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https://leetcode.com/problems/power-of-four/discuss/2667207/Python-solution-based-on-recursion | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 0:
return False
if n == 1 or n == 4:
return True
return self.isPowerOfFour(n/4) | power-of-four | Python solution based on recursion | MPoinelli | 0 | 4 | power of four | 342 | 0.458 | Easy | 5,900 |
https://leetcode.com/problems/power-of-four/discuss/2664980/Python-Easy-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n < 1:
return False
if n == 1:
return True
return self.isPowerOfFour(n / 4) | power-of-four | Python Easy Solution | kruzhilkin | 0 | 2 | power of four | 342 | 0.458 | Easy | 5,901 |
https://leetcode.com/problems/power-of-four/discuss/2662054/Python-solution-without-loopsrecursion | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1:
return True
bit = "{0:b}".format(n)
if bit[-1] == "1":
return False
if bit[0] == "1" and "1" not in bit[1:] and len(bit) % 2 == 1:
return True | power-of-four | Python solution without loops/recursion | maomao1010 | 0 | 7 | power of four | 342 | 0.458 | Easy | 5,902 |
https://leetcode.com/problems/power-of-four/discuss/2614089/Python-Solution-Interview-Approach | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<=0:
return False
while n>1:
if n%4!=0: return False
n//=4
return True | power-of-four | [Python Solution] Interview Approach | utsa_gupta | 0 | 9 | power of four | 342 | 0.458 | Easy | 5,903 |
https://leetcode.com/problems/power-of-four/discuss/2559569/Python-(Simple-Solution-and-Beginner-Friendly) | class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n>1:
n = n/4
return n == 1 | power-of-four | Python (Simple Solution and Beginner-Friendly) | vishvavariya | 0 | 39 | power of four | 342 | 0.458 | Easy | 5,904 |
https://leetcode.com/problems/power-of-four/discuss/2470113/Python3-bitwise-operator | class Solution(object):
def isPowerOfFour(self, num):
"""
:type n: int
:rtype: bool
"""
result = 0
if num <= 0:
return False
if num & (num - 1) != 0:
return False
for i in range(0, 31):
if i % 2 == 0:
result += 1 << i
return (num & result) == num
sol = Solution()
print(sol.isPowerOfFour(15)) | power-of-four | Python3 - bitwise operator | user2354hl | 0 | 4 | power of four | 342 | 0.458 | Easy | 5,905 |
https://leetcode.com/problems/power-of-four/discuss/2466684/Unique-Python-One-Liners | class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n > 0 \
and f'{n:032b}'.count('1') == 1 \
and f'{n:032b}'.find('1') & 1 | power-of-four | Unique Python One-Liners | timberg | 0 | 16 | power of four | 342 | 0.458 | Easy | 5,906 |
https://leetcode.com/problems/power-of-four/discuss/2466530/Python-Simple-Python-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
power = 0
number = 4
while True:
num = number**power
if num == n:
return True
if num > n:
return False
else:
power = power + 1
return False | power-of-four | [ Python ] ✅✅ Simple Python Solution 🥳✌👍 | ASHOK_KUMAR_MEGHVANSHI | 0 | 22 | power of four | 342 | 0.458 | Easy | 5,907 |
https://leetcode.com/problems/power-of-four/discuss/2464578/Python-Short-Faster-Solution-log-base-4-oror-Documented | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n > 0:
power = int(math.log(n,4)) # find power using log base 4
return 4 ** power == n # if 4^power is same as n, return True, else False
return False | power-of-four | [Python] Short, Faster Solution - log base 4 || Documented | Buntynara | 0 | 5 | power of four | 342 | 0.458 | Easy | 5,908 |
https://leetcode.com/problems/power-of-four/discuss/2463692/Python-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
result = 1
while n > result:
result *= 4
return result == n | power-of-four | Python Solution | hgalytoby | 0 | 14 | power of four | 342 | 0.458 | Easy | 5,909 |
https://leetcode.com/problems/power-of-four/discuss/2462341/Python-Fast-and-trivial-solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n <= 0:
return False
while n > 1:
n, m = divmod(n, 4)
if m > 0:
return False
return True | power-of-four | [Python] Fast and trivial solution | RegInt | 0 | 15 | power of four | 342 | 0.458 | Easy | 5,910 |
https://leetcode.com/problems/power-of-four/discuss/2462292/Three-Solution-using-recursion-iteration-and-Bit-Manipulation | class Solution:
def isPowerOfFour(self, n: int) -> bool:
# Solution 1 using recursion
while n % 4 == 0 and n > 0:
return self.isPowerOfFour(n/4)
return n == 1
# Solution 2 iteration
if n == 1:
return True
if n % 4:
return False
while n > 1:
if n % 4:
return False
n //= 4
return n == 1
# Solution 3 using bit manipulation
'''
Once we write numbers in it's binary representation, from there we can observe:=>
i. 000001 , power of 2 and 4
ii. 000010, power of only 2
iii. 000100 , power of 2 and 4
iv. 001000, power of only 2
v. 010000 , power of 2 and 4
vi. 100000, power of only 2
We can see if the set bit is at an odd position and is a power of 2, it's also power of 4.
'''
return n.bit_length() & 1 and not(n & (n-1)) | power-of-four | Three Solution using recursion, iteration and Bit-Manipulation | __Asrar | 0 | 25 | power of four | 342 | 0.458 | Easy | 5,911 |
https://leetcode.com/problems/power-of-four/discuss/2462065/30ms-or-fast-and-easy-approach | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n>0:
x=math.log(n,4)
return math.floor(x)==math.ceil(x)
return False | power-of-four | 30ms | fast and easy approach | ayushigupta2409 | 0 | 16 | power of four | 342 | 0.458 | Easy | 5,912 |
https://leetcode.com/problems/power-of-four/discuss/2461942/python3-with-binary-bit-manipulation | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<=0: return False
n_bin = bin(n)[2:]
return True if (len(n_bin) % 2 == 1) and (sum(map(int,n_bin)) == 1) else False | power-of-four | python3, with binary bit manipulation | pjy953 | 0 | 3 | power of four | 342 | 0.458 | Easy | 5,913 |
https://leetcode.com/problems/power-of-four/discuss/2461898/Python3-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
i = 0
curr = 0
while curr <= n:
curr = 4 ** i
if curr == n:
return True
i += 1
return False | power-of-four | Python3 Solution | creativerahuly | 0 | 19 | power of four | 342 | 0.458 | Easy | 5,914 |
https://leetcode.com/problems/power-of-four/discuss/2461708/Python-Recursion-O(logn)-time-O(logn)-space. | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1:
return True
if n < 1:
return False
return self.isPowerOfFour(n/4) | power-of-four | Python Recursion, O(logn) time, O(logn) space. | OsamