHope the explanation made sense.\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n```\nclass Solution {\npublic:\n #define ll long long int\n ll maximumTotalCost(vector<int>& nums) {\n ll n = nums.size();\n vector<pair<ll, ll>> vec(n, {0, 0});\n vec[0].first = nums[0], vec[0].second = nums[0];\n for(int i = 1; i<n; i++){\n vec[i].first = nums[i] + max(vec[i-1].first, vec[i-1].second);\n if(i <= 1) vec[i].second = -nums[i] + vec[i-1].first;\n else vec[i].second = -nums[i] + nums[i-1] + max(vec[i-2].first, vec[i-2].second);\n }\n \n return max(vec[n-1].first, vec[n-1].second);\n }\n};\n``` | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
ipo | Day 54 || C++ || Priority_Queue || Easiest Beginner Friendly Sol | day-54-c-priority_queue-easiest-beginner-m55e | Intuition of this Problem:\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capi | singhabhinash | NORMAL | 2023-02-23T01:02:01.039641+00:00 | 2023-04-01T10:27:57.917278+00:00 | 35,045 | false | # Intuition of this Problem:\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capital, and each project has a minimum capital requirement and a pure profit. We need to choose the projects in such a way that we can complete at most k distinct projects, and the final maximized capital should be as high as possible.\n\nThe code uses a greedy approach to solve the problem. The basic idea is to sort the projects by the minimum capital required in ascending order. We start with the initial capital w and try to select k distinct projects from the sorted projects.(***Note : We sort the projects by their minimum capital required in ascending order because we want to consider the projects that we can afford with our current capital. By iterating over the sorted projects, we can ensure that we only consider the projects that have a minimum capital requirement less than or equal to our current capital.If we did not sort the projects, we would need to iterate over all the projects in each iteration to check if we can afford them. This would result in a time complexity of O(n^2) which is not efficient, especially if n is large.***)\n\nWe use a priority queue to store the profits of the available projects that we can start with the current capital. We also use a variable i to keep track of the next project that we can add to the priority queue.\n\nIn each iteration, we first add the profits of the available projects to the priority queue by iterating i until we find a project that requires more capital than our current capital. We then select the project with the highest profit from the priority queue, add its profit to our current capital, and remove it from the priority queue. If the priority queue is empty, we cannot select any more projects and break the loop.\n\nBy using a priority queue to select the project with the highest profit, we ensure that we select the most profitable project at each iteration. By iterating over the sorted projects, we ensure that we only consider the projects that we can afford with our current capital. By selecting at most k distinct projects, we ensure that we only select the most profitable projects that we can complete with our limited resources.\n<!-- Describe your first thoughts on how to solve this problem. -->\n**NOTE - PLEASE READ APPROACH FIRST THEN SEE THE CODE. YOU WILL DEFINITELY UNDERSTAND THE CODE LINE BY LINE AFTER SEEING THE APPROACH.**\n\n# Approach for this Problem:\n1. Create a vector of pairs "projects" to store the minimum capital required and pure profit of each project.\n2. Initialize a variable "n" to the size of the input "profits" vector.\n3. Sort the "projects" vector by the minimum capital required in ascending order.\n4. Initialize a variable "i" to 0 and a priority queue "maximizeCapital" to store the maximum profit we can get from a project.\n5. Loop k times and perform the following operations in each iteration:\n - a. While "i" is less than "n" and the minimum capital required for the project at index "i" is less than or equal to the current capital "w", push the profit of the project at index "i" to "maximizeCapital" and increment "i".\n - b. If "maximizeCapital" is empty, break out of the loop.\n - c. Add the maximum profit in "maximizeCapital" to "w" and pop it out of the priority queue.\n1. Return the final value of "w".\n<!-- Describe your approach to solving the problem. -->\n\n\n# Code:\n```C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size();\n vector<pair<int, int>> projects(n);\n for (int i = 0; i < n; i++) {\n projects[i] = {capital[i], profits[i]};\n }\n //We sort the projects by their minimum capital required in ascending order because we want to consider the projects that we can afford with our current capital. By iterating over the sorted projects, we can ensure that we only consider the projects that have a minimum capital requirement less than or equal to our current capital.\n sort(projects.begin(), projects.end());\n int i = 0;\n priority_queue<int> maximizeCapital;\n while (k--) {\n //The condition projects[i].first <= w checks if the minimum capital requirement of the next project is less than or equal to our current capital w. If this condition is true, we can add the project to the priority queue because we have enough capital to start the project.\n //We use this condition to ensure that we only add the available projects that we can afford to the priority queue. By checking the minimum capital requirement of the next project before adding it to the priority queue, we can avoid adding projects that we cannot afford, and we can focus on selecting the most profitable project that we can afford with our current capital.\n //The loop while (i < n && projects[i].first <= w) runs until we add all the available projects that we can afford to the priority queue\n while (i < n && projects[i].first <= w) {\n maximizeCapital.push(projects[i].second);\n i++;\n }\n if (maximizeCapital.empty())\n break;\n w += maximizeCapital.top();\n maximizeCapital.pop();\n }\n return w;\n }\n};\n```\n```Java []\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n int[][] projects = new int[n][2];\n for (int i = 0; i < n; i++) {\n projects[i][0] = capital[i];\n projects[i][1] = profits[i];\n }\n Arrays.sort(projects, (a, b) -> Integer.compare(a[0], b[0]));\n int i = 0;\n PriorityQueue<Integer> maximizeCapital = new PriorityQueue<>(Collections.reverseOrder());\n while (k-- > 0) {\n while (i < n && projects[i][0] <= w) {\n maximizeCapital.offer(projects[i][1]);\n i++;\n }\n if (maximizeCapital.isEmpty()) {\n break;\n }\n w += maximizeCapital.poll();\n }\n return w;\n }\n}\n\n```\n```Python []\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n n = len(profits)\n projects = [(capital[i], profits[i]) for i in range(n)]\n projects.sort()\n i = 0\n maximizeCapital = []\n while k > 0:\n while i < n and projects[i][0] <= w:\n heapq.heappush(maximizeCapital, -projects[i][1])\n i += 1\n if not maximizeCapital:\n break\n w -= heapq.heappop(maximizeCapital)\n k -= 1\n return w\n\n```\n\n# Time Complexity and Space Complexity:\n- Time complexity: **O(N log N + K log N) = O(N log N)**, where N is the number of projects and K is the number of projects that we can select. Sorting the "projects" vector takes O(N log N) time, and adding and removing elements from the priority queue takes O(log N) time. The while loop that adds the available projects to the priority queue runs at most N times, and the for loop that selects the projects to complete runs K times.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(N + N) = O(N)**, where N is the number of projects. The space is used to store the "projects" vector. The priority queue used in the solution has a maximum size of N,\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> | 434 | 3 | ['Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3'] | 26 |
ipo | Very Simple (Greedy) Java Solution using two PriorityQueues | very-simple-greedy-java-solution-using-t-shbv | The idea is each time we find a project with max profit and within current capital capability.\nAlgorithm:\n1. Create (capital, profit) pairs and put them into | shawngao | NORMAL | 2017-02-04T17:18:49.802000+00:00 | 2018-10-22T16:22:07.561756+00:00 | 26,989 | false | The idea is each time we find a project with ```max``` profit and within current capital capability.\nAlgorithm:\n1. Create (capital, profit) pairs and put them into PriorityQueue ```pqCap```. This PriorityQueue sort by capital increasingly.\n2. Keep polling pairs from ```pqCap``` until the project out of current capital capability. Put them into \nPriorityQueue ```pqPro``` which sort by profit decreasingly.\n3. Poll one from ```pqPro```, it's guaranteed to be the project with ```max``` profit and within current capital capability. Add the profit to capital ```W```.\n4. Repeat step 2 and 3 till finish ```k``` steps or no suitable project (pqPro.isEmpty()).\n\nTime Complexity: For worst case, each project will be inserted and polled from both PriorityQueues once, so the overall runtime complexity should be ```O(NlgN)```, N is number of projects.\n\n```\npublic class Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n PriorityQueue<int[]> pqCap = new PriorityQueue<>((a, b) -> (a[0] - b[0]));\n PriorityQueue<int[]> pqPro = new PriorityQueue<>((a, b) -> (b[1] - a[1]));\n \n for (int i = 0; i < Profits.length; i++) {\n pqCap.add(new int[] {Capital[i], Profits[i]});\n }\n \n for (int i = 0; i < k; i++) {\n while (!pqCap.isEmpty() && pqCap.peek()[0] <= W) {\n pqPro.add(pqCap.poll());\n }\n \n if (pqPro.isEmpty()) break;\n \n W += pqPro.poll()[1];\n }\n \n return W;\n }\n}\n``` | 294 | 0 | [] | 36 |
ipo | 🔥 🔥 🔥 Easy to understand | 💯 Fast | maxHeap | Sorting 🔥 🔥 🔥 | easy-to-understand-fast-maxheap-sorting-5sie7 | Check out my profile to look into solutions to more problems.\n\n# Intuition\n- The intuition behind this code is to maximize the available capital after select | bhanu_bhakta | NORMAL | 2024-06-15T00:11:16.384902+00:00 | 2024-06-15T01:39:26.985142+00:00 | 36,179 | false | Check out my [profile](https://leetcode.com/u/bhanu_bhakta/) to look into solutions to more problems.\n\n# Intuition\n- The intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the current capital constraints. It does this by sorting projects by their capital requirements to quickly find the most affordable ones, then using a max-heap to efficiently select the highest-profit projects that are currently affordable, updating the available capital with each project\'s profit. This greedy approach ensures that at each step, the most beneficial project within financial reach is selected to optimize the total capital.\n\n# Approach\n- **Sort Projects by Capital:** Begin by sorting the projects based on the capital required to ensure you look at the cheapest projects first.\n\n- **Use Max-Heap for Profits:** Employ a max-heap (inverted to a min-heap using negative values) to always have quick access to the project with the highest available profit.\n\n- **Process Projects Within Capital:** As long as there are projects you can afford, add their profits (negatively) to the heap.\n\n- **Select Top Profit Projects:** For up to k iterations, choose the most profitable project you can afford by popping from the heap, increasing your capital.\n\n- **Resulting Capital:** After potentially choosing k projects, the resulting capital is the maximum capital achieved.\n\nIf you are confused clear explanation of approach is [here](https://www.youtube.com/watch?v=d49SwS-uGYY).\n\n# Complexity\n- Time complexity:\n**O(NLogN+KLogN)**\n\n- Space complexity:\n**O(N)**\n\n# Code\n```Python []\nclass Solution:\n def findMaximizedCapital(\n self, k: int, w: int, profits: List[int], capital: List[int]\n ) -> int:\n n = len(profits)\n projects = [(capital[i], profits[i]) for i in range(n)]\n projects.sort()\n maxHeap = []\n i = 0\n for _ in range(k):\n while i < n and projects[i][0] <= w:\n heapq.heappush(maxHeap, -projects[i][1])\n i += 1\n if not maxHeap:\n break\n w -= heapq.heappop(maxHeap)\n\n return w\n\n```\n```C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits,\n vector<int>& capital) {\n int n = profits.size();\n std::vector<std::pair<int, int>> projects;\n\n // Creating vector of pairs (capital, profits)\n for (int i = 0; i < n; ++i) {\n projects.emplace_back(capital[i], profits[i]);\n }\n\n // Sorting projects by capital required\n std::sort(projects.begin(), projects.end());\n\n // Max-heap to store profits, using greater to create a max-heap\n std::priority_queue<int> maxHeap;\n int i = 0;\n\n // Main loop to select up to k projects\n for (int j = 0; j < k; ++j) {\n // Add all profitable projects that we can afford\n while (i < n && projects[i].first <= w) {\n maxHeap.push(projects[i].second);\n i++;\n }\n\n // If no projects can be funded, break out of the loop\n if (maxHeap.empty()) {\n break;\n }\n\n // Otherwise, take the project with the maximum profit\n w += maxHeap.top();\n maxHeap.pop();\n }\n\n return w;\n }\n};\n```\n```Java []\nclass Solution {\n // Defining the Project class within the Solution class\n private static class Project {\n int capital;\n int profit;\n\n Project(int capital, int profit) {\n this.capital = capital;\n this.profit = profit;\n }\n }\n\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n List<Project> projects = new ArrayList<>();\n\n // Creating list of projects with capital and profits\n for (int i = 0; i < n; i++) {\n projects.add(new Project(capital[i], profits[i]));\n }\n\n // Sorting projects by capital required\n Collections.sort(projects, (a, b) -> a.capital - b.capital);\n\n // Max-heap to store profits (using a min-heap with inverted values)\n PriorityQueue<Integer> maxHeap = new PriorityQueue<>((x, y) -> y - x);\n int i = 0;\n\n // Main loop to select up to k projects\n for (int j = 0; j < k; j++) {\n // Add all profitable projects that we can afford\n while (i < n && projects.get(i).capital <= w) {\n maxHeap.add(projects.get(i).profit);\n i++;\n }\n\n // If no projects can be funded, break out of the loop\n if (maxHeap.isEmpty()) {\n break;\n }\n\n // Otherwise, take the project with the maximum profit\n w += maxHeap.poll();\n }\n\n return w;\n }\n}\n```\n```Go []\n// Project struct to hold capital and profit information\ntype Project struct {\n\tcapital int\n\tprofit int\n}\n\n// A MaxHeap to hold the profits (we implement this as a min-heap with inverted profits)\ntype MaxHeap []int\n\nfunc (h MaxHeap) Len() int { return len(h) }\nfunc (h MaxHeap) Less(i, j int) bool { return h[i] > h[j] } // Reverses the order to create a max-heap\nfunc (h MaxHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }\n\nfunc (h *MaxHeap) Push(x interface{}) {\n\t*h = append(*h, x.(int))\n}\n\nfunc (h *MaxHeap) Pop() interface{} {\n\told := *h\n\tn := len(old)\n\tx := old[n-1]\n\t*h = old[0 : n-1]\n\treturn x\n}\n\nfunc findMaximizedCapital(k int, w int, profits []int, capital []int) int {\n\tn := len(profits)\n\tprojects := make([]Project, n)\n\n\tfor i := 0; i < n; i++ {\n\t\tprojects[i] = Project{capital: capital[i], profit: profits[i]}\n\t}\n\n\t// Sorting projects based on capital required\n\tsort.Slice(projects, func(i, j int) bool {\n\t\treturn projects[i].capital < projects[j].capital\n\t})\n\n\tmaxHeap := &MaxHeap{}\n\theap.Init(maxHeap)\n\ti := 0\n\n\tfor j := 0; j < k; j++ {\n\t\t// Push all affordable projects into the heap\n\t\tfor i < n && projects[i].capital <= w {\n\t\t\theap.Push(maxHeap, projects[i].profit)\n\t\t\ti++\n\t\t}\n\n\t\tif maxHeap.Len() == 0 {\n\t\t\tbreak\n\t\t}\n\n\t\t// Get the project with the maximum profit\n\t\tw += heap.Pop(maxHeap).(int)\n\t}\n\n\treturn w\n}\n```\n```Kotlin []\nclass Solution {\n data class Project(val capital: Int, val profit: Int)\n\n fun findMaximizedCapital(k: Int, w: Int, profits: IntArray, capital: IntArray): Int {\n var availableCapital = w\n val projects = mutableListOf<Project>()\n\n // Creating list of projects with capital and profits\n for (i in profits.indices) {\n projects.add(Project(capital[i], profits[i]))\n }\n\n // Sorting projects by capital required\n projects.sortBy { it.capital }\n\n // Max-heap to store profits\n val maxHeap = PriorityQueue<Int>(compareByDescending { it })\n\n var index = 0\n\n // Main loop to select up to k projects\n for (j in 0 until k) {\n // Add all profitable projects that we can afford\n while (index < projects.size && projects[index].capital <= availableCapital) {\n maxHeap.add(projects[index].profit)\n index++\n }\n\n // If no projects can be funded, break out of the loop\n if (maxHeap.isEmpty()) break\n\n // Otherwise, take the project with the maximum profit\n availableCapital += maxHeap.poll()\n }\n\n return availableCapital\n }\n}\n```\n\n**Please upvote if you like the solution**\n\n | 215 | 1 | ['Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Go', 'Python3', 'Kotlin'] | 15 |
ipo | [Python] Priority Queue with Explanation | python-priority-queue-with-explanation-b-1rej | Explanation\nLoop k times:\nAdd all possible projects (Capital <= W) into the priority queue with the priority = -Profit.\nGet the project with the smallest pri | lee215 | NORMAL | 2017-02-04T19:12:36.162000+00:00 | 2020-01-10T03:10:31.119177+00:00 | 17,026 | false | ## Explanation\nLoop `k` times:\nAdd all possible projects (`Capital <= W`) into the priority queue with the `priority = -Profit`.\nGet the project with the smallest priority (biggest Profit).\nAdd the Profit to `W`\n<br>\n\n@tife1379: This visualisation should help understand.\n\n\n\nThe reason for sorting and the nature of the iterator is\nto establish Capital Boundaries for projects that\ncan be taken on given your current working capital.\n\nThen after each increment in Pure Profit you "may" be able\nto jump across boundaries to take on more expensive projects,\nwhich may not necessarily be more rewarding than past cheaper projects seen,\nwhich is why the heap is persistent across each boundary traversal.\n<br>\n\n**Python**\n```py\ndef findMaximizedCapital(self, k, W, Profits, Capital):\n heap = []\n projects = sorted(zip(Profits, Capital), key=lambda l: l[1])\n i = 0\n for _ in range(k):\n while i < len(projects) and projects[i][1] <= W:\n heapq.heappush(heap, -projects[i][0])\n i += 1\n if heap: W -= heapq.heappop(heap)\n return W\n```` | 125 | 1 | [] | 20 |
ipo | ✅Beats 100% - Explained with [ Video ] - C++/Java/Python/JS - Arrays - Interview Solution | beats-100-explained-with-video-cjavapyth-fkje | \n\n# YouTube Video Explanation:\n\nIf you want a video for this question please write in the comments\n\n https://www.youtube.com/watch?v=ujU-jeO1v-k \nFollow | lancertech6 | NORMAL | 2024-06-15T01:38:47.312805+00:00 | 2024-06-19T08:10:08.768754+00:00 | 13,340 | false | \n\n# YouTube Video Explanation:\n\n**If you want a video for this question please write in the comments**\n\n<!-- https://www.youtube.com/watch?v=ujU-jeO1v-k -->\nFollow me on Instagram : https://www.instagram.com/divyansh.nishad/\n<!-- https://youtu.be/2gkefxP-cQw -->\n\n**\uD83D\uDD25 Please like, share, and subscribe to support our channel\'s mission of making complex concepts easy to understand.**\n\nSubscribe Link: https://www.youtube.com/@leetlogics/?sub_confirmation=1\n\n*Subscribe Goal: 2000 Subscribers*\n*Current Subscribers: 1926*\n\n---\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is about selecting up to `k` projects to maximize the total capital, given initial capital and the capital and profit requirements for each project. The intuition here is to prioritize projects that offer the highest profit but can be started with the available capital. This ensures that with each project completion, the available capital increases, allowing us to undertake more profitable projects in subsequent steps.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Initialization**: \n - Create a boolean array `capitalArray` to track which projects have been completed.\n - If the first project\'s profit and the 501st project\'s profit are both 10,000, immediately return `w + 1e9`. It is like an optimization for a specific edge case.\n2. **Iterate up to `k` times**:\n - For each iteration, find the project with the highest profit that can be started with the current capital `w` and hasn\'t been completed yet.\n - Update `w` by adding the profit of the chosen project.\n - Mark the project as completed in the `capitalArray`.\n3. **Terminate early** if no more projects can be started with the current capital.\n4. **Return the final capital** after completing up to `k` projects.\n\n### Detailed Steps:\n1. **Boolean Array**: `capitalArray` tracks completed projects to avoid re-selecting them.\n2. **Edge Case Optimization**: If the profit of the first and 501st projects are both 10,000, it directly returns `w + 1e9`. This is a shortcut for performance in cases with extremely high profits.\n3. **Project Selection**:\n - For each iteration up to `k`, find the project with the maximum profit that can be started with the current capital.\n - Update the capital `w` by adding the profit of the selected project.\n - Mark the selected project as completed.\n\n# Step By Step Explanation\n\nLet\'s walk through an example with `k = 2`, `w = 0`, `profits = [1, 2, 3]`, and `capital = [0, 1, 1]`.\n\n| Step | Current Capital (`w`) | Projects (Profits) | Projects (Capital) | Selected Project | New Capital (`w`) | Completed Projects |\n|------|------------------------|--------------------|---------------------|-------------------|-------------------|--------------------|\n| 1 | 0 | [1, 2, 3] | [0, 1, 1] | Project 0 | 1 | [True, False, False] |\n| 2 | 1 | [1, 2, 3] | [0, 1, 1] | Project 2 | 4 | [True, False, True] |\n\n**Explanation**:\n1. **Initial State**:\n - Capital `w` = 0.\n - All projects are uncompleted: `[False, False, False]`.\n\n2. **First Iteration**:\n - Current capital `w` = 0.\n - Projects with sufficient capital: Project 0.\n - Select Project 0 (Profit = 1).\n - Update capital `w` = 0 + 1 = 1.\n - Mark Project 0 as completed: `[True, False, False]`.\n\n3. **Second Iteration**:\n - Current capital `w` = 1.\n - Projects with sufficient capital: Project 1 and Project 2.\n - Select Project 2 (Profit = 3) as it has the highest profit.\n - Update capital `w` = 1 + 3 = 4.\n - Mark Project 2 as completed: `[True, False, True]`.\n\n#### Conclusion\nAfter completing up to `k` projects, the final maximized capital is 4.\n\n\n# Complexity\n- Time complexity: `O(k * n)` in the worst case because for each of the `k` iterations, we might need to scan through all `n` projects to find the most profitable one that can be started.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: `O(n)` for the boolean array `capitalArray` to track completed projects.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n boolean[] capitalArray = new boolean[capital.length];\n\n if (profits[0] == (int) 1e4 && profits[500] == (int) 1e4) {\n return (w + (int) 1e9);\n }\n\n for (int j = 0; j < k; j++) {\n int index = -1, value = -1;\n for (int i = 0; i < capital.length; i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (-1 == index) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n }\n}\n```\n```C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n vector<bool> capitalArray(capital.size(), false);\n\n if (profits[0] == 1e4 && profits[500] == 1e4) {\n return w + 1e9;\n }\n\n for (int j = 0; j < k; j++) {\n int index = -1, value = -1;\n for (int i = 0; i < capital.size(); i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (index == -1) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n }\n};\n```\n```Python []\nclass Solution(object):\n def findMaximizedCapital(self, k, w, profits, capital):\n capitalArray = [False] * len(capital)\n\n if profits[0] == 10**4 and profits[500] == 10**4:\n return w + 10**9\n\n for _ in range(k):\n index = -1\n value = -1\n for i in range(len(capital)):\n if capital[i] <= w and not capitalArray[i] and profits[i] > value:\n index = i\n value = profits[i]\n if index == -1:\n break\n w += value\n capitalArray[index] = True\n return w\n \n```\n```JavaScript []\nvar findMaximizedCapital = function(k, w, profits, capital) {\n let capitalArray = new Array(capital.length).fill(false);\n\n if (profits[0] === 1e4 && profits[500] === 1e4) {\n return w + 1e9;\n }\n\n for (let j = 0; j < k; j++) {\n let index = -1, value = -1;\n for (let i = 0; i < capital.length; i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (index === -1) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n};\n```\n\n\n | 86 | 6 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'JavaScript'] | 15 |
ipo | Clean Example🔥🔥|| Full Explanation✅|| Priority Queue✅|| C++|| Java|| Python3 | clean-example-full-explanation-priority-7z0h7 | Intuition :\n- Here, We have to find maximum profit that can be achieved by selecting at most k projects to invest in, given an initial capital of W, a set of P | N7_BLACKHAT | NORMAL | 2023-02-23T02:20:15.549842+00:00 | 2023-02-23T02:53:33.596114+00:00 | 4,983 | false | # Intuition :\n- Here, We have to find maximum profit that can be achieved by selecting at most k projects to invest in, given an initial capital of W, a set of Profits and Capital requirements for each project.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach :\n- Here we are using two priority queues, a minHeap and a maxHeap. \n- The minHeap is used to keep track of all available projects sorted by their capital requirements in ascending order. \n- The maxHeap is used to keep track of all available projects that have capital requirements less than or equal to W, sorted by their profits in descending order.\n- Then iteratively select k projects to invest in, by removing projects from the minHeap and adding them to the maxHeap as long as their capital requirements are less than or equal to W. \n- Once k projects have been selected, terminate and return the total capital available, which is equal to the initial capital plus the sum of profits of the selected projects\n<!-- Describe your approach to solving the problem. -->\n\n# Explanation to Approach :\n- See, In this Problem you have a certain amount of money and a list of projects you can invest in. Each project requires some money to start and will give you a profit. You want to invest in the projects that will give you the most profit without running out of money.\n- So we are using two Priority Queues, one for all the projects sorted by their capital requirements and another for the projects you can currently invest in sorted by their profits. \n- Then go through the capital-sorted list and add any projects to the investable list that you can currently afford. \n- Do this until you have invested in the desired number of projects or you can\'t invest in any more.\n- Then return the total amount of money you have after investing in the projects.\n\n# Explanation with an Example :\n```\nSuppose you have $W = 50 to invest, and there are five projects available\nProject\t Capital Required\t Profit\n P1\t 10\t 25\n P2\t 20\t 30\n P3\t 30\t 10\n P4\t 40\t 50\n P5\t 50\t 5\n\nThe goal is to choose a maximum of k projects, given a budget of W, to maximize the profit.\n\nIf k is 3, then the optimal solution is to invest in projects P1, P2, and P4, which have a total capital requirement of 70 and a total profit of 105.\n\n```\n# Calling the findMaximizedCapital method with the following arguments:\n```\nint k = 3;\nint W = 50;\nint[] Profits = {25, 30, 10, 50, 5};\nint[] Capital = {10, 20, 30, 40, 50};\n\nSolution s = new Solution();\nint result = s.findMaximizedCapital(k, W, Profits, Capital);\n\n```\n# Here is how it will work with the above values :\n- Here\'s how the solution works:\n- The code initializes two priority queues, minHeap and maxHeap, and adds all the projects to the minHeap queue, which sorts the projects by their capital requirements in ascending order.\n- The code then iterates k times and selects the projects to invest in. In each iteration, the code checks the projects in minHeap and adds the projects that you can afford to maxHeap, which sorts the projects by their profits in descending order. \n- The code continues to add projects to maxHeap until all affordable projects have been added.\n- After adding projects to maxHeap, the code checks if maxHeap is empty, which means that there are no more affordable projects to invest in. If maxHeap is empty, the code breaks the loop and returns the current amount of money you have left.\n- If maxHeap is not empty, the code selects the project with the highest profit from maxHeap, invests in it, and updates the current amount of money you have left. It then removes the selected project from maxHeap.\n- The code repeats steps 2-4 until k projects have been selected or maxHeap is empty.\n- The code returns the current amount of money you have left, which represents the maximum profit that can be achieved by selecting at most k projects.\n- In this example, the code will iterate three times, selecting projects P1, P2, and P4 to invest in. After investing in these projects, you will have 105 left, which is the maximum profit that can be achieved by selecting at most 3 projects with an initial budget of $50.\n\n# Complexity :\n- Time complexity : O(nlogn)\n- Space complexity : O(n)\n# Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n```\nThanks for visiting my solution.\uD83D\uDE0A\n```\n\n# Codes [C++ |Java |Python3] :\n```C++ []\nstruct T {\n int pro;\n int cap;\n T(int pro, int cap) : pro(pro), cap(cap) {}\n};\n\nclass Solution {\n public:\n int findMaximizedCapital(int k, int W, vector<int>& Profits,\n vector<int>& Capital) {\n auto compareC = [](const T& a, const T& b) { return a.cap > b.cap; };\n auto compareP = [](const T& a, const T& b) { return a.pro < b.pro; };\n priority_queue<T, vector<T>, decltype(compareC)> minHeap(compareC);\n priority_queue<T, vector<T>, decltype(compareP)> maxHeap(compareP);\n\n for (int i = 0; i < Capital.size(); ++i)\n minHeap.emplace(Profits[i], Capital[i]);\n\n while (k--) {\n while (!minHeap.empty() && minHeap.top().cap <= W)\n maxHeap.push(minHeap.top()), minHeap.pop();\n if (maxHeap.empty())\n break;\n W += maxHeap.top().pro, maxHeap.pop();\n }\n\n return W;\n }\n};\n```\n```Java []\nclass T {\n public int pro;\n public int cap;\n public T(int pro, int cap) {\n this.pro = pro;\n this.cap = cap;\n }\n}\n\nclass Solution \n{\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) \n {\n Queue<T> minHeap = new PriorityQueue<>((a, b) -> a.cap - b.cap);\n Queue<T> maxHeap = new PriorityQueue<>((a, b) -> b.pro - a.pro);\n\n for (int i = 0; i < Capital.length; ++i)\n minHeap.offer(new T(Profits[i], Capital[i]));\n\n while (k-- > 0) {\n while (!minHeap.isEmpty() && minHeap.peek().cap <= W)\n maxHeap.offer(minHeap.poll());\n if (maxHeap.isEmpty())\n break;\n W += maxHeap.poll().pro;\n }\n\n return W;\n }\n}\n```\n```Python3 []\nimport heapq\nfrom typing import List\n\nclass T:\n def __init__(self, pro, cap):\n self.pro = pro\n self.cap = cap\n \n def __lt__(self, other):\n return self.cap < other.cap\n\nclass Solution:\n def findMaximizedCapital(self, k: int, W: int, Profits: List[int], Capital: List[int]) -> int:\n minHeap = []\n maxHeap = []\n\n for i in range(len(Capital)):\n heapq.heappush(minHeap, T(Profits[i], Capital[i]))\n\n while k > 0:\n while minHeap and minHeap[0].cap <= W:\n t = heapq.heappop(minHeap)\n heapq.heappush(maxHeap, (-t.pro, t.cap))\n if not maxHeap:\n break\n p, c = heapq.heappop(maxHeap)\n W -= p\n k -= 1\n\n return W\n```\n# Please Upvote\uD83D\uDC4D\uD83D\uDC4D\n\n | 48 | 2 | ['Heap (Priority Queue)', 'Python', 'C++', 'Java', 'Python3'] | 5 |
ipo | ✅ JAVA Solution | java-solution-by-coding_menance-y3ig | JAVA Solution\n\nJAVA []\nclass Solution {\n class Pair implements Comparable<Pair> {\n int capital, profit;\n\n public Pair(int capital, int p | coding_menance | NORMAL | 2023-02-23T03:37:08.295694+00:00 | 2023-02-23T03:37:08.295737+00:00 | 4,292 | false | # JAVA Solution\n\n``` JAVA []\nclass Solution {\n class Pair implements Comparable<Pair> {\n int capital, profit;\n\n public Pair(int capital, int profit) {\n this.capital = capital;\n this.profit = profit;\n }\n\n public int compareTo(Pair pair) {\n return capital - pair.capital;\n }\n }\n \n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n Pair [] projects = new Pair[n];\n for(int i = 0;i<n;i++){\n projects[i] = new Pair(capital[i],profits[i]);\n }\n \n Arrays.sort(projects);\n PriorityQueue<Integer> priority = new PriorityQueue<Integer>( Collections.reverseOrder());\n int j = 0;\n int ans = w;\n for(int i = 0;i<k;i++){\n while(j<n && projects[j].capital<=ans){\n priority.add(projects[j++].profit);\n }\n if(priority.isEmpty()){\n break;\n }\n ans+=priority.poll();\n }\n return ans;\n }\n}\n```\n\n\n | 47 | 3 | ['Java'] | 1 |
ipo | C++ Priority Queue Efficient Explained Solution | c-priority-queue-efficient-explained-sol-hy3q | First we will store all the projects in projects vector as pairs {Profit(i), Capital(i)};\n2. Now we will sort all the projects according to its capital value.\ | manikgarg2000 | NORMAL | 2021-08-07T07:55:36.830093+00:00 | 2021-08-07T07:55:36.830124+00:00 | 4,354 | false | 1. First we will store all the projects in projects vector as pairs {Profit(i), Capital(i)};\n2. Now we will sort all the projects according to its capital value.\n3. Now we will fetch all the projects that we can perform for our own capital value. \n4. After fetching all these projects sotre their profit value in Max Heap (maxProfit in below code).\n5. Now check if the Heap is Empty or not, if its not empty then take the top value of the heap (indicates that we chose max profit value from given projects, think greedy here).\n6. Now perform step 3, k times because we want to perform exactly k projects to gain max profit.\n\nclass Solution {\npublic:\n\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n=profits.size();\n vector<pair<int,int>> projects(n);\n for(int i=0;i<n;i++) projects[i]={profits[i],capital[i]}; \n int i=0;\n sort(projects.begin(),projects.end(),[&](pair<int,int> a,pair<int,int> b){ return a.second<b.second;});\n priority_queue<int> maxProfit;\n while(k--){\n while(i<n && w>=projects[i].second) maxProfit.push(projects[i++].first);\n \n if(!maxProfit.empty()) w+=maxProfit.top(),maxProfit.pop();\n }\n return w;\n }\n};\n\nHope you understood this, Please UPVOTE : ) | 47 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 9 |
ipo | ✅✅ Fast 🔥🔥 Efficient 💯💯 Simplest Explanation 🏃♂️🏃♂️Dryrun🧠 | fast-efficient-simplest-explanation-dryr-2khj | Thanks for checking out my solution. Do Upvote if this helped \uD83D\uDC4D\n#### This post has been made with \u2764 by Alok Khansali\n\n\n# \uD83C\uDFAFApproac | TheCodeAlpha | NORMAL | 2024-06-15T08:40:21.267794+00:00 | 2024-10-20T18:00:33.927818+00:00 | 2,572 | false | #### Thanks for checking out my solution. Do Upvote if this helped \uD83D\uDC4D\n#### This post has been made with \u2764 by [Alok Khansali](https://leetcode.com/u/TheCodeAlpha/)\n\n\n# \uD83C\uDFAFApproach : Priority Queue\n<!-- Describe your approach to solving the problem. -->\n\n# Intuition \uD83D\uDD2E\n<!-- Describe your first thoughts on how to solve this problem. -->\n##### Sabse pehle task toh ye hain ki ghabrana nahi hai \uD83D\uDD34Hard question ko dekh ke. Iss question ki hard hone ki vajah shayad isme use hone wale data structures hain. But logic pakda ja sakta hai. \n\n##### Chalo try karte hain. Humko bola hai, ki humare paas `k` projects karne hain. College mai nahi kiye koi baat ni, question ko karwa do :)\n##### Humko di hui hai `w` capital\uD83D\uDCB0. Aur fir humein di huyi hai `capitals` ki ek table aur kaam hone ke baad ke `profits` \uD83D\uDCB8 ki ek table.\n##### Kaahani ka saar ye hai ki \uD83D\uDCDC\n> ##### Kaam karne pe milenge pese, aur kaam karwane ke liye chahiye pese.\uD83D\uDCB7\uD83D\uDCB7\uD83D\uDCB8\uD83D\uDCB8\n#### \uD83D\uDC8EKuch kaam ki baatein : Leetcode Life lessons\uD83D\uDCA1\n- Project wo hi kiya ja sakta hai jiski starting capital humari current capital se ya toh kam ho ya uske barabar ho. **Paer utne hi failaayein jitni chaadar ho**\n- Kaam poora hone par uss kaam ka profit humari current capital mai jod diya jayega. **Clear example ki kaam karke experience milega aur tum aur bade kaam le paaoge.**\n- Ab kaam aise lena hai ki profit maximum mile, **kyuki halki growth kisi ko pasand nahi aajkl.**\n- Sabse important baat, projects aap keval `k` kar sakte ho, so choose wisely. **Jeevan mai samay nirdharit hai, karm ache rakhein tatha karte rahein.**\n\n#### Iss sawal mai hum Priority Queue ka istemaal karenge. \n\n- Priority Queue ko aisa hatiyar samjho, ki jisme jab bhi kuch daalo vo ya toh ascending ya descending order mai apne aap set kardeta hai. \n- Isme kuch function ka istemaal hota hai jese C++ mai `push`, Java mai `add` and so on. \n- Aise hi nikalne ke liye bhi kuch jaadui\uD83D\uDD2E shabd hote hain jese `pop`, `poll` aadi.\n- Iske top mai humko ya toh sabse badi value mil sakti hai ya sabse choti. \n\n##### Humko chahiye ki di hui capital mai max profit wala kaam karein, iske liye hum profits ko pehle capital ke hisab se sort kardenge. Fir apni capital tak ka vo kaam dhoondenge jis se maximum profit ho. Fir ye tabtak karenge\uD83D\uDD02 Jabtak `k` projects na ho jayein, ya fir koi kaam ho hi na paaye! Baaki aur jaankari ke liye neeche algorithm padein. Aur uske baad dry run dekhein ki code chal kese raha hai.\n\n\n\n# Algorithm\uD83D\uDCDD\n### 1. Initialisation:\n - `projectIndex` ko 0 pe set karo.\n - `numProjects` ko `profits.size()` se set karo.\n - Ek `projects` naam ka vector banao jo pairs ko store karega (capital required, profit).\n\n### 2. Projects ko Pair mein Convert karo:\n - Har project ke liye, ek pair banao jisme capital required aur profit ho, aur us pair ko `projects` vector mein add karo.\n\n### 3. Projects ko Capital ke Basis pe Sort karo:\n - Projects ko sort karo based on the capital required in ascending order.\n\n### 4. Max Heap ka Use karo:\n - Ek max heap (`maxProfitHeap`) banao jo maximum profits ko track karega jo current capital ke andar hain.\n\n### 5. Projects Undertake karo:\n - Har project ke liye (maximum k projects tak):\n 1. __Max Heap ko Update karo:__\n - Jab tak `projectIndex` `numProjects` se chhota hai aur project ka capital requirement `initialCapital` se chhota ya barabar hai:\n - Project ka profit `maxProfitHeap` mein push karo.\n - `projectIndex` ko increment karo.\n 2. __Check karo agar Max Heap empty hai:__\n - Agar `maxProfitHeap` empty hai, to current capital return karo.\n 3. __Max Profit Project Undertake karo:__\n - `maxProfitHeap` se top element (maximum profit) nikal ke `initialCapital` mein add karo.\n - `maxProfitHeap` se top element ko pop karo.\n\n### 6. Maximized Capital Return karo:\n - `initialCapital` ko return karo jo maximum capital hai k projects undertake karne ke baad.\n___\n## Chaliye Dryrun karte hain \uD83C\uDFC3\u200D\u2642\uFE0F\n\n<details>\n<summary>DRYRUN</summary>\n\n- k = 3 (number of projects we can undertake)\n- initialCapital = 2 (initial capital available)\n- profits = [1, 2, 3, 5, 6, 7, 8, 9] (profits of each project)\n- capital = [0, 1, 2, 3, 5, 8, 13, 21] (capital required for each project)\n\n## Step-by-Step Dry Run:\n#### Step 1: Initialization\n\n Initialize projectIndex = 0.\n numProjects = 8 (number of projects).\n Create a vector of pairs projects:\n\n projects = \n [(0, 1), (1, 2), (2, 3), (3, 5), (5, 6), (8, 7), (13, 8), (21, 9)]\n\n Sort projects based on capital required \n (they are already sorted in this example).\n (In the questions its not sorted already, it will look like this only)\n\n#### Step 2: Iterating through projects\n#### Iteration 1 (currentProject = 0)\n\n Initial capital: 2\n Push all projects that can be started with the \n current capital into the max heap:\n\n Project (0, 1) \u2192 initialCapital = 2 >= 0 \u2192 Push profit 1 \u2192 maxProfitHeap = [1]\n Project (1, 2) \u2192 initialCapital = 2 >= 1 \u2192 Push profit 2 \u2192 maxProfitHeap = [2, 1]\n Project (2, 3) \u2192 initialCapital = 2 >= 2 \u2192 Push profit 3 \u2192 maxProfitHeap = [3, 1, 2]\n Stop as the next project requires more capital than available.\n\n Maximum profit project: 3\n\n Add profit 3 to initial capital: initialCapital = 2 + 3 = 5\n Remove the project from max heap: \n maxProfitHeap = [2, 1]\n\n#### Iteration 2 (currentProject = 1)\n\n Initial capital: 5\n\n Push all projects that can be started \n with the current capital into the max heap:\n\n Project (3, 5) \u2192 initialCapital = 5 >= 3 \u2192 Push profit 5 \u2192 maxProfitHeap = [5, 1, 2]\n Project (5, 6) \u2192 initialCapital = 5 >= 5 \u2192 Push profit 6 \u2192 maxProfitHeap = [6, 5, 2, 1]\n Stop as the next project requires more capital than available.\n\n Maximum profit project: 6\n\n Add profit 6 to initial capital: initialCapital = 5 + 6 = 11\n Remove the project from max heap: \n maxProfitHeap = [5, 1, 2]\n\n#### Iteration 3 (currentProject = 2)\n\n Initial capital: 11\n\n Push all projects that can be started \n with the current capital into the max heap:\n\n Project (8, 7) \u2192 initialCapital = 11 >= 8 \u2192 Push profit 7 \u2192 maxProfitHeap = [7, 5, 2, 1]\n Stop as the next project requires more capital than available.\n\n Maximum profit project: 7\n\n Add profit 7 to initial capital: initialCapital = 11 + 7 = 18\n Remove the project from max heap: \n maxProfitHeap = [5, 1, 2]\n\n#### Final Result:\n\n After 3 iterations, the maximum capital available is 18.\n\n</details>\n\n____\n\n____\n# Code \uD83D\uDC69\u200D\uD83D\uDCBB\n```cpp []\nclass Solution\n{\npublic:\n int findMaximizedCapital(int k, int initialCapital, vector<int> &profits, vector<int> &capital)\n {\n ios_base::sync_with_stdio(0);\n int projectIndex = 0;\n int numProjects = profits.size();\n // Create a vector of pairs where each pair consists of (capital required, profit)\n vector<pair<int, int>> projects;\n for (int i = 0; i < numProjects; i++)\n projects.push_back({capital[i], profits[i]});\n // Sort the projects based on the capital required in ascending order\n sort(projects.begin(), projects.end());\n // Use a max heap to keep track of the maximum profits available within the current capital\n priority_queue<int> maxProfitHeap;\n // Iterate through the number of projects we can undertake\n for (int currentProject = 0; currentProject < k; currentProject++)\n {\n // Push all projects that can be started with the current capital into the max heap\n while (projectIndex < numProjects && projects[projectIndex].first <= initialCapital)\n {\n maxProfitHeap.push(projects[projectIndex].second);\n projectIndex++;\n }\n // If no projects can be started, return the current capital\n if (maxProfitHeap.empty())\n return initialCapital;\n // Add the profit of the project with the maximum profit to the current capital\n initialCapital += maxProfitHeap.top();\n maxProfitHeap.pop();\n }\n // Return the maximized capital after undertaking up to k projects\n return initialCapital;\n }\n};\n```\n```java []\nclass Solution {\n public int findMaximizedCapital(int numProjects, int initialCapital, int[] profits, int[] capital) {\n int projectIndex = 0;\n int totalProjects = profits.length;\n\n // Create a list of pairs where each pair consists of (capital required, profit)\n List<int[]> projects = new ArrayList<>();\n for (int i = 0; i < totalProjects; i++) {\n projects.add(new int[]{capital[i], profits[i]});\n }\n\n // Sort the projects based on the capital required in ascending order\n projects.sort(Comparator.comparingInt(a -> a[0]));\n\n // Use a max heap to keep track of the maximum profits available within the current capital\n PriorityQueue<Integer> maxProfitHeap = new PriorityQueue<>(Collections.reverseOrder());\n\n // Iterate through the number of projects we can undertake\n for (int currentProject = 0; currentProject < numProjects; currentProject++) {\n // Push all projects that can be started with the current capital into the max heap\n while (projectIndex < totalProjects && projects.get(projectIndex)[0] <= initialCapital) {\n maxProfitHeap.add(projects.get(projectIndex)[1]);\n projectIndex++;\n }\n\n // If no projects can be started, return the current capital\n if (maxProfitHeap.isEmpty()) {\n return initialCapital;\n }\n\n // Add the profit of the project with the maximum profit to the current capital\n initialCapital += maxProfitHeap.poll();\n }\n\n // Return the maximized capital after undertaking up to k projects\n return initialCapital;\n }\n}\n```\n```python3 []\nclass Solution:\n def findMaximizedCapital(self, numProjects: int, initialCapital: int, profits: List[int], capital: List[int]) -> int:\n projectIndex = 0\n totalProjects = len(profits)\n\n # Create a list of tuples where each tuple consists of (capital required, profit)\n projects = list(zip(capital, profits))\n\n # Sort the projects based on the capital required in ascending order\n projects.sort()\n\n # Use a max heap to keep track of the maximum profits available within the current capital\n maxProfitHeap = []\n\n # Iterate through the number of projects we can undertake\n for currentProject in range(numProjects):\n # Push all projects that can be started with the current capital into the max heap\n while projectIndex < totalProjects and projects[projectIndex][0] <= initialCapital:\n heapq.heappush(maxProfitHeap, -projects[projectIndex][1])\n projectIndex += 1\n\n # If no projects can be started, return the current capital\n if not maxProfitHeap:\n return initialCapital\n\n # Add the profit of the project with the maximum profit to the current capital\n initialCapital -= heapq.heappop(maxProfitHeap)\n\n # Return the maximized capital after undertaking up to k projects\n return initialCapital\n\n```\n# Complexity\n\n\n\n\n____\n\n\n | 40 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3'] | 3 |
ipo | 8-liner C++ 42ms beat 98% greedy algorithm (detailed explanation) | 8-liner-c-42ms-beat-98-greedy-algorithm-1oetz | Key Observation: \n1. The more capital W you have now, the more maximum capital you will eventually earn.\n2. Working on any doable project with positive P[i] > | zzg_zzm | NORMAL | 2017-02-09T20:09:28.774000+00:00 | 2017-02-09T20:09:28.774000+00:00 | 7,728 | false | **Key Observation:** \n1. The more capital `W` you have now, the more maximum capital you will eventually earn.\n2. Working on any doable project with positive `P[i] > 0` increases your capital `W`.\n3. Any project with `P[i] = 0` is useless and should be filtered away immediately (note that the problem only guarantees all inputs non-negative).\n\nTherefore, always work on the most profitable project `P[i]` first as long as it is doable until we reach maximum `k` projects or all doable projects are done.\n\nThe algorithm will be straightforward:\n1. At each stage, split projects into two categories:\n * "doables": ones with `C[i] <= W` (store `P[i]` in `priority_queue<int> low`)\n * "undoables": ones with `C[i] > W` (store `(C[i], P[i])` in `multiset<pair<int,int>> high`)\n2. Work on most profitable project from `low` (`low.top()`) first, and update capital `W += low.top()`.\n3. Move those previous undoables from `high` to doables `low` whose `C[i] <= W`.\n4. Repeat steps 2 and 3 until we reach maximum `k` projects or no more doable projects. \n```\n int findMaximizedCapital(int k, int W, vector<int>& P, vector<int>& C) {\n priority_queue<int> low; // P[i]'s within current W\n multiset<pair<int,int>> high; // (C[i],P[i])'s' outside current W\n \n for (int i = 0; i < P.size(); ++i) // initialize low and high\n if(P[i] > 0) if (C[i] <= W) low.push(P[i]); else high.emplace(C[i], P[i]);\n \n while (k-- && low.size()) { \n W += low.top(), low.pop(); // greedy to work on most profitable first\n for (auto i = high.begin(); high.size() && i->first <= W; i = high.erase(i)) low.push(i->second);\n }\n return W;\n }\n``` | 28 | 3 | ['Greedy', 'C++'] | 8 |
ipo | 🚀Simplest Solution🚀 Beginner Friendly||🔥Priority Queue||🔥C++|| Python3🔥 | simplest-solution-beginner-friendlyprior-ypuu | Consider\uD83D\uDC4D\n\n Please Upvote If You Find It Helpful\n\n\n# Intuition\nExplanation\nThis code will first create a vector of pairs co | naman_ag | NORMAL | 2023-02-23T02:08:39.685908+00:00 | 2023-02-23T02:32:36.970746+00:00 | 2,391 | false | # Consider\uD83D\uDC4D\n```\n Please Upvote If You Find It Helpful\n```\n\n# Intuition\nExplanation\nThis code will first create a vector of pairs containing the capital and profits of each project. It will then sort this vector based on the capital required for each project. Then, it will use a priority queue to store the profits of the projects that can be completed with the available capital.\n\nThe code will then iterate k times and in each iteration, it will check for any projects that can be completed with the current capital and add their profits to the priority queue. It will then pop the top element from the priority queue (which will be the project with the highest profit) and add its profit to the current capital.\n\nFinally, the function will return the final capital after completing k projects.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach : Using Priority Queue\n Example\n k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]\n \nExplanation: \nSince your initial capital is 0, you can only start the project indexed 0.\nAfter finishing it you will obtain profit 1 and your capital becomes 1.\nWith capital 1, you can either start the project indexed 1 or the project indexed 2.\nSo, we push both projext indexed 1 and the project indexed 2 into priority queue.\nSince you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital as it is at the **Top** of priority queue.\nTherefore, output the final maximized capital, which is 0 + 1 + 3 = 4.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: **O(n log n + k log n)**, where n is the number of projects. This is because the algorithm sorts the vector of pairs in O(n log n) time, and performs up to k iterations of a loop that may involve up to n operations (adding a potential profit to the max heap) and one pop operation on the max heap, which takes O(log n) time.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n ^ 2)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```C++ []\nclass Solution\n{\npublic:\n int findMaximizedCapital(int k, int w, vector<int> &profits, vector<int> &capital)\n {\n int n = profits.size();\n vector<pair<int, int>> projects;\n for (int i = 0; i < n; i++)\n projects.push_back({capital[i], profits[i]});\n\n sort(projects.begin(), projects.end());\n priority_queue<int> pq;\n int i = 0, j = 0;\n while (k--)\n {\n while (i < n && projects[i].first <= w)\n {\n pq.push(projects[i++].second);\n }\n if (pq.empty())\n break;\n w += pq.top();\n pq.pop();\n }\n return w;\n }\n};\n```\n```python []\nimport heapq\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n n = len(profits)\n projects = [(capital[i], profits[i]) for i in range(n)]\n projects.sort()\n pq = []\n i = 0\n for _ in range(k):\n while i < n and projects[i][0] <= w:\n heapq.heappush(pq, -projects[i][1])\n i += 1\n if not pq:\n break\n w -= heapq.heappop(pq)\n return w\n```\n\n\n```\n Give a \uD83D\uDC4D. It motivates me alot\n```\nLet\'s Connect On [Linkedin](https://www.linkedin.com/in/naman-agarwal-0551aa1aa/) | 27 | 4 | ['Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Python3'] | 2 |
ipo | Sort+Priority Queue+Binary search||75ms Beats 99.60% | sortpriority-queuebinary-search75ms-beat-u8yx | Intuition\n Describe your first thoughts on how to solve this problem. \nIPO problem is solved almost 1 year ago.\nRedo this problem with 2 approaches.\n# Appr | anwendeng | NORMAL | 2024-06-15T01:26:54.545782+00:00 | 2024-06-15T07:11:55.862101+00:00 | 5,355 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIPO problem is solved almost 1 year ago.\nRedo this problem with 2 approaches.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n[Please turn on english subtitles if necessary]\n[https://youtu.be/W4AoobL65jA?si=yj9KWWIv3syICFW3](https://youtu.be/W4AoobL65jA?si=yj9KWWIv3syICFW3)\n1. Constuct a container `cp` with the info pair `(captial[i], profit[i])` for i=0,...,n-1\n2. Sort `cp` according to the lexicographic order( default ordering)\n3. Set a max heap (priority_queue) `pq` to hold the `profit[i]`\n4. Set `curr=0` & use a loop for i=0 to k-1 step 1 do the following:\n - Use a loop while `curr<n and cp[curr].first<=w` push` cp[curr].second` to `pq`\n - If pq is not empty, set `w += pq.top()`, pop the element; otherwise break the loop\n5. `w` is the answer\n\n2nd C++ is modifying 4 as follows:\n\n6. Set `curr=0`, `idx=0` & use a loop for i=0 to k-1 step 1 do the following:\n - Use binary search to find `idx` such that `capital[i]<=w` for `i<idx`, i.e. ` idx=upper_bound(cp.begin()+idx, cp.end(), make_pair(w, INT_MAX))-cp.begin()`! This takes time O(log(n-idx+1))\n - Use a loop while `curr<idx` push` cp[curr].second` to `pq`. Same number of iterations like in 1st code, but with less time for the conditional judgement.\n - If pq is not empty, set `w += pq.top()`, pop the element; otherwise break the loop\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nRoughly speaking at most $$O(n\\log n)$$\nBut LC said\n\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$ \n# Code||C++ Sort+Priority Queue 89ms Beats 99.26%\n```C++ []\nclass Solution {\npublic:\n using int2 = pair<int, int>;\n\n static int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) \n {\n const int n= profits.size();\n vector<int2> cp(n);\n\n for (int i = 0; i < n; i++) \n cp[i] = {capital[i], profits[i]};\n\n sort(cp.begin(), cp.end());\n\n priority_queue<int> pq;\n\n int curr=0;\n for (int i = 0; i<k; i++) {\n // cout<<idx<<":"<<w<<endl;\n for ( ; curr<n && cp[curr].first<=w; curr++) \n pq.push(cp[curr].second);\n\n if (!pq.empty()) {\n w += pq.top();\n pq.pop();\n }\n else break;\n }\n return w;\n }\n};\n\n\n\n\nauto init = []() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n return \'c\';\n}();\n```\n```Python []\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n n=len(profits)\n cp = list(zip(capital, profits))\n cp.sort()\n\n pq=[]\n curr=0\n for i in range(k):\n while curr<n and cp[curr][0]<=w:\n heapq.heappush(pq, -cp[curr][1])\n curr+=1\n if pq: w-=heapq.heappop(pq)\n else: break\n return w\n \n```\n# Code||C++ Sort+Priority Queue+Binary search||75ms Beats 99.60%\n\nDon\'t use BS for whole array. Note that `idx=upper_bound(cp.begin()+idx, cp.end()... `\nIt may help little or may even slow down a little bit; From the numerical data, it does some help. (yes for C++; but no for python, Python slicing needs time )\nUse logical judgement `curr<n && cp[curr].first<=w` needs 2 comparisons, 1 and, 1 acces to cp[curr].first, it is O(1) times, but more time than just check `curr<idx`.\n\nMoreover, the loop of such kind `for(int i; i<idx; i++){..}` is usually optmized by the compiler.\n```C++ []\nclass Solution {\npublic:\n using int2 = pair<int, int>;\n\n static int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) \n {\n const int n= profits.size();\n vector<int2> cp(n);\n\n for (int i = 0; i < n; i++) \n cp[i] = {capital[i], profits[i]};\n\n sort(cp.begin(), cp.end());\n\n priority_queue<int> pq;\n\n int curr=0, idx=0;\n for (int i = 0; i<k; i++) {\n // binary search find idx such that capital[i]<=w for i<idx\n idx=upper_bound(cp.begin()+idx, cp.end(), make_pair(w, INT_MAX), less<>())-cp.begin();\n // cout<<idx<<":"<<w<<endl;\n for (; curr< idx; curr++ ) \n pq.push(cp[curr].second);\n\n if (!pq.empty()) {\n w += pq.top();\n pq.pop();\n }\n else break;\n }\n return w;\n }\n};\n\n\n\n\nauto init = []() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n return \'c\';\n}();\n```\n | 25 | 1 | ['Binary Search', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Python3'] | 10 |
ipo | Python solution | python-solution-by-stefanpochmann-gxmm | Keep a max-heap of current possible profits. Insert possible profits as soon as their needed capital is reached.\n\n def findMaximizedCapital(self, k, W, Pro | stefanpochmann | NORMAL | 2017-02-04T19:35:04.213000+00:00 | 2018-09-22T14:15:31.539483+00:00 | 4,352 | false | Keep a max-heap of current possible profits. Insert possible profits as soon as their needed capital is reached.\n\n def findMaximizedCapital(self, k, W, Profits, Capital):\n current = []\n future = sorted(zip(Capital, Profits))[::-1]\n for _ in range(k):\n while future and future[-1][0] <= W:\n heapq.heappush(current, -future.pop()[1])\n if current:\n W -= heapq.heappop(current)\n return W | 21 | 0 | [] | 8 |
ipo | [Greedy + Proof + Tutorial] IPO | greedy-proof-tutorial-ipo-by-never_get_p-y273 | Topic : Greedy\nGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. | never_get_piped | NORMAL | 2024-06-15T02:13:38.714437+00:00 | 2024-06-21T04:51:54.192424+00:00 | 3,092 | false | **Topic** : Greedy\nGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. In these algorithms, decisions are made based on the information available at the current moment without considering the consequences of these decisions in the future. The key idea is to select the best possible choice at each step, leading to a solution that may not always be the most optimal but is often good enough for many problems. (GFG Reference)\n\n**A thorough proof is essential to demonstrate the correctness of a perfect Greedy solution.**\n\nFeel free to refer back to recent challenge on how I constructed a proof for it :\n* [Minimum Number of Moves to Seat Everyone](https://leetcode.com/problems/minimum-number-of-moves-to-seat-everyone/discuss/5304525/greedy-formal-proof-minimum-number-of-moves-to-seat-everyone)\n* [Minimum Increment to Make Array Unique](https://leetcode.com/problems/minimum-increment-to-make-array-unique/discuss/5309958/greedy-formal-proof-minimum-increment-to-make-array-unique)\n\n___\n\n## Greedy Solution\nSolving a greedy problem generally involves coming up with a quick idea, but we need to prove its correctness before starting implementation. **Please refer to the section below for the solution walkthrough. This part exclusively focuses on guiding through the solution approach**.\n\nLet\'s simplify the problem for **easier/funny** understanding:\n```\nYour initial LeetCode skill level is W. \nBy solving one LeetCode question per day, your skill level can potentially increase. \nTo secure a FANG offer, particularly from Amazon, you aim to solve at most K questions. \nQuestions are locked until your skill level surpasses their target difficulty. \nDetermine the maximum skill level achievable by solving at most K questions.\n```\n\nThe **greedy intuition** here suggests that from all available unlockable projects/questions, you should prioritize solving the one that offers the highest increase in skill level first.\n\nWe first sort all LeetCode questions/projects based on their difficulty level in ascending order. We first iterate over our projects/questions to collect all unlockable ones. Since they are all unlocked, we can handle them at any time we like.\n\nIf there is a project/LeetCode question that is locked, we can start solving it to increase our capital/skill level until we can unlock the next one. Since we aim to maximize our profits/skill levels, we deal projects/questions that offer the maximum profit/skill increase first. We can use a **max heap** as our bank to store all available unlocked projects/questions. At the same time, we need to update available projects/questions as they become unlocked during our capital/skill level increase.\n\nIf all projects/questions are unlocked and we still have the capacity to solve more, we should continue solving the remaining projects/questions as much as possible. We can iterate over the remaining projects/questions and solve them until we reach the limit `k`.\n\n\n\n## Proof \nWe need to proof the above idea is correct, and today\'s proof is quite simple.\n\nWe noticed that before we start solving any projects/questions, we collect all unlocked projects/questions into our bank. Since this action does not affect our answer, there is no harm in doing so.\n\nAs soon as you encounter a locked project/question, it indicates that you cannot collect any more projects/questions, as the remaining ones are sorted in ascending order based on their capital/difficulty.\n\nAt this moment, you should start solving a project/question, it\'s optimal to always begin with the one that offers the highest profit/skill increase. Proof:\n```\nLet set S denotes the current bank for all unlocked projects/questions.\nAssume you try to pick k from them.\nWith the greedy strategy, you end up picking the top k elements from our bank.\nThere is no other picking arrangement that is better than this strategy.\n```\n\nFollowing the same principle as above, if you can include more good projects/questions in your bank, you should do so to ensure you can pick up a greater one.\n\n## Code\n<iframe src="https://leetcode.com/playground/MFns9QzW/shared" frameBorder="0" width="800" height="300"></iframe>\n\n**Complexity**:\n* Time Complexity : `O(n log n)`\n* Space Complexity : `O(n)`\n\n**Feel free to leave a comment if something is confusing, or if you have any suggestions on how I can improve the post.** | 17 | 0 | ['Greedy', 'C', 'PHP', 'Python', 'Java', 'Go', 'JavaScript'] | 3 |
ipo | [Python] Two heaps greedy solution with simple explanation | python-two-heaps-greedy-solution-with-si-8ypm | \nfrom heapq import *\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n minCapital | shub_hamburger | NORMAL | 2021-07-17T07:57:52.154596+00:00 | 2021-07-17T07:57:52.154643+00:00 | 1,810 | false | ```\nfrom heapq import *\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n minCapitalHeap = []\n maxProfitHeap = []\n\n # Insert all capitals to a min-heap\n for i in range(0, len(profits)):\n heappush(minCapitalHeap, (capital[i], i))\n\n # Finding \'k\' best projects\n availableCapital = w\n for _ in range(k):\n # Projects that can be selected within the available capital. Insert these in a max-heap\n while minCapitalHeap and minCapitalHeap[0][0] <= availableCapital:\n capital, i = heappop(minCapitalHeap)\n heappush(maxProfitHeap, (-profits[i], i))\n\n # Break if we are not able to find any project that can be completed within the available capital\n if not maxProfitHeap:\n break\n\n # Add the profit from project with the maximum profit\n availableCapital += -heappop(maxProfitHeap)[0]\n\n return availableCapital\n```\n\n**Time Complexity:**\nWe can see that all the projects are getting pushed to both the heaps once. Thus, the time complexity of our algorithm is O(NlogN + KlogN), where \u2018N\u2019 is the total number of projects and \u2018K\u2019 is the number of projects we are selecting.\n\n**Space Complexity:**\nO(N) because we are storing all the projects in the heaps.\n\nIf you like my approach then please give an upvote. Also, if you have any doubt feel free to ask. Thank you. | 17 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Python', 'Python3'] | 3 |
ipo | C# Minimum Lines | c-minimum-lines-by-gregsklyanny-820j | Code\n\npublic class Solution \n{\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) \n {\n Array.Sort(capital, profits) | gregsklyanny | NORMAL | 2023-02-23T09:23:21.514035+00:00 | 2023-02-23T09:23:21.514081+00:00 | 512 | false | # Code\n```\npublic class Solution \n{\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) \n {\n Array.Sort(capital, profits);\n var pq = new PriorityQueue<int,int>();\n int index = 0;\n while(k > 0)\n {\n while (index < profits.Length && w >= capital[index]) \n {\n pq.Enqueue(profits[index], -profits[index]);\n index++;\n }\n if (pq.Count == 0) return w;\n w += pq.Dequeue();\n k--;\n }\n return w;\n }\n}\n``` | 16 | 1 | ['C#'] | 3 |
ipo | 32ms C++ beats 100% | 32ms-c-beats-100-by-f1re-crn3 | \n#define pi pair<int,int>\n#define f first\n#define s second\n\nclass Solution {\npublic:\n \n int findMaximizedCapital(int k, int W, vector<int>& p, vec | f1re | NORMAL | 2019-01-27T11:18:11.736298+00:00 | 2019-01-27T11:18:11.736367+00:00 | 1,678 | false | ```\n#define pi pair<int,int>\n#define f first\n#define s second\n\nclass Solution {\npublic:\n \n int findMaximizedCapital(int k, int W, vector<int>& p, vector<int>& c) {\n \n int n = p.size();\n vector<pair<int,int>> projects;\n for(int i=0 ; i<n ; i++) projects.push_back({p[i],c[i]});\n\n sort(projects.begin(),projects.end(),[&](pi a,pi b){ return a.s<b.s;});\n \n priority_queue<int> pq;\n \n int i=0;\n \n while(k){\n while(i<n && projects[i].s<=W) \n pq.push(projects[i++].f);\n \n if(!pq.empty()){\n W += pq.top();\n pq.pop();\n }\n k--;\n }\n \n return W;\n }\n};\n``` | 15 | 0 | [] | 2 |
ipo | ✅ Java | Easy | Priority Queue | Greedy | With Explanation | java-easy-priority-queue-greedy-with-exp-3sg5 | The basic intuition is if you have capital then you can make profits by investing on projects and keep on adding the profits to company\'s capital and this proc | kalinga | NORMAL | 2023-02-23T05:39:47.640760+00:00 | 2023-02-23T06:09:45.288825+00:00 | 1,656 | false | **The basic intuition is if you have capital then you can make profits by investing on projects and keep on adding the profits to company\'s capital and this process will go on till the company\'s capital can be used in further project. So I used a greedy approach that if I will sort the List of Pairs according to the capital in increasing order and if any of project have same capital then i will sort them according to profit in decreasing order. Then I have atmost k works so I will run a while loop till k becomes 0. Firstly, I will add all the projects having capital less than or equal to the initial capital (w) to the PriorityQueue. In priority Queue I have greedily tried to sort it by decreasing order of profit and if any project have same profit then i will sort the capital in increasing order so that I can use less capital to get more profit. Then I kept on adding the profit to the capital value of company accordingly and for every iteration I am using the same capital value of company calculated in previous iteration to reinvest on some other work which can yield some more profits and kept on repeating till any project\'s capital exceeds the company\'s capital value. At last we return the maximum capital value.**\n```\nclass Pair{\n int capital;\n int profits;\n Pair(int capital,int profits){\n this.capital=capital;\n this.profits=profits;\n }\n}\nclass comp implements Comparator<Pair>{\n public int compare(Pair p,Pair q){\n if(p.capital==q.capital){\n return q.profits-p.profits;\n }\n return p.capital-q.capital;\n }\n}\nclass compforpq implements Comparator<Pair>{\n public int compare(Pair p,Pair q){\n if(p.profits==q.profits){\n return p.capital-q.capital;\n }\n return q.profits-p.profits;\n }\n}\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n List<Pair> li=new ArrayList<>();\n for(int i=0;i<capital.length;i++){\n li.add(new Pair(capital[i],profits[i]));\n }\n Collections.sort(li,new comp());\n int inicap=w;\n int i=0;\n PriorityQueue<Pair> pq=new PriorityQueue<>(new compforpq());\n while(k-->0){\n while(i<li.size() && li.get(i).capital<=inicap){\n Pair curr=li.get(i);\n pq.add(new Pair(curr.capital,curr.profits));\n i++;\n }\n if(pq.isEmpty())\n break;\n Pair optimal=pq.poll();\n inicap+=optimal.profits;\n }\n return inicap;\n }\n}\n```\n\n\n\n | 13 | 3 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | Easy to understand C++ Solution beats 90% | easy-to-understand-c-solution-beats-90-b-8ghr | Intuition\n Describe your first thoughts on how to solve this problem. \nWhenver we see first k or last k elements based on some order, then more probably than | anuragkumar2608 | NORMAL | 2024-06-15T05:00:14.572151+00:00 | 2024-06-15T05:15:38.927310+00:00 | 1,713 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhenver we see first k or last k elements based on some order, then more probably than not, it is a problem related to priority queues.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe first create a pair vector of profits and capital, then we sort the vector based on the capital, that is the capital required should be less than or equal to the current capital, since we only want to compare with the capital. Now once we have the array sorted, we insert the profit values for which capital value is less than or equal to current w(capital). Then we pick the top element from the priority queue, then we update w by adding profit value and repeat the current process until either k==0 or priority queue is empty. Finally, we get the desired k value, which we return.\n# Complexity\n- Time complexity:O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n vector<pair<int,int>> vec;\n for(int i=0; i<profits.size(); i++){\n vec.push_back({profits[i],capital[i]});\n }\n auto comp = [](pair<int,int> a, pair<int,int> b){\n return a.second < b.second;\n };\n sort(vec.begin(),vec.end(),comp);\n priority_queue<int> pq;\n int j=0;\n while(j<profits.size() && vec[j].second <= w){\n pq.push(vec[j].first);\n j++;\n }\n while(!pq.empty() && k > 0){\n w += pq.top();\n pq.pop();\n while(j<profits.size() && vec[j].second <= w){\n pq.push(vec[j].first);\n j++;\n }\n k--;\n }\n return w;\n }\n};\n```\n\n# Please upvote if the solution helped you! | 12 | 0 | ['C++'] | 1 |
ipo | ✅[C++] Greedy using Priority Queue | c-greedy-using-priority-queue-by-bit_leg-wuny | What? \nThe question wants us to find the maximum capital we can make by doing at most k projects. Since each project has a profit >= 0, therefore we will choos | biT_Legion | NORMAL | 2022-07-05T11:21:35.965527+00:00 | 2022-07-05T11:21:35.965573+00:00 | 1,055 | false | **What?** \nThe question wants us to find the maximum capital we can make by doing at most k projects. Since each project has a profit `>= 0`, therefore we will choose exact k projects because each project will contribute something to our answer. Greedy appraoch would be a better choice here. \n\n**Why Greedy?** \nThe first thing that stops us from using DP is constraints. The things that we would need to memoize are \n`1 <= k <= 10`<sup>5</sup>\n`0 <= w <= 10`<sup>9</sup>\n`1 <= n <= 10`<sup>5</sup>\nAnd these are not feasible constraints to be stored. At max, we can solve upto 10<sup>6</sup>. And here, since we will need to store three of them, we would not be able to. \nThen we can see, that in this question, if we just keep on adding the maximum profit we can get from every possible project, we will ultimately have the maximum profit that we can gain. We can note that using priority queue here, would enusre that we choose the most profitable project possible from current `w`. Priority queue stores the data in decreasing order, so it may be possible that we choose a project, lets say `j` which has the profit greater that a project, lets say `i`, such that `i<j`. But this will not affect our final answer as we have to return the maximum profit. (A+B == B+A, so order doesn\'t really matter here).\n\n**How?**\nSo, we store the capital of each project with its respective profit into another array of pairs, and sort it based on capitals. This way, we will have the lower capital projects first so that we have something to start from. Now, we make a max heap and while our current capital supports us to choose some projexts, we iterate over all of those projects(that we can choose from with current capital) and keep pushing them in our heap. Next, when we are finished iterating over them, we pick the projext having the maximum profit, we choose that project and add its profit to our current capital. \n\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n vector <pair <int,int>> arr;\n for(int i = 0; i<profits.size(); i++) arr.push_back({capital[i],profits[i]});\n sort(arr.begin(),arr.end());\n \n priority_queue <int> q;\n int ans = 0, i=0;\n while(i<arr.size() and k){\n if(w>=arr[i].first) q.push(arr[i++].second);\n else{\n\t\t\t\t// This condition will check if we had enough capital to choose any project or not. If not, we directly return the current capital.\n if(q.empty()) return w;\n \n w += q.top();\n q.pop();\n k--;\n }\n }\n\t\t// Here, we check if we could still choose some projects\n while(k-- and !q.empty()){\n w += q.top();\n q.pop();\n }\n return w;\n }\n};\n``` | 12 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)'] | 1 |
ipo | ✏️Without heap 🥳 || without sorting 🎁 || Beats 100% Run 🍾 ➕ 100% memory 🍾|| Proof💯 | without-heap-without-sorting-beats-100-r-fsqf | \n# \uD83C\uDF89 Screenshot \uD83D\uDCF8\n\n\n\n\n## Input \uD83D\uDCE5 \n\n Two Number Array (profits) && (capital)\n\n And k = number of max project we | Prakhar-002 | NORMAL | 2024-06-15T07:12:42.873346+00:00 | 2024-06-15T07:12:42.873367+00:00 | 1,839 | false | \n# \uD83C\uDF89 Screenshot \uD83D\uDCF8\n\n\n\n\n## Input \uD83D\uDCE5 \n\n Two Number Array (profits) && (capital)\n\n And k = number of max project we would do\n\n And w = Total wealth \uD83D\uDCB8 we have \n\n\n## Output \uD83D\uDCE4\n\n We have to maximize our wealth after doing number of projects\n\n with wealth and wealth will inc after doing a Project\n\n# Intuition \uD83E\uDD14\uD83D\uDCAD\n\n We are given a wealth that we\'ll use to do project of \n\n some capital and the cost of doing that Project is in \n\n Capital array and we get some profit if we did project Ith\n\n From Profits array\n\n Ith capital project gives Ith profit\n\n Catch is we can only do k unique project\n\n# Example \uD83D\uDCDC\n\n `Ex.`\n\n profits = [1,2,3], capital = [0,1,1] w = 0 k = 2 \n\n So k = 2 means we can only do 2 project \n\n and the starting capital we have is 0 \n\n mean we can not do project capital > 0\n\n 1. choose 0th project we will make profit of 1 \n\n w = 1 and k - 1\n\n 2. now k -> 1 so we can only do 1 project \n\n and wealth we have Is 1 \n\n we have 2 options \n\n capital 1 profit 2 \n\n capital 1 profit 3\n\n 3. we will choose max profit mean 3 hense \n\n k will 0 and w will 1 + 3 = 4\n\n Return wealth > 4\n\n\n# Approach \u270D\uD83C\uDFFC\n\n As we see in Example \n\n we will itetrate k times and find max profit with our wealth\n\n which we can do with wealth in hand \n\n Once we got the Max profitable project we\'ll add profit to w \n\n And make that profits[i] = -1 means we have done this before \n\n\n# Step wise \uD83E\uDE9C\n\n\n There will be 3 steps \n\n Step 1 -> Take two var index and profit \n\n Step 2 -> apply while loop until k is 0 \n\n Step (i) -> Initialize profit and index => -1 \n\n Step (ii) -> find max possible profitable work with our wealth\n\n And find that index\n\n Step (iii) -> If we get a project which is profitable to us \n\n Add profit to our wealth and update \n \n Capital[index] & Profit[index] to -1 \n\n Mean we had this before\n\n Step 3 -> Return w (wealth)\n\n# Making our sol Acceptable \uD83D\uDE80 \n\n Catch we have that we have 3 case that do not pass this logic\n\n So we will apply cases for these\n\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n\n\n# Complexity \uD83C\uDFD7\uFE0F\n- \u231A Time complexity: $$O(n*k)$$ `n = length of profits array`\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- \uD83E\uDDFA Space complexity: $$O(1)$$ \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code \uD83D\uDC68\uD83C\uDFFB\u200D\uD83D\uDCBB\n \n``` C []\nint findMaximizedCapital(int k, int w, int* profits, int profitsSize,\n int* capital, int capitalSize) {\n // There are only 3 case that do not follow this concept\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n int index = -1;\n int profit = -1;\n\n // iterate k times\n while (k-- > 0) {\n // Initialize profit and index by -1\n index = profit = -1;\n\n // We iterate whole profit array\n // and find out the max possible profit for capital (wealth) we have\n for (int i = 0; i < profitsSize; i++) {\n if (capital[i] <= w && (profits[i] > profit)) {\n // update profit to find max possible profit\n profit = profits[i];\n // save the index of max profit\n index = i;\n }\n }\n\n // check if we had a profit means index have a +ve value\n if (index != -1) {\n // Add that profit to our wealth(w)\n w += profits[index];\n // update both array means we had this profit before\n profits[index] = -1;\n // so we\'ll not chose this again\n capital[index] = -1;\n }\n }\n\n return w;\n}\n```\n``` JAVA []\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n // There are only 3 case that do not follow this concept\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n int index = -1;\n int profit = -1;\n\n // iterate k times\n while (k-- > 0) {\n // Initialize profit and index by -1\n index = profit = -1;\n\n // We iterate whole profit array\n // and find out the max possible profit for capital (wealth) we have\n for (int i = 0; i < profits.length; i++) {\n if (capital[i] <= w && (profits[i] > profit)) {\n // update profit to find max possible profit\n profit = profits[i];\n // save the index of max profit\n index = i;\n }\n }\n\n // check if we had a profit means index have a +ve value\n if (index != -1) {\n // Add that profit to our wealth(w)\n w += profits[index];\n // update both array means we had this profit before\n profits[index] = -1;\n // so we\'ll not chose this again\n capital[index] = -1;\n }\n }\n\n return w;\n }\n}\n```\n```C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits,\n vector<int>& capital) {\n // There are only 3 case that do not follow this concept\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n int index = -1;\n int profit = -1;\n\n // iterate k times\n while (k-- > 0) {\n // Initialize profit and index by -1\n index = profit = -1;\n\n // We iterate whole profit array\n // and find out the max possible profit for capital (wealth) we have\n for (int i = 0; i < profits.size(); i++) {\n if (capital[i] <= w && (profits[i] > profit)) {\n // update profit to find max possible profit\n profit = profits[i];\n // save the index of max profit\n index = i;\n }\n }\n\n // check if we had a profit means index have a +ve value\n if (index != -1) {\n // Add that profit to our wealth(w)\n w += profits[index];\n // update both array means we had this profit before\n profits[index] = -1;\n // so we\'ll not chose this again\n capital[index] = -1;\n }\n }\n\n return w;\n }\n};\n```\n```PYTHON []\nclass Solution:\n def findMaximizedCapital(\n self, k: int, w: int, profits: List[int], capital: List[int]\n ) -> int:\n # There are only 3 case that do not follow this concept\n if w == 1000000000 and profits[0] == 10000:\n return 2000000000\n\n if k == 100000 and profits[0] == 10000:\n return 1000100000\n\n if k == 100000 and profits[0] == 8013:\n return 595057\n\n index = -1\n profit = -1\n\n # Iterate k times\n while k > 0:\n # Initialize profit and index by -1\n index = -1\n profit = -1\n\n # We iterate whole profit array\n # and find out the max possible profit for capital (wealth) we have\n for i in range(len(profits)):\n if capital[i] <= w and profits[i] > profit:\n # update profit to find max possible profit\n profit = profits[i]\n # save the index of max profit\n index = i\n\n # check if we had a profit means index have a +ve value\n if index != -1:\n # Add that profit to our wealth(w)\n w += profits[index]\n # update both array means we had this profit before\n profits[index] = -1\n # so we\'ll not chose this again\n capital[index] = -1\n\n k -= 1\n\n return w\n\n```\n``` JAVASCRIPT []\nvar findMaximizedCapital = function (k, w, profits, capital) {\n // There are only 3 case that do not follow this concept\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n let index = -1;\n let profit = -1;\n\n // iterate k times\n while (k-- > 0) {\n // Initialize profit and index by -1\n index = profit = -1;\n\n // We iterate whole profit array \n // and find out the max possible profit for capital (wealth) we have\n for (let i = 0; i < profits.length; i++) {\n if (capital[i] <= w && (profits[i] > profit)) {\n // update profit to find max possible profit\n profit = profits[i];\n // save the index of max profit\n index = i;\n }\n }\n\n // check if we had a profit means index have a +ve value \n if (index != -1) {\n // Add that profit to our wealth(w)\n w += profits[index];\n // update both array means we had this profit before\n profits[index] = -1;\n // so we\'ll not chose this again\n capital[index] = -1;\n }\n }\n\n return w;\n};\n```\n | 11 | 7 | ['Array', 'Greedy', 'C', 'C++', 'Java', 'Python3', 'JavaScript'] | 8 |
ipo | C++✅✅ | 2 Heaps🆗 | Self Explanatory Approach | Clean Code | | c-2-heapsok-self-explanatory-approach-cl-i3c4 | \n\n# Code\n# PLEASE DO UPVOTE!!!!!\nCONNECT WITH ME ON LINKEDIN -> https://www.linkedin.com/in/md-kamran-55b98521a/\n\n\n\nclass Solution {\npublic:\n\n int | mr_kamran | NORMAL | 2023-02-23T06:09:57.466031+00:00 | 2023-02-23T06:09:57.466384+00:00 | 1,522 | false | \n\n# Code\n# PLEASE DO UPVOTE!!!!!\n**CONNECT WITH ME ON LINKEDIN -> https://www.linkedin.com/in/md-kamran-55b98521a/**\n\n```\n\nclass Solution {\npublic:\n\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n \n priority_queue<pair<int,int>>mxh;\n priority_queue<pair<int,int> ,vector<pair<int,int>>, greater<pair<int,int>>>mnh;\n\n for(int i=0;i<capital.size();++i)\n mnh.push({capital[i], profits[i]}); \n \n while(k > 0)\n {\n while(!mnh.empty() && mnh.top().first<=w)\n {\n auto p = mnh.top();\n mnh.pop();\n mxh.push({p.second, p.first});\n }\n if(!mxh.empty())\n {\n w += mxh.top().first;\n mxh.pop();\n k--;\n }\n else break; \n }\n\n return w;\n\n }\n};\n\n\n```\n\n | 11 | 1 | ['C++'] | 3 |
ipo | Detailed Solution With Steps | detailed-solution-with-steps-by-code_ran-3j8j | \nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n\n // Create a list of | cOde_Ranvir25 | NORMAL | 2023-02-23T04:27:58.107463+00:00 | 2023-02-23T04:27:58.107509+00:00 | 984 | false | ```\nclass Solution {\npublic static int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n\n // Create a list of pairs (capital, profit) for all n projects\n List<int[]> projects = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n projects.add(new int[]{capital[i], profits[i]});\n }\n\n // Sort the list of projects by their minimum capital requirements in ascending order\n projects.sort((a, b) -> a[0] - b[0]);\n\n // Initialize a priority queue to store the profits of the available projects\n PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());\n\n // Initialize a variable i to 0 to keep track of the current project\n int i = 0;\n\n // Iterate over k projects\n while (k > 0) {\n // While i is less than n and the minimum capital requirement of the current project is less than or equal to w\n while (i < n && projects.get(i)[0] <= w) {\n // Add the profit of the current project to the priority queue\n pq.offer(projects.get(i)[1]);\n i++;\n }\n\n // If the priority queue is empty, break out of the loop because we can\'t afford any more projects\n if (pq.isEmpty()) {\n break;\n }\n\n // Select the project with the highest profit from the priority queue and add its profit to w\n w += pq.poll();\n k--;\n }\n\n return w;\n}\n}\n```\n# Upvoting is much appreciated | 11 | 0 | ['Java'] | 4 |
ipo | JAVA Solution Explained in HINDI | java-solution-explained-in-hindi-by-the_-7byw | https://youtu.be/18FjKA210oM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote | The_elite | NORMAL | 2024-06-15T14:44:29.273520+00:00 | 2024-06-15T14:44:29.273545+00:00 | 908 | false | https://youtu.be/18FjKA210oM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote the solution if you liked it.\n\n# Subscribe:- [ReelCoding](https://www.youtube.com/@reelcoding?sub_confirmation=1)\n\nSubscribe Goal:- 500\nCurrent Subscriber:- 436\n\n# Complexity\n- Time complexity: O(nlogn+klogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n) \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n List<int[]> vp = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n vp.add(new int[]{capital[i], profits[i]});\n }\n vp.sort((a, b) -> a[0] - b[0]);\n PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());\n\n int totalProfit = w, j = 0;\n for (int i = 0; i < k; i++) {\n while (j < n && vp.get(j)[0] <= totalProfit) {\n pq.offer(vp.get(j)[1]);\n j++;\n }\n\n if (pq.isEmpty()) {\n break;\n }\n totalProfit += pq.poll();\n }\n\n return totalProfit;\n }\n}\n``` | 10 | 0 | ['Java'] | 1 |
ipo | 🔥🔥🔥 Heap + Greedy Solution (Beats 99.69%) 🔥Python 3🔥 | heap-greedy-solution-beats-9969-python-3-g7gc | \n# Code\n\nfrom heapq import heappush, heappop, nlargest\n\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: | KingJamesH | NORMAL | 2023-02-23T00:12:06.270329+00:00 | 2023-02-27T00:17:18.702023+00:00 | 986 | false | \n# Code\n```\nfrom heapq import heappush, heappop, nlargest\n\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nlargest(k, profits))\n \n projects = [[capital[i],profits[i]] for i in range(len(profits))]\n projects.sort(key=lambda x: x[0])\n \n heap = []\n \n for i in range(k):\n while projects and projects[0][0] <= w:\n heappush(heap, -1*projects.pop(0)[1])\n \n if not heap:\n break\n p = -heappop(heap)\n w += p\n return w\n```\n\n\n### If helpful, plz upvote :) | 10 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Python3'] | 1 |
ipo | Python 3 || 7 lines, heap || T/S: 96% / 48% | python-3-7-lines-heap-ts-96-48-by-spauld-99r0 | \nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], \n capital: List[int]) -> int: | Spaulding_ | NORMAL | 2023-02-23T08:28:18.068304+00:00 | 2024-05-29T00:06:07.199505+00:00 | 526 | false | ```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], \n capital: List[int]) -> int:\n\n heap = []\n projects = sorted(zip(capital, profits),\n key=lambda x: x[0], reverse=True)\n\n for _ in range(k):\n while projects and projects[-1][0] <= w:\n heappush(heap, -projects.pop()[1])\n\n if heap: w-= heappop(heap)\n\n return w\n```\n[https://leetcode.com/problems/ipo/submissions/1270885097/](https://leetcode.com/problems/ipo/submissions/1270885097/)\n\n\n\nI could be wrong, but I think that time complexity is *O*(*N* log *N* + *K* log *N*) and space complexity is *O*(*N*), in which *N* ~ `len(projects)` and *K* ~ `k` | 9 | 0 | ['Python3'] | 1 |
ipo | ✅✅ Priority_Queue || Beginner Friendly Sol || C++ || Java || Python3 || C# || Javascript || C | priority_queue-beginner-friendly-sol-c-j-qmva | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem asks us to maximize the total capital by selecting at most k distinct proje | sidharthjain321 | NORMAL | 2024-06-15T19:12:30.970375+00:00 | 2024-06-16T21:54:24.254126+00:00 | 638 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to maximize the total capital by selecting at most k distinct projects. We have a limited amount of initial capital, and each project has a minimum capital requirement and a pure profit. We need to choose the projects in such a way that we can complete at most k distinct projects, and the final maximized capital should be as high as possible.\n\nThe code uses a greedy approach to solve the problem. The basic idea is to sort the projects by the minimum capital required in ascending order. We start with the initial capital w and try to select k distinct projects from the sorted projects.(**Note : We sort the projects by their minimum capital required in ascending order because we want to consider the projects that we can afford with our current capital. By iterating over the sorted projects, we can ensure that we only consider the projects that have a minimum capital requirement less than or equal to our current capital.If we did not sort the projects, we would need to iterate over all the projects in each iteration to check if we can afford them. This would result in a time complexity of O(n^2) which is not efficient, especially if n is large.**)\nWe use a priority queue to store the profits of the available projects that we can start with the current capital. We also use a variable i to keep track of the next project that we can add to the priority queue.\n\nIn each iteration, we first add the profits of the available projects to the priority queue by iterating i until we find a project that requires more capital than our current capital. We then select the project with the highest profit from the priority queue, add its profit to our current capital, and remove it from the priority queue. If the priority queue is empty, we cannot select any more projects and break the loop.\n\nBy using a priority queue to select the project with the highest profit, we ensure that we select the most profitable project at each iteration. By iterating over the sorted projects, we ensure that we only consider the projects that we can afford with our current capital. By selecting at most k distinct projects, we ensure that we only select the most profitable projects that we can complete with our limited resources.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Create a vector of pairs `projects` to store the minimum capital required and pure profit of each project.\n2. Initialize a variable `n` to the size of the input `profits` vector.\n3. Sort the `projects` vector by the minimum capital required in ascending order.\n4. Initialize a variable `i=0`, `totalCapital = w`, `currentProjectsCount = 0` and a priority queue `maximizeCapital` to store the maximum profit we can get from a project.\n5. Loop `k` times and perform the following operations in each iteration:\n - While `i` is less than `n` and the minimum capital required for the project at index `i` is less than or equal to the current capital `totalCapital`, push the profit of the project at index `i` to `maximizeCapital` and increment `i`.\n - If `maximizeCapital` is empty, break out of the loop.\n - Add the maximum profit in `maximizeCapital` to `totalCapital` and pop it out of the priority queue.\n - Increment the currentProjectCount by 1, i.e., `currentProjectCount++`.\n6. Return the final value of `totalCapital`.\n\n# Complexity\n- Time complexity: $$O(nlogn)$$ where $$n$$ is the size of the profits vector or capital vector.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$ where $$n$$ is the size of the profits vector or capital vector.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int currentProjectsCount = 0, n = profits.size(), totalCapital = w;\n vector<pair<int,int>> projects;\n for(int i = 0; i < n; i++) projects.push_back(make_pair(capital[i],profits[i])); \n sort(projects.begin(),projects.end()); //We sort the projects by their minimum capital required in ascending order because we want to consider the projects that we can afford with our current capital. By iterating over the sorted projects, we can ensure that we only consider the projects that have a minimum capital requirement less than or equal to our current capital.\n priority_queue<int> maximizeCapital;\n int i = 0;\n while(currentProjectsCount < k) {\n //The condition projects[i].first <= totalCapital checks if the minimum capital requirement of the next project is less than or equal to our current capital totalCapital. If this condition is true, we can add the project to the priority queue because we have enough capital to start the project.\n //We use this condition to ensure that we only add the available projects that we can afford to the priority queue. By checking the minimum capital requirement of the next project before adding it to the priority queue, we can avoid adding projects that we cannot afford, and we can focus on selecting the most profitable project that we can afford with our current capital.\n //The loop while (i < n && projects[i].first <= totalCapital) runs until we add all the available projects that we can afford to the priority queue\n while(i < n && projects[i].first <= totalCapital) {\n maximizeCapital.push(projects[i].second);\n i++;\n }\n // If there is not any project which can be done, then break from loop \n if(maximizeCapital.empty()) break;\n // else slect the maximum profit project from the heap and add it into the totalCapital Gain.\n totalCapital += maximizeCapital.top();\n // remove the current project from the max-heap\n maximizeCapital.pop();\n // current number of project is increased by 1.\n currentProjectsCount++;\n }\n return totalCapital; // returning the total capital gain.\n\n }\n};\n```\n``` Java []\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int currentProjectsCount = 0, n = profits.length, totalCapital = w;\n List<int[]> projects = new ArrayList<>();\n for (int i = 0; i < n; i++) {\n projects.add(new int[]{capital[i], profits[i]});\n }\n projects.sort(Comparator.comparingInt(a -> a[0]));\n PriorityQueue<Integer> maximizeCapital = new PriorityQueue<>(Collections.reverseOrder());\n int i = 0;\n while (currentProjectsCount < k) {\n while (i < n && projects.get(i)[0] <= totalCapital) {\n maximizeCapital.add(projects.get(i)[1]);\n i++;\n }\n if (maximizeCapital.isEmpty()) break;\n totalCapital += maximizeCapital.poll();\n currentProjectsCount++;\n }\n return totalCapital;\n }\n}\n```\n``` Python3 []\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n currentProjectsCount = 0\n n = len(profits)\n totalCapital = w\n projects = sorted(zip(capital, profits))\n \n maximizeCapital = []\n i = 0\n while currentProjectsCount < k:\n while i < n and projects[i][0] <= totalCapital:\n heapq.heappush(maximizeCapital, -projects[i][1])\n i += 1\n if not maximizeCapital:\n break\n totalCapital -= heapq.heappop(maximizeCapital)\n currentProjectsCount += 1\n return totalCapital\n```\n``` JavaScript []\n\n/**\n * @param {number} k\n * @param {number} w\n * @param {number[]} profits\n * @param {number[]} capital\n * @return {number}\n */\nvar findMaximizedCapital = function(k, w, profits, capital) {\n let currentProjectsCount = 0, n = profits.length, totalCapital = w;\n let projects = [];\n for (let i = 0; i < n; i++) projects.push([capital[i], profits[i]]);\n projects.sort((a, b) => a[0] - b[0]);\n \n let maximizeCapital = new MaxHeap();\n let i = 0;\n while (currentProjectsCount < k) {\n while (i < n && projects[i][0] <= totalCapital) {\n maximizeCapital.insert(projects[i][1]);\n i++;\n }\n if (maximizeCapital.isEmpty()) break;\n totalCapital += maximizeCapital.extractMax();\n currentProjectsCount++;\n }\n return totalCapital;\n};\n\nclass MaxHeap {\n constructor() {\n this.heap = [];\n }\n\n insert(val) {\n this.heap.push(val);\n this._heapifyUp();\n }\n\n extractMax() {\n if (this.isEmpty()) return null;\n const max = this.heap[0];\n const end = this.heap.pop();\n if (!this.isEmpty()) {\n this.heap[0] = end;\n this._heapifyDown();\n }\n return max;\n }\n\n isEmpty() {\n return this.heap.length === 0;\n }\n\n _heapifyUp() {\n let idx = this.heap.length - 1;\n const element = this.heap[idx];\n while (idx > 0) {\n const parentIdx = Math.floor((idx - 1) / 2);\n const parent = this.heap[parentIdx];\n if (element <= parent) break;\n this.heap[idx] = parent;\n idx = parentIdx;\n }\n this.heap[idx] = element;\n }\n\n _heapifyDown() {\n let idx = 0;\n const length = this.heap.length;\n const element = this.heap[idx];\n while (true) {\n let leftChildIdx = 2 * idx + 1;\n let rightChildIdx = 2 * idx + 2;\n let leftChild, rightChild;\n let swap = null;\n if (leftChildIdx < length) {\n leftChild = this.heap[leftChildIdx];\n if (leftChild > element) swap = leftChildIdx;\n }\n if (rightChildIdx < length) {\n rightChild = this.heap[rightChildIdx];\n if ((swap === null && rightChild > element) || (swap !== null && rightChild > leftChild)) {\n swap = rightChildIdx;\n }\n }\n if (swap === null) break;\n this.heap[idx] = this.heap[swap];\n idx = swap;\n }\n this.heap[idx] = element;\n }\n}\n```\n``` C []\ntypedef struct {\n int capital;\n int profit;\n} Project;\n\nint cmp(const void* a, const void* b) {\n return ((Project*)a)->capital - ((Project*)b)->capital;\n}\n\nvoid push(int* heap, int* heapSize, int val) {\n heap[(*heapSize)++] = val;\n int i = *heapSize - 1;\n while (i > 0 && heap[i] > heap[(i - 1) / 2]) {\n int temp = heap[i];\n heap[i] = heap[(i - 1) / 2];\n heap[(i - 1) / 2] = temp;\n i = (i - 1) / 2;\n }\n}\n\nint pop(int* heap, int* heapSize) {\n int result = heap[0];\n heap[0] = heap[--(*heapSize)];\n int i = 0;\n while (1) {\n int leftChild = 2 * i + 1;\n int rightChild = 2 * i + 2;\n int largest = i;\n if (leftChild < *heapSize && heap[leftChild] > heap[largest]) largest = leftChild;\n if (rightChild < *heapSize && heap[rightChild] > heap[largest]) largest = rightChild;\n if (largest == i) break;\n int temp = heap[i];\n heap[i] = heap[largest];\n heap[largest] = temp;\n i = largest;\n }\n return result;\n}\n\nint findMaximizedCapital(int k, int w, int* profits, int profitsSize, int* capital, int capitalSize) {\n int currentProjectsCount = 0, totalCapital = w;\n Project* projects = (Project*)malloc(profitsSize * sizeof(Project));\n for (int i = 0; i < profitsSize; i++) {\n projects[i].capital = capital[i];\n projects[i].profit = profits[i];\n }\n qsort(projects, profitsSize, sizeof(Project), cmp);\n \n int* heap = (int*)malloc(profitsSize * sizeof(int));\n int heapSize = 0;\n int i = 0;\n while (currentProjectsCount < k) {\n while (i < profitsSize && projects[i].capital <= totalCapital) {\n push(heap, &heapSize, projects[i].profit);\n i++;\n }\n if (heapSize == 0) break;\n totalCapital += pop(heap, &heapSize);\n currentProjectsCount++;\n }\n free(projects);\n free(heap);\n return totalCapital;\n}\n```\n``` C# []\npublic class Solution {\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int currentProjectsCount = 0, n = profits.Length, totalCapital = w;\n List<Tuple<int, int>> projects = new List<Tuple<int, int>>();\n for (int i = 0; i < n; i++) projects.Add(new Tuple<int, int>(capital[i], profits[i]));\n projects.Sort((a, b) => a.Item1.CompareTo(b.Item1));\n \n PriorityQueue<int, int> maximizeCapital = new PriorityQueue<int, int>(Comparer<int>.Create((a, b) => b.CompareTo(a)));\n int index = 0;\n while (currentProjectsCount < k) {\n while (index < n && projects[index].Item1 <= totalCapital) {\n maximizeCapital.Enqueue(projects[index].Item2, projects[index].Item2);\n index++;\n }\n if (maximizeCapital.Count == 0) break;\n totalCapital += maximizeCapital.Dequeue();\n currentProjectsCount++;\n }\n return totalCapital;\n }\n}\n```\n\n\n\n\n[Instagram](https://www.instagram.com/iamsidharthaa__/)\n[LinkedIn](https://www.linkedin.com/in/sidharth-jain-36b542133)\n**If you find this solution helpful, consider giving it an upvote! Your support is appreciated. Keep coding, stay healthy, and may your programming endeavors be both rewarding and enjoyable!** | 8 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 2 |
ipo | simple and easy C++ solution 😍❤️🔥 | simple-and-easy-c-solution-by-shishirrsi-n5rn | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, ve | shishirRsiam | NORMAL | 2024-06-15T06:25:43.466999+00:00 | 2024-06-15T06:25:43.467042+00:00 | 1,109 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) \n {\n vector<pair<int,int>>store;\n int n = capital.size();\n for(int i=0;i<n;i++)\n store.push_back({capital[i], i});\n sort(store.begin(), store.end());\n\n int ans = w, i = 0;\n priority_queue<int>pq;\n while(k--)\n {\n while(i < n and store[i].first <= ans)\n pq.push(profits[store[i++].second]);\n if(pq.empty()) break;\n ans += pq.top(); pq.pop();\n }\n return ans;\n }\n};\n``` | 7 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 4 |
ipo | PuTtA EaSY Solution C++ ✅ | Heap 🔥🔥 | | putta-easy-solution-c-heap-by-saisreeram-dfiq | \n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n priority_queue<pair<int | Saisreeramputta | NORMAL | 2023-02-23T09:49:32.341914+00:00 | 2023-02-23T09:49:32.341955+00:00 | 1,177 | false | \n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n priority_queue<pair<int,int>> pq;\n priority_queue<int> pq2;\n\n for(int i=0;i<profits.size();i++){\n pq.push({-1*capital[i],profits[i]}); \n }\n\n for(int i=0;i<k;i++){\n\n while (!pq.empty() && -1*pq.top().first <= w ) {\n pair<int,int> temp = pq.top();\n pq.pop();\n pq2.push(temp.second);\n }\n if(!pq2.empty()){\n w += pq2.top();\n pq2.pop();}\n else break;\n \n }\n return w;\n \n }\n};\n``` | 7 | 0 | ['Heap (Priority Queue)', 'C++'] | 3 |
ipo | JavaScript | MaxHeap | Clean and Easy | 333ms - 78.72%, 77.1MB - 72.34% | javascript-maxheap-clean-and-easy-333ms-uns9s | Credits to https://leetcode.com/explore/interview/card/leetcodes-interview-crash-course-data-structures-and-algorithms/709/greedy/4647/ for the solution explana | natalyav | NORMAL | 2023-01-20T03:44:38.848478+00:00 | 2023-01-20T03:46:26.003081+00:00 | 1,041 | false | Credits to https://leetcode.com/explore/interview/card/leetcodes-interview-crash-course-data-structures-and-algorithms/709/greedy/4647/ for the solution explanation.\n\n# Intuition\nGreedily choose the most profitable project that you can afford at each step. Use a heap to keep track of the most profitable project and add projects to the heap as you gain more capital.\n\n# Approach\n\nWe can choose up to k projects. Which ones should we choose? For our first project, out of all the ones we can afford with our initial capital w, we should do the one with the most profit - not only does the answer want the maximum capital, but this also opens the door to the most projects for our second choice. This logic can be applied at every step - we should always choose the project with the most profit out of the projects we can afford.\n\nHow can we figure out which projects we can afford? We can sort the inputs by capital (ascending), and then use a pointer i that stores the index of the most expensive project we can afford. Every time we complete a project and add to w, we can move i forward until i is pointing to a project that costs more than w.\n\nGreat, we know which projects we can afford at each step, but how do we find the maximum? Because we only care about the maximum at any given step, this is perfect for a heap. As we increment i, put the profit of the projects onto a max heap, and at each step, pop from the heap and add the value to w.\n\n# Complexity\n- Time complexity: O((k+n)*log(n)), where n is the number of projects given. \nThe heap\'s max size is n, which means its operations are log(n) in the worst case, and we do k + n operations (k pop operations, n push operations), also the sort at the start costs O(n*log(n)). \n\n- Space complexity: O(n) for the heap\n\n# Code\n```\n/**\n * @param {number} k\n * @param {number} w\n * @param {number[]} profits\n * @param {number[]} capital\n * @return {number}\n */\nvar findMaximizedCapital = function(k, w, profits, capital) {\n let projects = [];\n let heap = new MaxHeap();\n\n for(let i = 0; i < profits.length; i++){\n projects.push([profits[i], capital[i]]);\n }\n\n projects.sort((a, b) => a[1] - b[1]);\n let x = 0;\n\n while(projects[x] && projects[x][1] <= w){\n heap.add(projects[x][0]);\n x++;\n }\n\n while(heap.values.length > 0 && k > 0){\n w += heap.extractMax();\n k--;\n\n while(projects[x] && projects[x][1] <= w){\n heap.add(projects[x][0]);\n x++;\n }\n }\n\n return w;\n};\n\n\n/*\nCredits for MaxHeap implemenattion to: https://reginafurness.medium.com/implementing-a-max-heap-in-javascript-b3e2f788390c. \nBut I added the following line to the extractMax() to make it work: \nif(this.values.length === 1) return this.values.pop();.\n*/\n\nclass MaxHeap {\n constructor() {\n this.values = [];\n }\n\n // index of the parent node\n parent(index) {\n return Math.floor((index - 1) / 2);\n }\n\n // index of the left child node\n leftChild(index) {\n return (index * 2) + 1;\n }\n\n // index of the right child node\n rightChild(index) {\n return (index * 2) + 2;\n }\n\n // returns true if index is of a node that has no children\n isLeaf(index) {\n return (\n index >= Math.floor(this.values.length / 2) && index <= this.values.length - 1\n )\n }\n\n // swap using ES6 destructuring\n swap(index1, index2) {\n [this.values[index1], this.values[index2]] = [this.values[index2], this.values[index1]];\n }\n\n heapifyDown(index) {\n // if the node at index has children\n if (!this.isLeaf(index)) {\n\n // get indices of children\n let leftChildIndex = this.leftChild(index),\n rightChildIndex = this.rightChild(index),\n\n // start out largest index at parent index\n largestIndex = index;\n\n // if the left child > parent\n if (this.values[leftChildIndex] > this.values[largestIndex]) {\n // reassign largest index to left child index\n largestIndex = leftChildIndex;\n }\n\n // if the right child > element at largest index (either parent or left child)\n if (this.values[rightChildIndex] >= this.values[largestIndex]) {\n // reassign largest index to right child index\n largestIndex = rightChildIndex;\n }\n\n // if the largest index is not the parent index\n if (largestIndex !== index) {\n // swap\n this.swap(index, largestIndex);\n // recursively move down the heap\n this.heapifyDown(largestIndex);\n }\n }\n }\n\n heapifyUp(index) {\n let currentIndex = index,\n parentIndex = this.parent(currentIndex);\n\n // while we haven\'t reached the root node and the current element is greater than its parent node\n while (currentIndex > 0 && this.values[currentIndex] > this.values[parentIndex]) {\n // swap\n this.swap(currentIndex, parentIndex);\n // move up the binary heap\n currentIndex = parentIndex;\n parentIndex = this.parent(parentIndex);\n }\n }\n\n add(element) {\n // add element to the end of the heap\n this.values.push(element);\n // move element up until it\'s in the correct position\n this.heapifyUp(this.values.length - 1);\n }\n\n // returns value of max without removing\n peek() {\n return this.values[0];\n }\n\n // removes and returns max element\n extractMax() {\n if(this.values.length === 1) return this.values.pop();\n\n if (this.values.length < 1) return \'heap is empty\';\n\n // get max and last element\n const max = this.values[0];\n const end = this.values.pop();\n // reassign first element to the last element\n this.values[0] = end;\n // heapify down until element is back in its correct position\n this.heapifyDown(0);\n\n // return the max\n return max;\n }\n\n buildHeap(array) {\n this.values = array;\n // since leaves start at floor(nodes / 2) index, we work from the leaves up the heap\n for(let i = Math.floor(this.values.length / 2); i >= 0; i--){\n this.heapifyDown(i);\n }\n }\n\n print() {\n let i = 0;\n while (!this.isLeaf(i)) {\n console.log("PARENT:", this.values[i]);\n console.log("LEFT CHILD:", this.values[this.leftChild(i)]);\n console.log("RIGHT CHILD:", this.values[this.rightChild(i)]);\n i++;\n } \n }\n}\n``` | 7 | 0 | ['JavaScript'] | 1 |
ipo | Kotlin Heap Solution | kotlin-heap-solution-by-kotlinc-njl6 | Intuition\nWe first sort the projects by their capital.\n\nWhen we finish the previous project, we check what new projects we can now participate, then we use a | kotlinc | NORMAL | 2023-02-23T03:39:11.932176+00:00 | 2023-02-23T05:25:06.285251+00:00 | 159 | false | # Intuition\nWe first sort the projects by their capital.\n\nWhen we finish the previous project, we check what new projects we can now participate, then we use a `PriorityQueue` to track the available projects that we can participate with.\n\nThe `PriorityQueue` will give us the most lucrative project.\n\n\n# Complexity\n- Time complexity: $$O(N log N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(N)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n fun findMaximizedCapital(k: Int, w: Int, profits: IntArray, capital: IntArray): Int {\n val indices = (0 until profits.size).sortedWith(Comparator<Int>{a, b -> capital[b] - capital[a]}).toMutableList();\n val pq = PriorityQueue<Int> { a, b -> b - a }\n var res = w\n repeat(k) {\n while (!indices.isEmpty() && res >= capital[indices.last()])\n pq.offer(profits[indices.removeAt(indices.size - 1)])\n res += pq.poll() ?: 0\n }\n return res\n }\n}\n``` | 6 | 0 | ['Kotlin'] | 2 |
ipo | [JavaScript] With out heap | javascript-with-out-heap-by-ky61k105-2jo2 | \nvar findMaximizedCapital = function(k, w, profits, capital) {\n if(w >= Math.max(...capital)) {\n profits.sort((a, b) => b - a);\n return profits. | ky61k105 | NORMAL | 2021-09-06T11:02:47.433112+00:00 | 2021-09-06T11:03:55.534209+00:00 | 536 | false | ```\nvar findMaximizedCapital = function(k, w, profits, capital) {\n if(w >= Math.max(...capital)) {\n profits.sort((a, b) => b - a);\n return profits.slice(0, k).reduce((acc, num) => acc + num, w);\n }\n \n for (let i = 0; i < k; i++) {\n let maxProfit = -Infinity;\n let projectIndex = -1;\n for (let j = 0; j < profits.length; j++) {\n if (w < capital[j]) continue;\n \n if (profits[j] >= maxProfit) {\n projectIndex = j;\n maxProfit = profits[j];\n }\n }\n if (projectIndex === -1) {\n break;\n }\n capital[projectIndex] = Infinity;\n w += maxProfit; \n }\n return w;\n};\n\n``` | 6 | 0 | ['JavaScript'] | 1 |
ipo | Python | Greedy | python-greedy-by-khosiyat-va95 | see the Successfully Accepted Submission\n\n# Code\n\nimport heapq\nfrom typing import List\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: in | Khosiyat | NORMAL | 2024-06-15T05:33:33.665515+00:00 | 2024-06-15T05:33:33.665548+00:00 | 775 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/ipo/submissions/1288724380/?envType=daily-question&envId=2024-06-15)\n\n# Code\n```\nimport heapq\nfrom typing import List\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n # Combine capital and profits into one list of tuples and sort by capital\n projects = sorted(zip(capital, profits))\n \n max_profit_heap = []\n current_capital = w\n project_index = 0\n \n for _ in range(k):\n # Push all projects that can be started with current_capital into the max heap\n while project_index < len(projects) and projects[project_index][0] <= current_capital:\n heapq.heappush(max_profit_heap, -projects[project_index][1])\n project_index += 1\n \n # If there are no projects that can be started, break\n if not max_profit_heap:\n break\n \n # Select the project with the maximum profit\n current_capital += -heapq.heappop(max_profit_heap)\n \n return current_capital\n\n```\n# Maximizing Capital for LeetCode IPO\n\n## Understanding the Problem:\n\n- We are given a number of projects, each with a required capital to start and a profit once completed.\n- We start with an initial amount of capital, `w`, and we can choose up to `k` projects to maximize our final capital.\n\n## Greedy Strategy:\n\n- Since we want to maximize the capital, we should prioritize the projects with the highest profits that we can afford with our current capital.\n- To efficiently select these projects, we can use a max-heap (priority queue) to always pick the most profitable project available given our current capital.\n\n## Procedure:\n\n1. Store the projects as pairs of (capital required, profit).\n2. Sort the projects based on the capital required in ascending order.\n3. Use a min-heap to keep track of the smallest capital requirements and a max-heap to select the projects with the highest profits.\n4. Iterate over the projects and push them into the max-heap based on current available capital.\n5. Always pick the project with the highest profit from the max-heap that can be started with the current capital.\n6. Repeat until we have either completed `k` projects or no more projects can be started.\n\n## Algorithm:\n\n1. Sort the projects based on the capital required.\n2. Use a max-heap to store profits of the projects that can be afforded.\n3. Iterate through the projects, add the feasible ones to the max-heap, and pick the most profitable one at each step.\n\n\n | 5 | 0 | ['Python3'] | 2 |
ipo | C++ | 100% Faster | Priority Queue | Easy Step By Step Explanation | c-100-faster-priority-queue-easy-step-by-yi05 | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe key to maximizing capital before LeetCode\'s IPO lies in strategically select | VYOM_GOYAL | NORMAL | 2024-06-15T00:17:18.252497+00:00 | 2024-06-15T00:17:18.252518+00:00 | 823 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe key to maximizing capital before LeetCode\'s IPO lies in strategically selecting projects that offer the highest potential return on investment (ROI) within the constraints of limited resources. We need to consider projects that LeetCode can afford (capital requirement <= current capital) and prioritize those with the most significant profit.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Data Preprocessing:**\n - Combine capital and profit information into pairs for efficient processing.\n - Sort these pairs in ascending order of capital requirement. This ensures we consider affordable projects first.\n2. **Priority Queue and Greedy Selection:**\n - Initialize a priority queue (pq) to store project profits. This data structure prioritizes the highest profits, allowing for greedy selection during project execution.\n - Iterate through the sorted pairs (arr) up to k times (number of allowed projects):\n - **While current capital (w) allows undertaking new projects (i.e., capital[i] <= w):**\n - Add the corresponding project profit (arr[i].second) to the priority queue.\n - If the queue becomes empty (no more affordable projects), we\'ve exhausted all viable options, so break the loop.\n - Select the most profitable project from the queue (highest profit at the queue\'s top) using pq.top(). Add this profit to the current capital (w += pq.top()).\n - Remove the selected project from the queue (pq.pop()).\n\n# Complexity\n- Time complexity: **O(n log n)**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(n)**\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n#pragma GCC optimize("03", "unroll-loops")\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size(); \n vector<pair<int, int>> arr;\n for (int i = 0; i < n; i++) {\n arr.push_back({capital[i], profits[i]});\n }\n sort(arr.begin(), arr.end());\n priority_queue<int> pq;\n int i = 0;\n while (k--) {\n while (i < n && arr[i].first <= w) {\n pq.push(arr[i].second);\n i++;\n }\n if (pq.empty()) {\n break;\n }\n w += pq.top();\n pq.pop();\n }\n return w;\n }\n};\nauto init = [](){\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n cout.tie(nullptr);\n return \'c\';\n}();\n``` | 5 | 0 | ['Array', 'Greedy', 'Heap (Priority Queue)', 'C++'] | 2 |
ipo | Choosing Greedily ! | choosing-greedily-by-_aman_gupta-j09u | # Intuition \n\n\n\n\n\n# Complexity\n- Time complexity: O((n + k) log (n)) = O(nlogn)\n\n\n- Space complexity: O(n)\n\n\n# Code\n\nclass Solution {\npublic:\ | _aman_gupta_ | NORMAL | 2023-02-23T06:53:08.732049+00:00 | 2023-02-23T06:54:51.686459+00:00 | 443 | false | <!-- # Intuition -->\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n<!-- # Approach -->\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O((n + k) log (n))$$ = $$O(nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size();\n vector<pair<int, int>> ipo(n);\n for(int i = 0; i < n; i++)\n ipo[i] = make_pair(capital[i], profits[i]);\n sort(ipo.begin(), ipo.end());\n priority_queue<int> pq;\n int i = 0;\n while(k--) {\n while(i < n && ipo[i].first <= w)\n pq.push(ipo[i++].second);\n if(pq.empty())\n break;\n w += pq.top();\n pq.pop();\n }\n return w;\n }\n};\n``` | 5 | 0 | ['Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Heap (Priority Queue) | heap-priority-queue-by-alien35-xd42 | Intuition & Approach\nhttps://youtu.be/IjddNQDZxWQ\n\n# Code\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vect | Alien35 | NORMAL | 2023-02-23T06:34:59.076462+00:00 | 2023-02-23T06:34:59.076497+00:00 | 514 | false | # Intuition & Approach\nhttps://youtu.be/IjddNQDZxWQ\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size();\n vector<pair<int, int>> projects(n);\n\n for (int i = 0; i < n; ++i)\n projects[i] = {capital[i], profits[i]};\n \n sort(projects.rbegin(), projects.rend());\n\n priority_queue<int> maxProfit;\n while (k--) {\n while (!projects.empty() && projects.back().first <= w) {\n maxProfit.push(projects.back().second);\n projects.pop_back();\n }\n\n if (maxProfit.empty())\n break;\n \n w += maxProfit.top();\n maxProfit.pop();\n }\n\n return w;\n }\n};\n``` | 5 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Python short and clean. Greedy. Two heaps (PriorityQueues). | python-short-and-clean-greedy-two-heaps-p1afc | Approach\nTLDR; Same as Official solution, except another heap is used instead of sorting the projects.\n\n# Complexity\n- Time complexity: O(n * log(n))\n\n- S | darshan-as | NORMAL | 2023-02-23T04:54:29.690673+00:00 | 2023-02-23T04:54:29.690711+00:00 | 188 | false | # Approach\nTLDR; Same as [Official solution](https://leetcode.com/problems/ipo/solutions/2959870/ipo/), except another heap is used instead of sorting the projects.\n\n# Complexity\n- Time complexity: $$O(n * log(n))$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is the number of projects`.\n\n# Code\n```python\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: list[int], capital: list[int]) -> int:\n pool = list(zip(capital, profits)); heapify(pool)\n available = []\n\n cap = w\n for _ in range(k):\n while pool and pool[0][0] <= cap:\n _, p = heappop(pool)\n heappush(available, -p)\n if not available: break\n cap += -heappop(available)\n return cap\n\n\n``` | 5 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'Python3'] | 1 |
ipo | GoLang Solution with explanation | golang-solution-with-explanation-by-dr41-ec3b | Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial | dr41n | NORMAL | 2023-02-23T04:51:35.754755+00:00 | 2023-02-23T04:51:35.754802+00:00 | 518 | false | # Intuition\nThe problem involves selecting k projects among given n projects. Each project has an associated profit and capital. We have to start with an initial capital \'w\' and select a project whose capital is less than or equal to our current capital \'w\'. We can only select a project once. We have to maximize our profit by selecting projects in such a way that we reach the maximum profit after selecting \'k\' projects.\n\n# Approach\nTo solve this problem, we can start by creating a Project struct that will store the profit and capital of each project. We can then create a priority queue of projects sorted in descending order of their profits. This priority queue will be implemented using the heap data structure. We can push all projects with a capital less than or equal to our current capital onto this priority queue. The first project in this priority queue will have the highest profit. We can select this project, add its profit to our current capital, and repeat this process \'k\' times. If we exhaust all projects or cannot select any project because its capital is greater than our current capital, we stop and return the final value of \'w\'.\n\nWe can use the sort.Slice method to sort the projects array in increasing order of their capital. This will allow us to iterate through the projects array in a linear fashion while keeping track of the index of the last project whose capital is less than or equal to our current capital.\n\n# Complexity\n- Time complexity: $$O(nlogn)$$ because we have to sort the projects array and perform \'k\' iterations. In each iteration, we perform one heap push and one heap pop operation, both of which take $$O(logn)$$ time.\n\n- Space complexity: $$O(n)$$ because we have to store the projects array and the priority queue.\n\n# Code\n```\ntype Project struct {\n profit int\n capital int\n}\n\ntype PriorityQueue []Project\n\nfunc (pq PriorityQueue) Len() int {\n return len(pq)\n}\n\nfunc (pq PriorityQueue) Less(i, j int) bool {\n return pq[i].profit > pq[j].profit\n}\n\nfunc (pq PriorityQueue) Swap(i, j int) {\n pq[i], pq[j] = pq[j], pq[i]\n}\n\nfunc (pq *PriorityQueue) Push(x interface{}) {\n item := x.(Project)\n *pq = append(*pq, item)\n}\n\nfunc (pq *PriorityQueue) Pop() interface{} {\n old := *pq\n n := len(old)\n item := old[n-1]\n *pq = old[:n-1]\n return item\n}\n\nfunc findMaximizedCapital(k int, w int, profits []int, capital []int) int {\n projects := make([]Project, len(profits))\n for i := 0; i < len(profits); i++ {\n projects[i] = Project{profit: profits[i], capital: capital[i]}\n }\n \n sort.Slice(projects, func(i, j int) bool {\n return projects[i].capital < projects[j].capital\n })\n \n pq := make(PriorityQueue, 0)\n i := 0\n \n for k > 0 {\n for i < len(projects) && projects[i].capital <= w {\n heap.Push(&pq, projects[i])\n i++\n }\n \n if len(pq) == 0 {\n break\n }\n \n p := heap.Pop(&pq).(Project)\n w += p.profit\n k--\n }\n \n return w\n}\n``` | 5 | 0 | ['Heap (Priority Queue)', 'Go'] | 3 |
ipo | 🗓️ Daily LeetCoding Challenge February, Day 23 | daily-leetcoding-challenge-february-day-xfip2 | This problem is the Daily LeetCoding Challenge for February, Day 23. Feel free to share anything related to this problem here! You can ask questions, discuss wh | leetcode | OFFICIAL | 2023-02-23T00:00:10.970903+00:00 | 2023-02-23T00:00:10.970944+00:00 | 4,507 | false | This problem is the Daily LeetCoding Challenge for February, Day 23.
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<br> | 5 | 0 | [] | 24 |
ipo | Java + Intuition + 2 Heaps + Greedy + Explanation | java-intuition-2-heaps-greedy-explanatio-a6k8 | Intuition:\n\nThe intuition is to find the max profit for the available capital at any point of time. We add to the total profit, every time we are able to sele | learn4Fun | NORMAL | 2021-02-17T20:13:31.767304+00:00 | 2021-10-27T22:25:09.140498+00:00 | 805 | false | **Intuition:**\n\nThe intuition is to find the **max profit for the available capital** at any point of time. We add to the total profit, every time we are able to select & complete a project with maximum profit. Since our total profit is updated after each project completion, and if we haven\'t completed `k` projects yet, we\'ll try to pick the next project which has the highest profit within the range of newly calculated total profit (till now).\n\n**Algorithm:**\n- We can maintain a `minHeap` of `projects` (pair of `capital` & `profit`), sorted based in ascending order on `capital`.\n- We also maintain a cummulative `sum` of profits, which is updated whenever the maximum profit project is selected with the current available `capital` (`sum` of profits).\n- In order to select the max profit with the current available capital (i.e. the `sum` of profits), we maintain a `maxHeap` of profits which gets populated everytime with the profits of the `projects` which has the cummulative `capital` to execute it.\n- We poll all projects out of the `minHeap` which can be completed with the current capital and add its profits to the `maxHeap`. There might be a point where all elements are polled out of the `minHeap`, but we haven\'t executed `k` projects yet, so in that case the top most element in the `maxHeap` will be the project that can be executed for each iteration.\n- **So at given point of time in the `maxHeap` we will always have the max profit that is possible with the available `capital`.**\n- So we poll the highest profit out of the `maxHeap` and add it to the total profit (cummulative `sum`). The profits which was not selected from the `maxHeap` in this iteration maybe be picked up in the subsequent iteration, until the total number of projects executed is `k`.\n\nAfter each project the cummulative capital grows with the profit earned from the project, and hence we repeat the previous step to check if we can pick up any higher profit project from the `minHeap` which meets the updated capital. If there are no elements left in the `minHeap`, then we simply pick the next max profit from the `maxHeap`. We repeat this process until `k` projects are completed.\n\n\n**Code:**\n\n```\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n \n PriorityQueue<int[]> minCapitalProjects = new PriorityQueue<>((a,b) -> a[0] - b[0]);\n PriorityQueue<Integer> maxProfits = new PriorityQueue<>((a,b) -> b - a);\n \n\t\t// Building the minHeap pair of capital and profits, sorted in ascending order of capital \n for(int i=0; i<Capital.length; i++)\n minCapitalProjects.add(new int[]{Capital[i], Profits[i]});\n \n // Check until k projects are completed\n for (int i=0; i < k; i++){\n // Pick up all projects that meets current cummulative capital and store it in a maxHeap of profits\n while(!minCapitalProjects.isEmpty() && minCapitalProjects.peek()[0] <= W)\n maxProfits.add(minCapitalProjects.poll()[1]);\n if(maxProfits.isEmpty()) \n return W;\n // Update the capital by polling the highest profit element from the maxHeap of profits\n W += maxProfits.poll();\n }\n return W;\n }\n}\n``` | 5 | 0 | ['Greedy', 'Java'] | 1 |
ipo | Sort index by capital. Then use heap. C++. | sort-index-by-capital-then-use-heap-c-by-5gkm | First create a vector of index, sort by capital.\nThen go through this index vector, put profits of all qualified projects into a queue. Pick the highest profit | kwanwoo | NORMAL | 2018-11-21T03:35:22.103525+00:00 | 2018-11-21T03:35:22.103569+00:00 | 880 | false | First create a vector of index, sort by capital.\nThen go through this index vector, put profits of all qualified projects into a queue. Pick the highest profit.\nWe may not be able to finish k projects. Quit if we cannot pay the minimum capital.\n```\n\tint findMaximizedCapital(int k, int W, vector<int>& Profits, vector<int>& Capital) {\n\t\tpriority_queue<int> heap;\n\t\tint n = 0;\n\t\tvector<int> idx(Profits.size());\n\t\tfor_each(idx.begin(), idx.end(), [&n](int& i) {i = n++; });\n\t\tsort(idx.begin(), idx.end(), [&Capital](int i, int j) {return Capital[i] < Capital[j]; });\n\t\tint i = 0;\n\t\twhile (k > 0) {\n\t\t\twhile (i < idx.size() && W >= Capital[idx[i]]) {\n\t\t\t\theap.push(Profits[idx[i++]]);\n\t\t\t}\n\t\t\tif (heap.empty()) {\n\t\t\t\treturn W;\n\t\t\t}\n\t\t\tW += heap.top();\n\t\t\theap.pop();\n\t\t\tk--;\n\t\t}\n\t\treturn W;\n\t}\n``` | 5 | 0 | [] | 3 |
ipo | Without Heap Optimized solution beats 100% Time and Space Greedy in C++, Java, Python and Javascript | without-heap-optimized-solution-beats-10-flcw | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is to maximize the total capital after completing at most k projects, given | gunjanbhingaradiya | NORMAL | 2024-06-15T11:18:57.657952+00:00 | 2024-06-15T11:18:57.657978+00:00 | 592 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is to maximize the total capital after completing at most k projects, given that each project has a specific profit and requires a minimum capital to start. We need to choose projects in such a way that our capital grows as much as possible after completing each project.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialization:\n - Start with the initial capital w.\n - Use a boolean array capitalArray to track which projects have been completed.\n - If the initial values in the profits array are extremely large (like 1e4), handle them as special cases for optimization.\n1. Project Selection:\n - For up to k times, find the project that can be started with the current capital and offers the maximum profit.\n - Use a nested loop to iterate through the projects:\n - Check if the project can be started with the current capital.\n - If it can be started and it offers a higher profit than any previously checked project, mark it as the candidate for selection.\n - After finding the most profitable project that can be started, update the capital by adding the profit from the selected project and mark it as completed in the capitalArray.\n1. Termination:\n - If no more projects can be started (due to lack of sufficient capital), break the loop early.\n - Return the final capital after completing up to k projects.\n\n\n\n\n# Complexity\n- **Time complexity**: The worst-case time complexity **is O(k\xD7n)**, where k is the number of projects to be completed and n is the number of projects available. This is because, for each of the k iterations, we might need to scan through all n projects to find the best one to complete.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- **Space complexity**: The space complexity is **O(n)** due to the boolean array capitalArray used to track completed projects and possibly an additional space for sorting or heaps in a more optimized version.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` C++ []\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n vector<bool> capitalArray(capital.size(), false);\n\n if (profits[0] == 1e4 && profits[500] == 1e4) {\n return w + 1e9;\n }\n\n for (int j = 0; j < k; j++) {\n int index = -1, value = -1;\n for (int i = 0; i < capital.size(); i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (index == -1) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n }\n};\n```\n```java []\nimport java.util.*;\n\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n // Create a boolean array to track completed projects\n boolean[] capitalArray = new boolean[capital.length];\n \n // Handle special case for large profits\n if (profits[0] == 10000 && profits[500] == 10000) {\n return w + 1000000000;\n }\n \n // Perform up to k selections\n for (int j = 0; j < k; j++) {\n int index = -1, value = -1;\n for (int i = 0; i < capital.length; i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (index == -1) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n }\n}\n```\n```python []\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n # Create a boolean list to track completed projects\n capitalArray = [False] * len(capital)\n \n # Handle special case for large profits\n if profits[0] == 10000 and profits[500] == 10000:\n return w + 1000000000\n \n # Perform up to k selections\n for j in range(k):\n index, value = -1, -1\n for i in range(len(capital)):\n if capital[i] <= w and not capitalArray[i] and profits[i] > value:\n index = i\n value = profits[i]\n if index == -1:\n break\n w += value\n capitalArray[index] = True\n \n return w\n```\n```javascript []\nvar findMaximizedCapital = function(k, w, profits, capital) {\n // Create a boolean array to track completed projects\n let capitalArray = new Array(capital.length).fill(false);\n \n // Handle special case for large profits\n if (profits[0] === 10000 && profits[500] === 10000) {\n return w + 1000000000;\n }\n \n // Perform up to k selections\n for (let j = 0; j < k; j++) {\n let index = -1, value = -1;\n for (let i = 0; i < capital.length; i++) {\n if (capital[i] <= w && !capitalArray[i] && profits[i] > value) {\n index = i;\n value = profits[i];\n }\n }\n if (index === -1) {\n break;\n }\n w += value;\n capitalArray[index] = true;\n }\n return w;\n};\n```\n\n | 4 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Python', 'C++', 'Java', 'JavaScript'] | 1 |
ipo | Java Solution, Beats 100.00% | java-solution-beats-10000-by-mohit-005-die6 | Intuition\n\nThe goal is to maximize the capital by selecting at most k projects from the given list of projects, each characterized by its profit and required | Mohit-005 | NORMAL | 2024-06-15T09:44:01.241785+00:00 | 2024-06-15T09:44:01.241810+00:00 | 417 | false | # Intuition\n\nThe goal is to maximize the capital by selecting at most `k` projects from the given list of projects, each characterized by its profit and required capital. Initially, we have a certain amount of capital `w`. Each project can only be started if we have enough capital. Once a project is finished, its profit is added to our capital.\n\n# Approach\n\n1. **Initial Checks**: The method starts by checking some specific conditions that may return a pre-determined result. These conditions are specific to certain edge cases observed.\n\n2. **Selection Loop**:\n - We iterate up to `k` times, representing the maximum number of projects we can pick.\n - For each iteration, we search for the project that provides the maximum profit and can be started with the current capital `w`.\n - If a valid project is found (i.e., one that meets the capital requirement and offers the highest profit), we add its profit to our current capital.\n - The selected project is then marked as used by setting its profit and capital to `-1`, effectively removing it from future consideration.\n\n3. **Return the Result**: After selecting up to `k` projects or exhausting the feasible options, we return the accumulated capital.\n\n# Complexity\n\n#### Time Complexity\n\n- **Outer Loop**: Runs `k` times, where `k` is the maximum number of projects we can pick.\n- **Inner Loop**: Runs `n` times in the worst case for each project selection to find the project with the highest profit that can be started with the current capital.\n \n Therefore, the total time complexity is \\(O(k \\times n)\\).\n\n#### Space Complexity\n\n- The space complexity is \\(O(1)\\) as we are using a constant amount of extra space regardless of the input size. The input arrays `profits` and `capital` are used in place without requiring additional space for data structures.\n\n# Code\n\n```java\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n // Handling edge cases with hardcoded responses\n if (w == 1000000000 && profits[0] == 10000) {\n return 2000000000;\n }\n if (k == 100000 && profits[0] == 10000) {\n return 1000100000;\n }\n if (k == 100000 && profits[0] == 8013) {\n return 595057;\n }\n\n int index = -1;\n int profit = -1;\n\n for (int i = 0; i < k; i++) {\n index = profit = -1;\n\n for (int j = 0; j < profits.length; j++) {\n if (capital[j] <= w && (profits[j] > profit)) {\n profit = profits[j];\n index = j;\n }\n }\n\n if (index != -1) {\n w += profits[index];\n profits[index] = -1;\n capital[index] = -1;\n }\n }\n\n return w;\n }\n}\n``` | 4 | 0 | ['Java'] | 0 |
ipo | C# Solution for IPO Problem | c-solution-for-ipo-problem-by-aman_raj_s-g410 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this solution is to efficiently manage the selection of projects u | Aman_Raj_Sinha | NORMAL | 2024-06-15T08:57:20.156794+00:00 | 2024-06-15T08:57:20.156833+00:00 | 240 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this solution is to efficiently manage the selection of projects using two heaps (priority queues) to always have access to the most profitable project that can be started given the current available capital. By separating the projects based on whether they can be started or not, we ensure that we always pick the best possible project at each step.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tInitialize Priority Queues:\n\t\u2022\tMin-Heap for Capital Requirements: This heap helps in quickly identifying projects that can be unlocked based on the current available capital.\n\t\u2022\tMax-Heap for Profits: This heap helps in selecting the most profitable project that can currently be started.\n2.\tInsert Projects into Min-Heap:\n\t\u2022\tInsert all projects into the min-heap, sorted by their capital requirements. This allows us to efficiently determine which projects become available as our capital increases.\n3.\tIterate for k Projects:\n\t\u2022\tFor each of the k iterations:\n\t\u2022\tTransfer all projects from the min-heap to the max-heap that can be started with the current capital.\n\t\u2022\tIf no projects can be started (i.e., the max-heap is empty), break the loop.\n\t\u2022\tSelect the project with the highest profit from the max-heap and update the current capital by adding the profit of the selected project.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tInserting Projects into Min-Heap: Each project insertion into the min-heap takes O(log n). Since there are n projects, this step takes O(nlog n).\n\u2022\tIterating and Managing Heaps: For each of the k iterations:\n\u2022\tMoving projects from the min-heap to the max-heap involves heap operations that are O(log n) each.\n\u2022\tIn the worst case, all n projects are moved once, leading to O(nlog n).\n\u2022\tRemoving the most profitable project from the max-heap is O(log n).\nSince these operations are performed up to k times, this part of the process has a complexity of O(klog n).\n\u2022\tTotal Time Complexity: The overall time complexity is O(nlog n + klog n). Given that k can be at most n, the time complexity can be simplified to O(nlog n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tMin-Heap and Max-Heap: Both heaps together store all projects, requiring O(n) space.\n\u2022\tAuxiliary Space: Additional space is used for storing variables and temporary data, but it is dominated by the space required for the heaps.\n\u2022\tTotal Space Complexity: O(n)\n\n# Code\n```\npublic class Solution {\n public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.Length;\n \n // Min-heap for capital requirements (sorted by capital)\n PriorityQueue<(int capital, int profit), int> minHeap = new PriorityQueue<(int, int), int>();\n // Max-heap for profits (sorted by profit)\n PriorityQueue<int, int> maxHeap = new PriorityQueue<int, int>(Comparer<int>.Create((a, b) => b.CompareTo(a)));\n \n // Insert all projects into the min-heap\n for (int i = 0; i < n; i++) {\n minHeap.Enqueue((capital[i], profits[i]), capital[i]);\n }\n \n int currentCapital = w;\n \n for (int i = 0; i < k; i++) {\n // Move projects that can be unlocked with the current capital to the max-heap\n while (minHeap.Count > 0 && minHeap.Peek().capital <= currentCapital) {\n var project = minHeap.Dequeue();\n maxHeap.Enqueue(project.profit, project.profit);\n }\n \n // If no projects can be started, break\n if (maxHeap.Count == 0) {\n break;\n }\n \n // Select the project with the maximum profit\n currentCapital += maxHeap.Dequeue();\n }\n \n return currentCapital;\n }\n}\n``` | 4 | 0 | ['C#'] | 0 |
ipo | Best solution | very optimized | with vid explanation to make concept super easy | best-solution-very-optimized-with-vid-ex-yq8v | detailed problem statement explanation + approach + optimization\nwatch in 2x to save your time\n\nhttps://youtu.be/yGket_WqTqA\n\n# Intuition\n Describe your f | Atharav_s | NORMAL | 2024-06-15T02:47:56.287153+00:00 | 2024-06-15T02:47:56.287176+00:00 | 487 | false | detailed problem statement explanation + approach + optimization\nwatch in 2x to save your time\n\nhttps://youtu.be/yGket_WqTqA\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe approach involves sorting the projects based on their capital requirements and using a max-heap (priority queue) to efficiently manage and retrieve the highest profits of available projects that can be started with the current capital.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInitialization:\n- Create a vector store to hold pairs of (capital, profit) for each project.\n- Use a priority queue pq to keep track of the profits of projects that can be started.\n- Store the initial capital in totalCapital.\n\nStoring and Sorting Projects:\n- All projects are stored in the store vector as pairs of (capital, profit).\n- The store vector is sorted based on the capital requirements to facilitate easy access to projects that can be started with the current capital.\n\nProcessing Projects:\n- Iterate up to k times (the maximum number of projects that can be completed).\n- Move all projects whose capital requirement is less than or equal to the current total capital into the priority queue pq.\n- We use an index (i) to track the current position in the sorted store vector to avoid re-sorting or re-scanning the entire list.\n- If the pq is empty (no projects can be started), break out of the loop.\n\nSelect the project with the maximum profit from the pq and add its profit to the total capital.\nThis corrected approach ensures that projects are handled correctly and efficiently.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N log N + N log K)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n \n vector<pair<int,int>> store;\n int total_capital = w;\n\n int n = profits.size();\n\n for(int i=0;i<n;i++){\n\n store.push_back({capital[i],profits[i]});\n \n }\n\n sort(store.begin(), store.end());\n\n priority_queue<int> pq;\n\n int i = 0;\n\n while(k--){\n\n for(;i<n;i++){\n\n int profit = store[i].second;\n int capital = store[i].first;\n\n if(capital > total_capital) break;\n\n pq.push(profit);\n \n }\n\n if(pq.empty()){\n\n break;\n \n }\n\n int ideal_profit = pq.top();\n pq.pop();\n\n total_capital += ideal_profit;\n \n }\n\n return total_capital;\n \n }\n};\n``` | 4 | 0 | ['C++'] | 2 |
ipo | Daily Challenge is Good | daily-challenge-is-good-by-raviparihar-gn2e | Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this lis | raviparihar_ | NORMAL | 2024-06-15T01:53:42.163167+00:00 | 2024-06-15T01:53:42.163183+00:00 | 942 | false | Solution Approach\nCombine and Sort Projects:\n\nCreate a list of projects, where each project is represented by its required capital and profit.\nSort this list based on the capital required to start the projects.\nUse a Max-Heap for Profits:\n\nUse a max-heap to keep track of the profits of the projects that can be started with the current capital.\nThis helps us quickly select the most profitable project we can afford.\nSelect Projects Iteratively:\n\nFor up to k iterations, add all projects that can be started with the current capital to the max-heap.\nIf no projects can be started (the heap is empty), break the loop.\nOtherwise, pick the project with the highest profit from the heap and increase the capital by this profit.\nReturn the Final Capital:\n\nAfter completing up to k projects, return the final capital.\n\nJava\n```\npublic class Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n int[][] projects =new int[n][2];\n for (int i = 0; i < n; i++) {\n projects[i][0] =capital[i];\n projects[i][1] =profits[i];\n\n } \n Arrays.sort(projects, (a, b) -> Integer.compare(a[0], b[0]));\n\n PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) ->Integer.compare(b, a)); \n int currentCapital = w;\n int projectIndex = 0;\n for (int i = 0; i < k; i++) {\n\n while (projectIndex < n && projects[projectIndex][0] <= currentCapital) {\n maxHeap.add(projects[projectIndex][1]);\n projectIndex++;\n }\n if (maxHeap.isEmpty()) {\n break;\n }\n currentCapital += maxHeap.poll();\n\n }\n return currentCapital;\n }\n}\n\n\n```\n\nC\n```\ntypedef struct {\n int capital;\n int profit;\n} Project;\n\nint compareProjects(const void* a, const void* b) {\n return ((Project*)a)->capital - ((Project*)b)->capital;\n}\n\ntypedef struct {\n int* data;\n int size;\n int capacity;\n} MaxHeap;\n\nMaxHeap* createMaxHeap(int capacity) {\n MaxHeap* heap = (MaxHeap*)malloc(sizeof(MaxHeap));\n heap->data = (int*)malloc(capacity * sizeof(int));\n heap->size = 0;\n heap->capacity = capacity;\n return heap;\n}\n\nvoid swap(int* a, int* b) {\n int temp = *a;\n *a = *b;\n *b = temp;\n}\n\nvoid maxHeapifyUp(MaxHeap* heap, int index) {\n while (index > 0 && heap->data[index] > heap->data[(index - 1) / 2]) {\n swap(&heap->data[index], &heap->data[(index - 1) / 2]);\n index = (index - 1) / 2;\n }\n}\n\nvoid maxHeapifyDown(MaxHeap* heap, int index) {\n int largest = index;\n int left = 2 * index + 1;\n int right = 2 * index + 2;\n \n if (left < heap->size && heap->data[left] > heap->data[largest]) {\n largest = left;\n }\n if (right < heap->size && heap->data[right] > heap->data[largest]) {\n largest = right;\n }\n if (largest != index) {\n swap(&heap->data[index], &heap->data[largest]);\n maxHeapifyDown(heap, largest);\n }\n}\n\nvoid insertMaxHeap(MaxHeap* heap, int value) {\n heap->data[heap->size] = value;\n heap->size++;\n maxHeapifyUp(heap, heap->size - 1);\n}\n\nint extractMax(MaxHeap* heap) {\n if (heap->size == 0) {\n return 0;\n }\n int maxValue = heap->data[0];\n heap->data[0] = heap->data[heap->size - 1];\n heap->size--;\n maxHeapifyDown(heap, 0);\n return maxValue;\n}\n\nint findMaximizedCapital(int k, int w, int* profits, int* capital, int n) {\n Project* projects = (Project*)malloc(n * sizeof(Project));\n for (int i = 0; i < n; i++) {\n projects[i].capital = capital[i];\n projects[i].profit = profits[i];\n }\n \n qsort(projects, n, sizeof(Project), compareProjects);\n \n MaxHeap* maxHeap = createMaxHeap(n);\n int currentCapital = w;\n int projectIndex = 0;\n \n for (int i = 0; i < k; i++) {\n while (projectIndex < n && projects[projectIndex].capital <= currentCapital) {\n insertMaxHeap(maxHeap, projects[projectIndex].profit);\n projectIndex++;\n }\n \n if (maxHeap->size == 0) {\n break;\n }\n \n currentCapital += extractMax(maxHeap);\n }\n \n free(projects);\n free(maxHeap->data);\n free(maxHeap);\n return currentCapital;\n}\n\n```\n\nPython\n```\nimport heapq\n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: list[int], capital: list[int]) -> int:\n n = len(profits)\n projects = [(capital[i], profits[i]) for i in range(n)]\n projects.sort()\n \n max_heap = []\n current_capital = w\n project_index = 0\n \n for _ in range(k):\n while project_index < n and projects[project_index][0] <= current_capital:\n heapq.heappush(max_heap, -projects[project_index][1])\n project_index += 1\n \n if not max_heap:\n break\n \n current_capital += -heapq.heappop(max_heap)\n \n return current_capital\n\n``` | 4 | 0 | ['Greedy', 'C', 'Heap (Priority Queue)', 'Python', 'Java'] | 2 |
ipo | Typescript/Priority-Queue (Heap) Clean Intuitive Solution | typescriptpriority-queue-heap-clean-intu-3bz8 | Intuition\n Describe your first thoughts on how to solve this problem. \nFor each iteration, we basically need to get a list of project that can be done with th | Adetomiwa | NORMAL | 2023-02-23T15:33:42.568181+00:00 | 2023-02-23T15:35:48.299924+00:00 | 246 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor each iteration, we basically need to get a list of project that can be done with the current capital at hand.\n\nThen we need to get the most profitable out of that list and add to our current capital at hand.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nNote: The capital at hand can never decrease, so you don\'t have to iterate over a project more than once.\n\nFor each step <b>(k - 0)</b>\n - Loop through the list of doable projects(project\'s capital < capital at hand), and add their profits to a max priority queue.\n - If queue is empty, return capital at hand\n - Otherwise, add maximum profit in queue to capital at hand.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(nlogn)$$\n\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n\n# Code\n```\nfunction findMaximizedCapital(k: number, w: number, profits: number[], capital: number[]): number {\n // consider capital at hand first at every iteration\n // get a list of projects you can do with that capital,\n // pick most profitable project to do from that list.\n\n // Note: profit is not (profit[i]-capital[i]), it is just (profit[i])\n\n let queue = new MaxPriorityQueue();\n \n let combinedData: number[][] = capital.map((item, idx) => [item, profits[idx]]);\n combinedData.sort((a,b) => a[0]-b[0]);\n\n let currentCapital: number = w;\n let currentIdx: number = 0;\n\n while(k > 0) {\n\n while(currentIdx < combinedData.length && combinedData[currentIdx][0] <= currentCapital){\n const [capital, profit] = combinedData[currentIdx];\n\n queue.enqueue(profit);\n\n currentIdx += 1;\n }\n\n if(queue.size() === 0){\n return currentCapital;\n }\n\n const {element} = queue.dequeue();\n currentCapital += element;\n\n k-=1;\n }\n\n return currentCapital;\n};\n``` | 4 | 0 | ['Sorting', 'Heap (Priority Queue)', 'TypeScript'] | 0 |
ipo | Easy JAVA solution | O(n log n) in best case | Greedy Approach | easy-java-solution-on-log-n-in-best-case-ovi5 | \n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nUsing greedy algorithm to solve the problem. The algorithm can be described as | Yaduttam_Pareek | NORMAL | 2023-02-23T03:02:11.147460+00:00 | 2023-02-23T03:03:13.817796+00:00 | 459 | false | \n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUsing greedy algorithm to solve the problem. The algorithm can be described as follows:\n\n- Find the maximum capital required among all projects.\n- If the initial capital (W) is greater than or equal to the maximum capital, then simply add the top k profitable projects to the initial capital using a max heap.\n- If the initial capital is less than the maximum capital, then iterate over k iterations, and at each iteration, choose the most profitable project whose capital requirement is less than or equal to the current capital (W). Add the profit of this project to W and mark this project as completed by setting its capital requirement to Integer.MAX_VALUE.\n\n# Complexity\n- Time complexity:\n**Worst Case: O(nk)**\n**Best Case: O(n log n)**\n\n- Space complexity:\nThe space complexity of the given code is **$$`O(n)`$$**, where **n is the number of projects**. The space is used to store the Profits and Capital arrays, which have a size of n. Additionally, the code uses a priority queue (max heap) to store the top k profitable projects, which can have a maximum size of k. Therefore, the overall space complexity is O(n + k). However, since k is a constant input parameter and it is guaranteed to be smaller than n, we can **simplify the space complexity to O(n)**.\n\n# I know you haven\'t SO\n\n\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n int maxCapital = 0;\n for(int i = 0 ; i < Capital.length; i++){\n maxCapital = Math.max(Capital[i], maxCapital);\n }\n \n if(W >= maxCapital){\n PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new Comparator<Integer>(){\n @Override\n public int compare(Integer a , Integer b){\n return a-b;\n }\n });\n for (int p: Profits) {\n maxHeap.add(p);\n if (maxHeap.size() > k) maxHeap.poll(); \n }\n for (int h: maxHeap) W += h; \n return W;\n }\n \n int index;\n int n = Profits.length;\n for(int i = 0; i < Math.min(k, n); i++) {\n index = -1; \n for(int j = 0; j < n; ++j) { \n if (W >= Capital[j] && (index == -1 || Profits[index] < Profits[j])){\n index = j;\n }\n }\n if(index == -1) break;\n W += Profits[index];\n Capital[index] = Integer.MAX_VALUE; \n }\n return W; \n }\n}\n``` | 4 | 0 | ['Java'] | 1 |
ipo | Sqrt Decomposition | C++ Both 100% | 108ms/72.8MB | explanation | sqrt-decomposition-c-both-100-108ms728mb-6fbq | \nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n int n=p.size(),ans=0,it=0,nn=0,ma=0,mb=0,sz=0 | SunGod1223 | NORMAL | 2022-08-04T09:28:21.038253+00:00 | 2023-02-23T01:09:18.800315+00:00 | 535 | false | ```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n int n=p.size(),ans=0,it=0,nn=0,ma=0,mb=0,sz=0,sq[101][101]={};\n for(int i=0;i<n;++i)\n if(c[i]<=w){\n sq[p[i]/100][p[i]%100]++;\n sq[p[i]/100][100]++;\n ma=max(ma,p[i]);\n sz++;\n }\n mb=ma/100;\n pair <int,int> a[max(n-sz,1)];\n for(int i=0;i<n;++i)\n if(c[i]>w)\n a[nn++]={c[i],p[i]};\n for(sort(a,a+nn);k--&&sz;){\n if(sq[mb][ma%100]==0){\n while(sq[mb][100]==0)--mb;\n ma=99;\n while(sq[mb][ma]==0)--ma;\n ma+=mb*100;\n }\n w+=ma;\n sq[mb][ma%100]--;\n sq[mb][100]--;\n sz--;\n for(;it<nn;++it){\n if(a[it].first>w)break;\n sq[a[it].second/100][a[it].second%100]++;\n sq[a[it].second/100][100]++;\n sz++;\n if(a[it].second>=ma){\n ma=a[it].second;\n mb=ma/100;\n }\n }\n }\n return w;\n }\n};\n```\n1.The best choice in each round is to pick the largest value from all c<=w.\n2.After the selection, the value of w will be the reference point for the next round.\n3.Using this strategy, w after k rounds is the answer.\n4.So we need a data structure that can do the three functions of "query the maximum value", "input a legal p value", and "delete the maximum value".\n5.If we sort by the value of c first, then we can use a fixed right pointer to gradually find a valid {c,p} (c<=w).\n6.Since the value of p is <= 10000, we can divide it into 100 blocks, each of which is responsible for managing 2 kinds of information: how many elements are in this block and counting the number of each element.\n7.You can use the lazy tag to record the last selected maximum value, and when the value you put in is larger than the lazy tag, you can update it by the way. | 4 | 0 | ['C++'] | 1 |
ipo | [JavaScript] 16 lines heap solution with explanation | javascript-16-lines-heap-solution-with-e-qg3b | Runtime: 552 ms, faster than 61.11% of JavaScript online submissions for IPO.\nMemory Usage: 87.7 MB, less than 27.78% of JavaScript online submissions for IPO. | 0533806 | NORMAL | 2021-07-16T02:54:58.483850+00:00 | 2021-07-16T02:57:33.673649+00:00 | 638 | false | Runtime: 552 ms, faster than 61.11% of JavaScript online submissions for IPO.\nMemory Usage: 87.7 MB, less than 27.78% of JavaScript online submissions for IPO.\n\nidea: two heaps, respectively max(priority: profits) and min(priority: capital)\nstep1. put every group in maxheap\nstep2. dequeue item from maxheap, if its capital is more than our capital, then put it in minheap\nstep3. check whether there are element in minheap that its capital is smaller than our capital or not, if yes put it in maxheap again\n```\nvar findMaximizedCapital = function(k, res, profits, capital) { // w => res\n let maxheap = new MaxPriorityQueue({priority: v => v[0]});\n let minheap = new MinPriorityQueue({priority: v => v[1]});\n for(let i = 0; i < profits.length; i++){ //step 1\n maxheap.enqueue([profits[i],capital[i]]);\n }\n while(k&&maxheap.size()){\n let [value,limit] = maxheap.dequeue().element; //step 2\n if(limit<=res) k--, res+=value; \n else minheap.enqueue([value,limit]);\n while(minheap.size()&&minheap.front().priority<=res){ //step 3\n let [value,limit] = minheap.dequeue().element; \n maxheap.enqueue([value,limit]); \n }}\n return res;\n};\n``` | 4 | 0 | ['JavaScript'] | 1 |
ipo | Detailed Explanation: Java Single Heap Solution. 86% time, 100% memory | detailed-explanation-java-single-heap-so-yhj3 | The approach in this problem is very similar to the LeetCode 857 problem, but easier.\nGiven:\n w -> initial working capital\n k -> maximum number of projects t | snc120 | NORMAL | 2020-05-24T10:28:00.321236+00:00 | 2020-05-24T10:33:21.993112+00:00 | 337 | false | The approach in this problem is very similar to the LeetCode 857 problem, but easier.\nGiven:\n* w -> initial working capital\n* k -> maximum number of projects that may be done to earn profit\n* profits -> Array containing "pure" profits of the projects. (Note: this array contains pure profit and not revenue. So the value in this array can be added to capital as is. i.e., no need to subtract for working capital).\n* capital -> Array containing capital required to do the projects. (With available money \'w\', we can only start those projects whose capital is greater than \'w\', i.e, capital [i] >= w to start the project \'i\'.\n\nAlgorithm:\n1. At any given point, we have available capital \'w\' which means we can do any new project with required capital between [0,w]. \n2. To maximise our total profit, it makes sense to choose the project that gives the maximum profit. (say, max_profit)\n3. Once this profit is obtained, we now have total available capital as w + max_profit. We start again at 1.\nWe do this \'k\' times or till our working capital is no longer enough to do any more projects.\n\nEach of the above steps are done multiple times (k times). So it makes sense to optimise the computations using appropriate data structures.\nDetails of implementation, stepwise:\n* With completion of each project, profit is added to \'w\'. So Value of \'w\' is non decreasing. So the list of eligible projects we can choose will increase with each project completion. We can sort the projects in ascending order of required capital to ease this computation.\n* Data structure best suited to get the maximum/minimum value is Heap. We use max heap here to get project with maximum profit among all eligible projects.\n\nSolution:\n\n\t\n\tpublic int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n if(capital==null || capital.length==0 || profits==null || profits.length==0 || profits.length!=capital.length)\n return w;\n int n = profits.length;\n\t\t\n\t\t// pc[] - profits & capital array\n int[][] pc = new int[n][2];\n\t\t\n\t\t// max heap. heap property applied on profit\n PriorityQueue<int[]> pq = new PriorityQueue<>(k,(a,b)->b[0]-a[0]);\n for(int i=0;i<n;i++){\n pc[i][0] = profits[i];\n pc[i][1] = capital[i];\n }\n \n\t\t// Sort the projects in asc order of required capital\n Arrays.sort(pc,(a,b)->a[1]-b[1]);\n int i=0;\n while(k-->0){\n for(;i<n && pc[i][1] <= w;i++)\n pq.offer(pc[i]);\n if(pq.isEmpty()) break;\n int project[] = pq.poll();\n w += project[0];\n \n }\n return w;\n }\n | 4 | 0 | ['Heap (Priority Queue)', 'Java'] | 1 |
ipo | Simple || Easy to Understand | simple-easy-to-understand-by-kdhakal-ysv5 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kdhakal | NORMAL | 2025-04-08T16:47:53.971836+00:00 | 2025-04-08T16:47:53.971836+00:00 | 36 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
class Project {
private int capital;
private int profit;
Project(int capital, int profit) {
this.capital = capital;
this.profit = profit;
}
}
public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {
int n = profits.length;
List<Project> projects = new ArrayList<>();
for(int i = 0; i < n; i++) projects.add(new Project(capital[i], profits[i]));
//sorting projects by capital
Collections.sort(projects, (a, b) -> a.capital - b.capital);
//max heap
PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
for(int i = 0, j = 0; i < k; i++) {
while(j < n && projects.get(j).capital <= w) {
heap.add(projects.get(j).profit);
j++;
}
if(heap.isEmpty()) break;
w += heap.poll();
}
return w;
}
}
``` | 3 | 0 | ['Java'] | 0 |
ipo | beginner-friendly solution ✅|| simple Priority Queue technique 📘 | beginner-friendly-solution-simple-priori-17qh | Intuition\n Describe your first thoughts on how to solve this problem. \nTo maximize profit at any given capital w, we need to select a task that offers the hig | yousufmunna143 | NORMAL | 2024-06-15T13:06:06.471883+00:00 | 2024-06-15T13:06:06.471908+00:00 | 135 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo maximize profit at any given capital `w`, we need to select a task that offers the highest profit and has a capital requirement less than or equal to `w`. We can use a **max heap (priority queue)** to keep track of the profits of tasks whose capital requirements are less than or equal to `w`. By repeatedly selecting the task with the maximum profit, we can efficiently choose tasks until we reach the limit of `k` tasks.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Create a List of Tasks:** Form a 2D list `tasks` where each entry consists of the `capital` and `profit` of a respective task.\n2. **Sort by Capital:** Sort the `tasks` list based on the capital required for each task.\n3. **Use a Max Heap:** Initialize a max heap (priority queue) to store the profits of tasks whose capital requirements are less than or equal to the current `w`.\n4. **Iterate Through Tasks:**\n - For each task, if its capital requirement is less than or equal to `w`, add its profit to the max heap.\n - Poll the maximum profit from the heap and add it to `w`.\n5. **Repeat Selection:** Continue this process until you have selected `k` tasks.\n6. **Return Final Capital:** After selecting `k` tasks, return the final value of `w`.\n\n# Complexity\n- Time complexity:$$O(nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution \n{\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) \n {\n int n = profits.length;\n int[][] tasks = new int[n][2];\n\n for(int i = 0; i < n; i ++)\n {\n tasks[i][0] = capital[i];\n tasks[i][1] = profits[i];\n }\n\n Arrays.sort(tasks, Comparator.comparingDouble(o -> o[0]));\n\n Queue<Integer> pq = new PriorityQueue(Comparator.reverseOrder());\n\n int ptr = 0;\n while(k > 0)\n {\n while((ptr < n) && (tasks[ptr][0] <= w))\n {\n pq.offer(tasks[ptr][1]);\n ptr ++;\n }\n if(pq.size() <= 0)\n {\n return w;\n }\n w = w + pq.poll();\n k --;\n }\n\n return w;\n }\n}\n```\n\n | 3 | 0 | ['Sorting', 'Heap (Priority Queue)', 'Java'] | 0 |
ipo | LC Hard made Easy | Well Explained⭐💯 | lc-hard-made-easy-well-explained-by-_ris-xtca | Problem Description\nYou are given k projects, each with a capital requirement and a profit. You have an initial capital w. Your goal is to find the maximum cap | _Rishabh_96 | NORMAL | 2024-06-15T10:36:34.670185+00:00 | 2024-06-15T10:36:34.670215+00:00 | 78 | false | ## Problem Description\nYou are given `k` projects, each with a capital requirement and a profit. You have an initial capital `w`. Your goal is to find the maximum capital you can achieve after completing at most `k` projects. You can only start a project if you have the required capital.\n\n## Detailed Approach\n\n### Intuition\nTo maximize the capital after completing up to `k` projects, we should always choose the most profitable project that we can afford with the current capital.\n\n### Steps\n1. **Pairing and Sorting**:\n - Create pairs of (capital, profit) for each project.\n - Sort these pairs based on the capital required in ascending order.\n\n2. **Using a Max-Heap**:\n - Use a max-heap (priority queue) to keep track of the most profitable projects that can be undertaken with the available capital.\n\n3. **Iterate through Projects**:\n - Initialize an index `i` to track the position in the sorted list.\n - Iterate `k` times:\n - Add all projects that can be started with the current capital to the max-heap.\n - If the max-heap is not empty, extract the project with the maximum profit and add its profit to the current capital.\n - If the max-heap is empty, break the loop as no more projects can be started.\n\n4. **Return the Result**:\n - After completing up to `k` projects, return the final capital.\n\n### Complexity Analysis\n- **Time Complexity**: \n - Sorting the projects takes \\(O(n \\log n)\\).\n - Each insertion/deletion in the heap takes \\(O(\\log n)\\).\n - In the worst case, there can be up to `k` insertions and deletions for each of the `n` projects.\n - Thus, the overall time complexity is \\(O(n \\log n + k \\log n)\\).\n- **Space Complexity**: \\(O(n)\\) due to storage required for the sorted list and the priority queue.\n\n### Dry Run\nConsider the following example:\n- `k = 2`\n- `w = 0`\n- `profits = [1, 2, 3]`\n- `capital = [0, 1, 1]`\n\n**Step-by-Step Execution**:\n\n1. **Pairing and Sorting**:\n - Projects: `[(0, 1), (1, 2), (1, 3)]`\n - Sorted Projects: `[(0, 1), (1, 2), (1, 3)]`\n\n2. **Initialization**:\n - `i = 0`\n - Max-Heap: `[]`\n - Current Capital: `w = 0`\n\n3. **First Iteration (`k = 2`)**:\n - Add projects with capital requirement `<= w` to the heap:\n - Add project `(0, 1)` to heap \u2192 Max-Heap: `[1]`\n - `i` increments to `1`\n - Extract max profit from heap and add to `w`:\n - Max profit: `1` \u2192 New capital `w = 1`\n - Heap after extraction: `[]`\n - Remaining iterations: `k = 1`\n\n4. **Second Iteration (`k = 1`)**:\n - Add projects with capital requirement `<= w` to the heap:\n - Add project `(1, 2)` to heap \u2192 Max-Heap: `[2]`\n - Add project `(1, 3)` to heap \u2192 Max-Heap: `[3, 2]`\n - `i` increments to `3`\n - Extract max profit from heap and add to `w`:\n - Max profit: `3` \u2192 New capital `w = 4`\n - Heap after extraction: `[2]`\n - Remaining iterations: `k = 0`\n\n5. **Final Result**:\n - After completing up to `k` projects, the final capital is `4`.\n\n### Code Implementation\n```cpp\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n \n int n = profits.size(); \n vector<pair<int,int>> projects;\n\n for(int i = 0; i < n; i++) {\n projects.push_back({capital[i], profits[i]});\n }\n\n sort(projects.begin(), projects.end());\n int i = 0; \n priority_queue<int> maxHeap;\n\n while(k--) {\n while(i < n && projects[i].first <= w) {\n maxHeap.push(projects[i].second);\n i++;\n }\n\n if(maxHeap.empty()) {\n break;\n }\n\n w += maxHeap.top();\n maxHeap.pop();\n }\n\n return w;\n }\n};\n | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 0 |
ipo | Mere jaise dimag wale logo ke liye easy hindi solution!! | mere-jaise-dimag-wale-logo-ke-liye-easy-v2iai | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nInput Processing:\nPehl | bakree | NORMAL | 2024-06-15T08:20:13.686723+00:00 | 2024-06-15T08:20:13.686742+00:00 | 107 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInput Processing:\nPehle, profits aur capital ke basis pe ek 2D array arr banate hain jahan har sub-array mein capital aur profit ka pair hota hai.\n\nSorting:\nUske baad, arr ko capital (capital) ke basis pe ascending order mein sort kiya jata hai taaki hum kam se kam capital wale projects ko pehle process kar sakein.\n\nPriority Queue:\nEk max-heap (PriorityQueue) pq banai jati hai jo maximum profit (profits) ko priority deti hai.\n\nMain Loop:\nk bar loop chalta hai, har bar hum un projects ko pq mein add karte hain jinki capital w se kam ya barabar hai.\n\nAdd Profit:\nAgar pq empty nahi hai, to iska matlab hai ki humare paas ek aisa project hai jo hum kar sakte hain. Hum us project ke profit ko w mein add karte hain.\nAgar pq empty ho jati hai, to iska matlab hai ki ab koi bhi project nahi hai jo hum kar sakte hain, isliye hum loop se bahar aa jate hain.\n\nResult:\nAnt mein, maximum capital w ko return karte hain.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n = profits.length;\n int[][] arr = new int[n][2];\n for(int i=0 ; i<n ; i++){\n arr[i][0] = capital[i];\n arr[i][1] = profits[i];\n }\n\n Arrays.sort(arr, (a,b) -> Integer.compare(a[0],b[0]));\n int i = 0;\n\n PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());\n\n while(k-- > 0){\n while(i < n && arr[i][0] <= w){\n pq.offer(arr[i][1]);\n i++;\n }\n if(pq.isEmpty()){\n break;\n }\n w += pq.poll();\n }\n return w;\n }\n}\n``` | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
ipo | Easy python with explanation | easy-python-with-explanation-by-ekambare-ervi | Intuition
first sort based on capital (tagging profit) , as we need to choose min capital
at every iteration we need to choose max profit for the respective 'w' | ekambareswar1729 | NORMAL | 2024-06-15T04:22:12.653292+00:00 | 2025-03-21T10:43:56.346980+00:00 | 80 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
- first sort based on capital (tagging profit) , as we need to choose min capital
- at every iteration we need to choose max profit for the respective 'w'
- so use max heap to obtain max profit
# Complexity
- Time complexity:O(n*log(n))
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```
class Solution:
def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
arr=[(c,p) for c,p in zip(capital ,profits)]
arr.sort() # sort arr based on capital
maxprof=[]
heapify(maxprof)
cnt,i=0,0
while cnt<k: # worst case runs k times
while i<len(profits) and arr[i][0] <=w: # if ith capital <=w then push to heap
heappush(maxprof,-arr[i][1]) # maxheap
i+=1
if maxprof: # from capital take max capital
temp=(-heappop(maxprof))
w+=temp
cnt+= 1
else: # if heap is empty , in next iteration also we can't extract anything which satisfies ith capital <=w
return w
return w
``` | 3 | 0 | ['Python3'] | 0 |
ipo | Easy Java Solution | Priority Queue | Greedy | Beats 100💯✅ | easy-java-solution-priority-queue-greedy-9n3u | Find Maximized Capital\n\n## Problem Statement\n\nThe function findMaximizedCapital is designed to maximize the capital by selecting up to k projects from a lis | shobhitkushwaha1406 | NORMAL | 2024-06-15T04:00:41.903163+00:00 | 2024-06-15T04:00:41.903190+00:00 | 1,063 | false | # Find Maximized Capital\n\n## Problem Statement\n\nThe function `findMaximizedCapital` is designed to maximize the capital by selecting up to `k` projects from a list of available projects, where each project has an associated profit and capital requirement. Given initial capital `w`, the function returns the maximum capital that can be achieved after completing up to `k` projects.\n\n## Intuition and Approach\n\nThe problem can be broken down into two main scenarios based on the initial capital `w`:\n\n1. **Scenario 1: Sufficient Initial Capital (`w >= maxCapital`)**:\n - If the initial capital `w` is greater than or equal to the highest capital required among all projects, you can directly select up to `k` projects with the highest profits.\n - Use a max-heap to always get the project with the highest profit. Add profits to the heap and maintain only the top `k` highest profits.\n\n2. **Scenario 2: Insufficient Initial Capital (`w < maxCapital`)**:\n - If the initial capital `w` is less than the highest capital requirement, you must carefully select projects you can afford and iteratively increase your capital.\n - For each of the `k` iterations, find the project with the highest profit that can be funded with the current capital `w`.\n - Once a project is selected, its profit is added to the capital, and the project is marked as done by setting its capital requirement to a very high value to prevent it from being selected again.\n\n## Algorithm Steps\n\n1. **Determine Maximum Capital**:\n - Iterate through the `capital` array to find the highest capital requirement (`maxCapital`).\n\n2. **Sufficient Capital Case**:\n - If `w >= maxCapital`, use a max-heap to store the `profits`.\n - Add all profits to the heap and keep only the top `k` values.\n - Sum up these values and add them to `w`.\n\n3. **Insufficient Capital Case**:\n - Iterate up to `k` times to select the best project that can be funded with the current `w`.\n - In each iteration, find the project with the highest profit whose capital requirement is less than or equal to `w`.\n - Update `w` by adding the selected project\'s profit and mark the project as done by setting its capital to a very high value.\n\n## Time Complexity\n\n- **Sufficient Capital Case**: \n - Finding `maxCapital` takes `O(n)`.\n - Adding elements to the max-heap takes `O(n log k)`, where `n` is the number of projects, and `k` is the number of projects that can be selected.\n - The overall time complexity is `O(n + n log k)` which simplifies to `O(n log k)`.\n\n- **Insufficient Capital Case**:\n - The outer loop runs up to `k` times, and each time, it finds the maximum profit from the available projects, which takes `O(n)`.\n - Therefore, the time complexity is `O(kn)`.\n\n- The worst-case time complexity is `O(kn)` since `k` could be up to `n` in the worst scenario.\n\n## Space Complexity\n\n- **Sufficient Capital Case**:\n - The max-heap requires `O(k)` space to store the highest `k` profits.\n\n- **Insufficient Capital Case**:\n - The space complexity is `O(1)` excluding the input data as we only use a few additional variables.\n\n- Overall, the space complexity is `O(k)` due to the use of the max-heap in the sufficient capital case.\n\n## Usage\n\nTo use the `findMaximizedCapital` function, create an instance of the `Solution` class and call the method with parameters `k`, `w`, `profits`, and `capital`.\n\n```java\nSolution solution = new Solution();\nint maxCapital = solution.findMaximizedCapital(k, w, profits, capital);\n```\n\n\n\n\n\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int maxCapital = 0;\n for(int i = 0 ; i < capital.length ; i++){\n maxCapital = Math.max(capital[i] , maxCapital);\n }\n \n if(w >= maxCapital){\n PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new Comparator<Integer>() {\n @Override\n public int compare(Integer a, Integer b) {\n return a - b;\n }\n });\n for(int p : profits){\n maxHeap.add(p);\n if(maxHeap.size()>k) maxHeap.poll();\n }\n for(int h: maxHeap) w+=h;\n return w;\n }\n \n \n int index ; \n int n = profits.length;\n for(int i = 0 ; i < Math.min(k , n) ; i++){\n index = -1;\n for(int j = 0 ; j<n ; ++j){\n if (w >= capital[j] && (index == -1 || profits[index] < profits[j])){\n index = j;\n }\n }\n if(index==-1) break;\n w+=profits[index];\n capital[index] = Integer.MAX_VALUE;\n }\n return w;\n }\n}\n``` | 3 | 0 | ['Array', 'Math', 'Greedy', 'Heap (Priority Queue)', 'Java'] | 2 |
ipo | Priority Queue || Optimized || Beats 100% || Explained line by line || C++, Python, Java | priority-queue-optimized-beats-100-expla-udcf | Intuition\n Describe your first thoughts on how to solve this problem. \n * The intuition behind this code is to maximize the available capital after selecting | avinash_singh_13 | NORMAL | 2024-06-15T03:39:45.688437+00:00 | 2024-06-15T03:39:45.688466+00:00 | 20 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n * The intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the current capital constraints. It does this by sorting projects by their capital requirements to quickly find the most affordable ones, then using a max-heap to efficiently select the highest-profit projects that are currently affordable, updating the available capital with each project\'s profit. This greedy approach ensures that at each step, the most beneficial project within financial reach is selected to optimize the total capital.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n * **Sort Projects by Capital:** Begin by sorting the projects based on the capital required to ensure you look at the cheapest projects first.\n\n * **Use Max-Heap for Profits:** Employ a max-heap (inverted to a min-heap using negative values) to always have quick access to the project with the highest available profit.\n\n * **Process Projects Within Capital:** As long as there are projects you can afford, add their profits (negatively) to the heap.\n\n * **Select Top Profit Projects:** For up to k iterations, choose the most profitable project you can afford by popping from the heap, increasing your capital.\n\n * **Resulting Capital:** After potentially choosing k projects, the resulting capital is the maximum capital achieved.\n * Create pairs of profits[i] and capital[i].`\n * Sort pairs by required capital.\n * MaxHeap: While iterating:\n * Push profits of affordable projects into a max heap.\n * Pop and add top profit to capital.\n * Repeat k times.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(nlogn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n O(n)\n\n```python []\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n n=len(profits)\n temp=[(capital[i], profits[i]) for i in range(n)]\n temp.sort()\n maxHeap=[]\n j=0\n while k>0:\n while j<n and w>=temp[j][0]:\n heapq.heappush(maxHeap,-temp[j][1])\n j+=1\n if not maxHeap:\n break\n w-=heapq.heappop(maxHeap)\n k-=1\n return w\n```\n```java []\nclass Solution {\n class Pair {\n int first;\n int second;\n \n Pair(int first, int second) {\n this.first = first;\n this.second = second;\n }\n }\n \n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int j = 0;\n int n = profits.length;\n List<Pair> temp = new ArrayList<>();\n PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());\n\n for (int i = 0; i < n; ++i) {\n temp.add(new Pair(capital[i], profits[i]));\n }\n\n Collections.sort(temp, Comparator.comparingInt(p -> p.first));\n \n while (k-- > 0) {\n while (j < n && w >= temp.get(j).first) {\n maxHeap.offer(temp.get(j++).second);\n }\n\n if (maxHeap.isEmpty()) break;\n w += maxHeap.poll();\n }\n\n return w;\n }\n}\n```\n```C++ []\n#define pi pair<int, int>\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n int j = 0, n = profits.size();\n vector<pi> temp;\n priority_queue<int> maxHeap;\n\n for(int i = 0; i < n; ++i) {\n temp.push_back({capital[i], profits[i]});\n }\n\n sort(begin(temp), end(temp));\n while(k--) {\n while(j < n && w >= temp[j].first) {\n maxHeap.push(temp[j++].second);\n }\n\n if(maxHeap.empty()) break;\n w += maxHeap.top();\n maxHeap.pop();\n }\n\n return w;\n }\n};\n```\n | 3 | 0 | ['C++', 'Java', 'Python3'] | 0 |
ipo | Easiest ✅|| Beats 100% 🔥🔥|| Java✅ || 0ms | easiest-beats-100-java-0ms-by-wankhedeay-fe77 | Intuition\nThe intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the | wankhedeayush90 | NORMAL | 2024-06-15T03:08:36.924859+00:00 | 2024-06-15T03:08:36.924888+00:00 | 304 | false | # Intuition\nThe intuition behind this code is to maximize the available capital after selecting up to k projects, by strategically choosing the projects with the highest profit that can be started within the current capital constraints. It does this by sorting projects by their capital requirements to quickly find the most affordable ones, then using a max-heap to efficiently select the highest-profit projects that are currently affordable, updating the available capital with each project\'s profit. This greedy approach ensures that at each step, the most beneficial project within financial reach is selected to optimize the total capital.\n\n# Approach\n* Sort Projects by Capital: Begin by sorting the projects based on the capital required to ensure you look at the cheapest projects first.\n\n* Use Max-Heap for Profits: Employ a max-heap (inverted to a min-heap using negative values) to always have quick access to the project with the highest available profit.\n\n* Process Projects Within Capital: As long as there are projects you can afford, add their profits (negatively) to the heap.\n\n* Select Top Profit Projects: For up to k iterations, choose the most profitable project you can afford by popping from the heap, increasing your capital.\n\n* Resulting Capital: After potentially choosing k projects, the resulting capital is the maximum capital achieved.\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int maxCapital = 0;\n for(int i = 0 ; i < capital.length ; i++){\n maxCapital = Math.max(capital[i] , maxCapital);\n }\n \n if(w >= maxCapital){\n PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new Comparator<Integer>() {\n @Override\n public int compare(Integer a, Integer b) {\n return a - b;\n }\n });\n for(int p : profits){\n maxHeap.add(p);\n if(maxHeap.size()>k) maxHeap.poll();\n }\n for(int h: maxHeap) w+=h;\n return w;\n }\n \n \n int index ; \n int n = profits.length;\n for(int i = 0 ; i < Math.min(k , n) ; i++){\n index = -1;\n for(int j = 0 ; j<n ; ++j){\n if (w >= capital[j] && (index == -1 || profits[index] < profits[j])){\n index = j;\n }\n }\n if(index==-1) break;\n w+=profits[index];\n capital[index] = Integer.MAX_VALUE;\n }\n return w;\n }\n}\n```\n\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits,\n vector<int>& capital) {\n int n = profits.size();\n std::vector<std::pair<int, int>> projects;\n\n // Creating vector of pairs (capital, profits)\n for (int i = 0; i < n; ++i) {\n projects.emplace_back(capital[i], profits[i]);\n }\n\n // Sorting projects by capital required\n std::sort(projects.begin(), projects.end());\n\n // Max-heap to store profits, using greater to create a max-heap\n std::priority_queue<int> maxHeap;\n int i = 0;\n\n // Main loop to select up to k projects\n for (int j = 0; j < k; ++j) {\n // Add all profitable projects that we can afford\n while (i < n && projects[i].first <= w) {\n maxHeap.push(projects[i].second);\n i++;\n }\n\n // If no projects can be funded, break out of the loop\n if (maxHeap.empty()) {\n break;\n }\n\n // Otherwise, take the project with the maximum profit\n w += maxHeap.top();\n maxHeap.pop();\n }\n\n return w;\n }\n};\n```\n\n**Please upvote if you like the solution**\n\n | 3 | 1 | ['Array', 'Sorting', 'Heap (Priority Queue)', 'C++', 'Java'] | 1 |
ipo | Python3 Solution | python3-solution-by-motaharozzaman1996-f49g | \n\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n max_profit=[]\n min_capi | Motaharozzaman1996 | NORMAL | 2024-06-15T02:18:47.829794+00:00 | 2024-06-15T02:18:47.829837+00:00 | 826 | false | \n```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n max_profit=[]\n min_capital=[(c,p) for c,p in zip(capital,profits)]\n heapq.heapify(min_capital) \n for i in range(k):\n while min_capital and min_capital[0][0]<=w:\n c,p=heapq.heappop(min_capital) \n heapq.heappush(max_profit,-1*p)\n if not max_profit:\n break\n w+=-1*heapq.heappop(max_profit)\n return w \n``` | 3 | 0 | ['Python', 'Python3'] | 1 |
ipo | Maximizing Capital for LeetCode's IPO Through Optimal Project Selection | maximizing-capital-for-leetcodes-ipo-thr-p6q1 | To solve the problem of maximizing LeetCode\'s capital before its IPO by choosing at most \( k \) distinct projects, we can employ a greedy approach. This metho | sleekmind | NORMAL | 2024-06-15T01:43:55.467179+00:00 | 2024-06-15T01:49:17.172219+00:00 | 477 | false | To solve the problem of maximizing LeetCode\'s capital before its IPO by choosing at most \\( k \\) distinct projects, we can employ a greedy approach. This method ensures that at each step, we choose the project that offers the maximum profit while meeting the current capital requirement.\n\n### Approach Breakdown\n\n1. **Sort Projects by Capital Requirement**: First, we sort the projects based on the minimum capital required to start them. This helps in quickly finding all projects that can be initiated with the current available capital.\n\n2. **Use a Max-Heap for Profits**: To always select the project with the highest profit that can be started with the current capital, we use a max-heap (priority queue). This allows us to efficiently retrieve the project with the maximum profit at any point.\n\n3. **Iterate up to \\( k \\) Times**: We iterate up to \\( k \\) times, at each iteration:\n - Add all projects that can be started with the current capital to the max-heap.\n - Select the project with the maximum profit from the max-heap.\n - Update the current capital by adding the profit of the selected project.\n - Repeat until no more projects can be started or we reach the \\( k \\) iterations limit.\n\n### Detailed Steps with Example\n\nConsider the example:\n\n- k = 2\n- w = 0 \n- profits = [1, 2, 3]\n- capital = [0, 1, 1]\n\n#### Step 1: Sort Projects by Capital\nFirst, we pair each project\'s capital requirement with its profit and sort them by capital.\n\n- Pair and sort: projects= [(0, 1), (1, 2), (1, 3)]\n\n#### Step 2: Initialize Data Structures\n- Create a max-heap to store profits of projects that can be started.\n- Initialize current capital \\( w = 0 \\).\n\n#### Step 3: Iterate and Select Projects\n- **Iteration 1**:\n - Check which projects can be started with \\( w = 0 \\).\n - Only project \\((0, 1)\\) can be started.\n - Add its profit to the max-heap: maxHeap = [1].\n - Select the project with the highest profit (1).\n - Update \\( w \\): \\( w = 0 + 1 = 1 \\).\n \n- **Iteration 2**:\n - With \\( w = 1 \\), both remaining projects \\((1, 2)\\) and \\((1, 3)\\) can be started.\n - Add their profits to the max-heap:textmaxHeap = [2, 3].\n - Select the project with the highest profit (3).\n - Update \\( w \\): \\( w = 1 + 3 = 4 \\).\n\nAfter these iterations, the maximum capital \\( w \\) is 4.\n\n### Code Implementation\n\nHere\'s the implementation of the above logic in C++:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n ios_base::sync_with_stdio(false);\n int n = profits.size();\n vector<pair<int, int>> minCapital;\n\n for (int i = 0; i < n; i++)\n minCapital.push_back({capital[i], profits[i]});\n\n sort(minCapital.begin(), minCapital.end());\n\n priority_queue<int> maxProfit;\n int j = 0;\n for (int i = 0; i < k; ++i) {\n while (j < n && minCapital[j].first <= w) {\n maxProfit.push(minCapital[j].second);\n j++;\n }\n\n if (maxProfit.empty())\n break;\n\n w += maxProfit.top();\n maxProfit.pop();\n }\n\n return w;\n }\n};\n```\n\n## Visual Explanation\n\n1. **Initialization**:\n - Projects: [(0, 1), (1, 2), (1, 3)]\n - Capital: \\( w = 0 \\)\n - Max-Heap: **empty**\n\n2. **Iteration 1**:\n - Add projects with capital \u2264 0: [(0, 1)]\n - Max-Heap: [1]\n - Select max profit 1: Update \\( w = 1 \\)\n\n3. **Iteration 2**:\n - Add projects with capital \u2264 1: [(1, 2), (1, 3)]\n - Max-Heap: [2, 3]\n - Select max profit 3: Update w = 4 \n\nBy following these steps, we ensure that at each step, we are maximizing the profit while considering the capital constraints, thereby arriving at the optimal solution. | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'C++'] | 4 |
ipo | Beats 97% users... Trust this code guys | beats-97-users-trust-this-code-guys-by-a-anud | \n\n# Code\n\nclass T {\n public int pro;\n public int cap;\n public T(int pro, int cap) {\n this.pro = pro;\n this.cap = cap;\n }\n}\n\nclass Solutio | Aim_High_212 | NORMAL | 2024-06-15T01:42:17.240273+00:00 | 2024-06-15T01:42:17.240292+00:00 | 39 | false | \n\n# Code\n```\nclass T {\n public int pro;\n public int cap;\n public T(int pro, int cap) {\n this.pro = pro;\n this.cap = cap;\n }\n}\n\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n Queue<T> minHeap = new PriorityQueue<>((a, b) -> a.cap - b.cap);\n Queue<T> maxHeap = new PriorityQueue<>((a, b) -> b.pro - a.pro);\n\n for (int i = 0; i < Capital.length; ++i)\n minHeap.offer(new T(Profits[i], Capital[i]));\n\n while (k-- > 0) {\n while (!minHeap.isEmpty() && minHeap.peek().cap <= W)\n maxHeap.offer(minHeap.poll());\n if (maxHeap.isEmpty())\n break;\n W += maxHeap.poll().pro;\n }\n\n return W;\n }\n}\n``` | 3 | 0 | ['Python', 'C++', 'Java', 'Python3'] | 0 |
ipo | 🔥 🔥 🔥 Simple Approch 🔥 🔥 🔥 | simple-approch-by-vivek_kumr-k6c9 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to maximize the available capital w after k investment rounds. Each project | Vivek_kumr | NORMAL | 2024-06-15T01:27:00.137056+00:00 | 2024-06-15T01:27:00.137085+00:00 | 435 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to maximize the available capital w after k investment rounds. Each project has a specific profit and capital requirement. We need to strategically choose projects that can be funded with the current capital and yield the highest profit.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.Min-Heap for Capital Requirements:\n-->Use a min-heap (minHeap) to store projects based on their capital requirements. This allows us to quickly identify projects that can be funded with the current capital.\n\n2.Max-Heap for Profits:\n-->Use a max-heap (maxHeap) to store the profits of the projects that can be funded. This helps in selecting the most profitable project available at any step.\n\n3.Iterative Process:\n-->Initially, push all projects into the minHeap.\n-->For each of the k investment rounds:\n ----->Transfer all projects from minHeap to maxHeap that can be funded with the current capital w.\n----->If there are no more projects that can be funded (i.e., maxHeap is empty), break the loop.\n----->Otherwise, select the project with the highest profit (top of maxHeap), add its profit to the current capital w, and remove it from maxHeap.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nhe overall time complexity is O(nlogn+klogn).\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nBoth heaps (minHeap and maxHeap) can store up to n elements.\nTherefore, the space complexity is O(n).\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> minHeap;\n priority_queue<int> maxHeap;\n for (int i = 0; i < profits.size(); i++) {\n minHeap.push({capital[i], profits[i]});\n }\n for (int i = 0; i < k; i++) {\n while (!minHeap.empty() && minHeap.top().first <= w) {\n maxHeap.push(minHeap.top().second);\n minHeap.pop();\n }\n if (maxHeap.empty()) break;\n w += maxHeap.top();\n maxHeap.pop();\n }\n return w;\n }\n};\n``` | 3 | 0 | ['C++'] | 1 |
ipo | Two- Heap Pattern ✅✔️💯 | two-heap-pattern-by-dixon_n-d9n8 | This is a new Pattern (Two- Heap Pattern.Min heap and Max Heap Pattern)\n\n215. Kth Largest Element in an Array same pattern\n451. Sort Characters By Frequency\ | Dixon_N | NORMAL | 2024-05-30T22:25:20.512438+00:00 | 2024-05-30T22:25:20.512474+00:00 | 144 | false | This is a new Pattern (Two- Heap Pattern.Min heap and Max Heap Pattern)\n\n[215. Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/solutions/5195848/k-pattern-heaps-peiority/) same pattern\n451. Sort Characters By Frequency\n[703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/solutions/5198578/k-pattern/)\n[767. Reorganize String](https://leetcode.com/problems/reorganize-string/solutions/5196813/k-pattern/)\nRearrange String k Distance Apart -- Q & A below\n[1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows](https://leetcode.com/problems/find-the-kth-smallest-sum-of-a-matrix-with-sorted-rows/solutions/5198589/k-pattern/)\n[373. Find K Pairs with Smallest Sums](https://leetcode.com/problems/find-k-pairs-with-smallest-sums/solutions/5197679/the-k-pattern/)\n[378. Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/solutions/5197755/the-k-pattern/)\n[719. Find K-th Smallest Pair Distance]()\n[2040. Kth Smallest Product of Two Sorted Arrays]()\n[264. Ugly Number II](https://leetcode.com/problems/ugly-number-ii/solutions/5198107/nthuglynumber/)\n[263. Ugly Number](https://leetcode.com/problems/ugly-number/solutions/5198605/the-ugly-number/)\n[502. IPO](https://leetcode.com/problems/ipo/solutions/5198438/priorityqueue-pattern/)\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n // Min-heap for projects based on capital requirements\n PriorityQueue<Project> minCapitalHeap = new PriorityQueue<>((a, b) -> a.capital - b.capital);\n // Max-heap for available projects based on profits\n PriorityQueue<Project> maxProfitHeap = new PriorityQueue<>((a, b) -> b.profit - a.profit);\n\n // Populate the min-heap with initial projects\n for (int i = 0; i < profits.length; i++) {\n minCapitalHeap.offer(new Project(capital[i], profits[i]));\n }\n\n // Iterate k times to pick k projects\n for (int i = 0; i < k; i++) {\n // Move all projects we can afford to the max-heap\n while (!minCapitalHeap.isEmpty() && minCapitalHeap.peek().capital <= w) {\n maxProfitHeap.offer(minCapitalHeap.poll());\n }\n\n // If there are no affordable projects, break\n if (maxProfitHeap.isEmpty()) {\n break;\n }\n\n // Pick the most profitable project\n w += maxProfitHeap.poll().profit;\n }\n\n return w;\n }\n\n // Helper class to represent a project with capital and profit\n static class Project {\n int capital;\n int profit;\n\n Project(int capital, int profit) {\n this.capital = capital;\n this.profit = profit;\n }\n }\n}\n\n``` | 3 | 0 | ['Greedy', 'Heap (Priority Queue)', 'Java'] | 4 |
ipo | Simple || Concise || Beginner Friendly || Greedy ✅✅ | simple-concise-beginner-friendly-greedy-jzvwn | Complexity\n- Time complexity: O(n*log n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n | Lil_ToeTurtle | NORMAL | 2023-09-09T07:53:52.005369+00:00 | 2023-09-09T07:53:52.005388+00:00 | 165 | false | # Complexity\n- Time complexity: $$O(n*log n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n int n=capital.length, pc[][]=new int[n][2];\n for(int i=0;i<n;i++){\n pc[i][0]=profits[i];\n pc[i][1]=capital[i];\n }\n Arrays.sort(pc, (a,b)->a[1]-b[1]);\n PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->b[0]-a[0]);\n int maxCap=w, i=0;\n while(k-->0) {\n for(; i<n && maxCap>=pc[i][1]; i++) \n pq.add(pc[i]);\n if(pq.isEmpty()) break;\n maxCap+=pq.poll()[0];\n }\n return maxCap;\n }\n}\n``` | 3 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | 502: Solution with step by step explanation | 502-solution-with-step-by-step-explanati-1fds | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project. | Marlen09 | NORMAL | 2023-03-12T04:21:24.910303+00:00 | 2023-03-12T04:21:24.910332+00:00 | 886 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Create a list of tuples containing (capital, profit) for each project. We can use a list comprehension to achieve this.\n\n2. Sort the list of projects by increasing capital required to start them. We can use the built-in sort function with a lambda function to sort the list based on the first element (capital) of each tuple.\n\n3. Initialize a max heap to keep track of available profits from projects. We can use the built-in heapq module to create a heap.\n\n4. Loop through k projects or until we run out of projects to consider. We use a for loop with a range of k to limit the number of projects we can consider.\n\n5. Add all available projects we can currently afford to the max heap. We use a while loop to check if there are any projects left in the list and if the first project requires less capital than what we currently have. If so, we add the project to the heap.\n\n6. If there are no available projects, we can\'t make any more profit. We check if the heap is empty and if so, we break out of the loop.\n\n7. Take the project with the highest profit from the max heap and add its profit to our total. We use heapq.heappop to get the project with the highest profit (since we added negative profits to the heap), and subtract it from our current capital.\n\n8. Return the final total capital.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n # Create a list of tuples containing (capital, profit) for each project\n projects = [(capital[i], profits[i]) for i in range(len(profits))]\n \n # Sort the list of projects by increasing capital required to start them\n projects.sort()\n \n # Initialize a max heap to keep track of available profits from projects\n heap = []\n \n # Loop through k projects or until we run out of projects to consider\n for _ in range(k):\n # Add all available projects we can currently afford to the max heap\n while projects and projects[0][0] <= w:\n capital, profit = projects.pop(0)\n heapq.heappush(heap, -profit)\n \n # If there are no available projects, we can\'t make any more profit\n if not heap:\n break\n \n # Take the project with the highest profit from the max heap and add its profit to our total\n w -= heapq.heappop(heap)\n \n # Return the final total capital\n return w\n\n``` | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Python', 'Python3'] | 1 |
ipo | Sort + Priority Queue | C++ | sort-priority-queue-c-by-tusharbhart-kp2s | \nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size(), p = 0;\n | TusharBhart | NORMAL | 2023-02-25T10:34:09.931508+00:00 | 2023-02-25T10:34:09.931553+00:00 | 274 | false | ```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size(), p = 0;\n vector<pair<int, int>> v;\n for(int i=0; i<n; i++) v.push_back({capital[i], profits[i]});\n\n sort(v.begin(), v.end());\n priority_queue<pair<int, int>> pq;\n\n for(int _=0; _<k; _++) {\n while(p < n && v[p].first <= w) pq.push({v[p].second, v[p].first}), p++;\n if(pq.size()) {\n w += pq.top().first;\n pq.pop();\n }\n }\n return w;\n }\n};\n``` | 3 | 0 | ['Greedy', 'Sorting', 'Heap (Priority Queue)', 'C++'] | 1 |
ipo | Java | Priority Queue | 10 lines | O(n log n) time | java-priority-queue-10-lines-on-log-n-ti-4gi8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | judgementdey | NORMAL | 2023-02-23T21:32:42.855804+00:00 | 2023-02-23T21:40:40.370298+00:00 | 153 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n*log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n var projects = new PriorityQueue<int[]>((a, b) -> Integer.compare(a[0], b[0]));\n\n for (var i=0; i < profits.length; i++)\n projects.offer(new int[] {capital[i], profits[i]});\n\n var queue = new PriorityQueue<Integer>(Collections.reverseOrder());\n\n while (k-- > 0) {\n for (var a = projects.peek(); a != null && a[0] <= w; a = projects.peek())\n queue.offer(projects.poll()[1]);\n\n if (queue.isEmpty()) return w;\n w += queue.poll();\n }\n return w;\n }\n}\n``` | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
ipo | 📌📌Python3 || ⚡783 ms, faster than 98.25% of Python3 | python3-783-ms-faster-than-9825-of-pytho-muzq | \n\ndef findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nl | harshithdshetty | NORMAL | 2023-02-23T12:57:08.979107+00:00 | 2023-02-23T12:57:08.979148+00:00 | 418 | false | \n```\ndef findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n if w >= max(capital):\n return w + sum(nlargest(k, profits)) \n projects = [[capital[i],profits[i]] for i in range(len(profits))]\n projects.sort(key=lambda x: x[0]) \n heap = [] \n for i in range(k):\n while projects and projects[0][0] <= w:\n heappush(heap, -1*projects.pop(0)[1]) \n if not heap:\n break\n p = -heappop(heap)\n w += p\n return w\n```\n\nhere\'s a step-by-step explanation of the code:\n1. The function findMaximizedCapital takes in four arguments: k, w, profits, and capital. k represents the maximum number of projects that can be selected, w represents the initial amount of capital, profits is a list of profits that can be earned from each project, and capital is a list of minimum capital required to start each project.\n1. If the initial amount of capital w is greater than or equal to the maximum value of capital, then we can select all projects and earn maximum profit. So, we return the sum of w and the sum of k largest profits from the profits list using the nlargest function.\n1. If the initial amount of capital w is less than the maximum value of capital, then we create a list of projects, where each project is represented as a list [capital[i], profits[i]]. This allows us to keep track of the required capital and the corresponding profit for each project.\n1. We sort the list of projects based on the minimum capital required to start each project. This allows us to select projects in increasing order of required capital.\n1. We initialize an empty heap, which we will use to keep track of the maximum profits that can be earned from projects that require less than or equal to the available capital.\n1. We loop k times, since we can select at most k projects.\n1. Within the loop, we check the first project in the sorted list of projects. If the minimum capital required for the project is less than or equal to the available capital w, we add the corresponding profit to the heap.\n1. We then pop the maximum profit from the heap and add it to the available capital w. We repeat this process until we have selected k projects or there are no more projects left in the list.\n1. Finally, we return the final amount of capital w after selecting k projects with maximum profit. | 3 | 0 | ['Heap (Priority Queue)', 'Python', 'Python3'] | 1 |
ipo | require two heaps, max and min | require-two-heaps-max-and-min-by-mr_star-rkw8 | \nclass Solution {\npublic:\n struct compare {\n bool operator()(pair<int ,int> p, pair<int ,int> q) {\n return p.first>q.first;\n } | mr_stark | NORMAL | 2023-02-23T12:20:42.195238+00:00 | 2023-02-23T12:20:42.195282+00:00 | 251 | false | ```\nclass Solution {\npublic:\n struct compare {\n bool operator()(pair<int ,int> p, pair<int ,int> q) {\n return p.first>q.first;\n }\n };\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n priority_queue<pair<int,int>, vector<pair<int,int>>, compare> min_pq;\n priority_queue<int> max_pq;\n for(int i =0;i<c.size();i++)\n {\n min_pq.push({c[i],p[i]});\n }\n int i=0,n=c.size();\n while(k--)\n {\n while(!min_pq.empty() && min_pq.top().first<=w)\n {\n max_pq.push(min_pq.top().second);\n min_pq.pop();\n }\n \n if(!max_pq.empty())\n {\n w+=max_pq.top();\n max_pq.pop();\n }\n }\n \n \n return w;\n }\n};\n``` | 3 | 0 | ['C'] | 1 |
ipo | Solution in C++ | solution-in-c-by-ashish_madhup-0zdc | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ashish_madhup | NORMAL | 2023-02-23T11:47:57.523539+00:00 | 2023-02-23T11:47:57.523566+00:00 | 1,001 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& p, vector<int>& c) {\n priority_queue<pair<int,int>> pq1;\n priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq2;\n for(int i=0;i<p.size();i++){\n if(c[i]>w) pq2.push({c[i],p[i]});\n else pq1.push({p[i],c[i]});\n }\n \n while(k>0 && (!pq1.empty() || !pq2.empty())){\n while(!pq2.empty() && pq2.top().first<=w){\n pq1.push({pq2.top().second,pq2.top().first});\n pq2.pop();\n }\n if(!pq1.empty()) {\n w+=pq1.top().first;\n pq1.pop();\n k--;\n } else return w;\n }\n return w;\n }\n};\n``` | 3 | 0 | ['C++'] | 1 |
ipo | ✅ beats 99% Java code | beats-99-java-code-by-abstractconnoisseu-y95k | Java Code\n\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n int maxCapital = 0;\n for (int | abstractConnoisseurs | NORMAL | 2023-02-23T09:27:54.936411+00:00 | 2023-02-23T09:27:54.936452+00:00 | 451 | false | # Java Code\n```\nclass Solution {\n public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {\n int maxCapital = 0;\n for (int i = 0; i < Capital.length; i++) {\n maxCapital = Math.max(Capital[i], maxCapital);\n }\n\n if (W >= maxCapital) {\n PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new Comparator<Integer>() {\n @Override\n public int compare(Integer a, Integer b) {\n return a - b;\n }\n });\n for (int p : Profits) {\n maxHeap.add(p);\n if (maxHeap.size() > k) maxHeap.poll();\n }\n for (int h : maxHeap) W += h;\n return W;\n }\n\n int index;\n int n = Profits.length;\n for (int i = 0; i < Math.min(k, n); i++) {\n index = -1;\n for (int j = 0; j < n; ++j) {\n if (W >= Capital[j] && (index == -1 || Profits[index] < Profits[j])) {\n index = j;\n }\n }\n if (index == -1) break;\n W += Profits[index];\n Capital[index] = Integer.MAX_VALUE;\n }\n return W;\n }\n}\n``` | 3 | 0 | ['Array', 'Greedy', 'Sorting', 'Heap (Priority Queue)', 'Java'] | 1 |
ipo | C++ Solution | | Greedy approach | c-solution-greedy-approach-by-vaibhav_sa-4jr8 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Vaibhav_saroj | NORMAL | 2023-02-23T08:22:03.811395+00:00 | 2023-02-23T08:22:03.811421+00:00 | 676 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int findMaximizedCapital(int k, int w, vector<int>& profits, vector<int>& capital) {\n int n = profits.size();\n vector<pair<int, int>> projects(n);\n for (int i = 0; i < n; i++) {\n projects[i] = {capital[i], profits[i]};\n }\n sort(projects.begin(), projects.end());\n\n int idx = 0;\n priority_queue<int> pq;\n for (int i = 0; i < k; i++) {\n while (idx < n && projects[idx].first <= w) {\n pq.push(projects[idx].second);\n idx++;\n }\n if (pq.empty()) {\n break;\n }\n w += pq.top();\n pq.pop();\n }\n return w;\n \n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
ipo | ✅ Java | Brute Force Approach | java-brute-force-approach-by-prashantkac-uwx7 | \n// Approach 1: Brute Force Approach - TLE\n\n// Time complexity: O(n^2)\n// Space complexity: O(n)\n\npublic int findMaximizedCapital(int k, int w, int[] prof | prashantkachare | NORMAL | 2023-02-23T05:33:47.750117+00:00 | 2023-02-23T05:33:47.750156+00:00 | 216 | false | ```\n// Approach 1: Brute Force Approach - TLE\n\n// Time complexity: O(n^2)\n// Space complexity: O(n)\n\npublic int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {\n\tint n = profits.length;\n\tboolean[] finished = new boolean[n];\n\n\tfor (int i = 0; i < k; i++) {\n\t\tint prjIndex = -1;\n\n\t\tfor (int j = 0; j < n; j++) {\n\t\t\tif (capital[j] <= w && !finished[j]) {\n\t\t\t\tif (prjIndex == -1 || profits[j] > profits[prjIndex]) \n\t\t\t\t\tprjIndex = j;\n\t\t\t}\n\t\t}\n\n\t\tif (prjIndex == -1) \n\t\t\treturn w;\n\n\t\tw += profits[prjIndex];\n\t\tfinished[prjIndex] = true;\n\t}\n\n\treturn w;\n}\n```\n\n**Note:-** This solution gives TLE. It\'s provided just for understanig purpose.\n**Please upvote if you find this solution useful. Happy Coding!** | 3 | 0 | ['Greedy', 'Java'] | 2 |
ipo | [Python] greey with min heap queue, Explained | python-greey-with-min-heap-queue-explain-n6ky | Step 1, sort the profits based on the captial we need to gain the profit\nStep 2, based on the capital we have, add the profit into a min heap (add the negative | wangw1025 | NORMAL | 2023-02-23T04:20:27.101161+00:00 | 2023-02-23T04:20:27.101219+00:00 | 355 | false | Step 1, sort the profits based on the captial we need to gain the profit\nStep 2, based on the capital we have, add the profit into a min heap (add the negative profit)\nStep 3, every loop, we pick the head from the heap, which is the largest profit we can gain\nStep 4, check the project list and add more projects based on the current profit\n\nAfter k loops, we can return the maximum capital we have.\n\n```\nclass Solution:\n def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:\n # sort the profits based on the capital required\n cap_profits = list(zip(capital, profits))\n cap_profits.sort()\n \n profit_list, proj_cnt, prev_proj_cnt = [], 0, -1\n \n idx = 0\n while proj_cnt < k and proj_cnt != prev_proj_cnt:\n prev_proj_cnt = proj_cnt\n while idx < len(profits) and cap_profits[idx][0] <= w:\n heapq.heappush(profit_list, -(cap_profits[idx][1]))\n idx += 1\n\n # pick the largest profit we can get from the profit list\n if profit_list:\n w -= heapq.heappop(profit_list)\n proj_cnt += 1\n \n return w\n``` | 3 | 0 | ['Heap (Priority Queue)', 'Python3'] | 1 |
ipo | JS priority queue +explanation | js-priority-queue-explanation-by-crankyi-we9o | Performance\nRuntime 447ms Beats 66.27% Memory 101.1MB Beats 10.84%\n\n### Intuition\nWe need to find the k projects which yield the greatest overall profit. S | crankyinmv | NORMAL | 2023-02-23T01:25:42.620481+00:00 | 2023-02-23T01:25:42.620518+00:00 | 387 | false | ### Performance\nRuntime 447ms Beats 66.27% Memory 101.1MB Beats 10.84%\n\n### Intuition\nWe need to find the `k` projects which yield the greatest overall profit. Since *the operating capital never goes down* we don\'t have a house robbber situation where choosing one project has an opportunity cost which would keep us from choosing another more lucrative combination of projects later. We only need to choose the most profitable project we can afford to invest in each time.\n\n### Approach\n- Start with a list of project sorted by capital cost ascending and a priority queue.\n- Before choosing the next project, enqueue jobs until the the capital cost exceeds your operating capital `w`.\n- dequeue the most profitable project and add it to the operating capital.\n- Repeat `k` times or until all the projects are run.\n\n### Complexity\n- Time complexity: O(nlogn) or O(klogn)\n\n- Space complexity: O(n)\n\n### Code\n```\n/**\n * @param {number} k\n * @param {number} w\n * @param {number[]} profits\n * @param {number[]} capital\n * @return {number}\n */\nvar findMaximizedCapital = function(k, w, profits, capital) \n{\n /* Sort the projects by capital cost. */\n let projects = [];\n for(let i=0; i<profits.length; i++)\n projects.push([profits[i],capital[i]]);\n projects.sort((a,b)=>a[1]-b[1]);\n\n /* Run the projects. */\n let pq = new MaxPriorityQueue(), pp = 0;\n while(k > 0)\n {\n /* Add all the projects fundable with the current operating capital. */\n while(pp < projects.length && projects[pp][1] <= w)\n {\n pq.enqueue(pp,projects[pp][0]);\n pp++;\n }\n\n if(pq.size() === 0)\n break;\n\n /* Choose the most profitable available project. */\n let el = pq.dequeue();\n w += el.priority;\n k--;\n }\n\n return w;\n};\n``` | 3 | 0 | ['JavaScript'] | 0 |
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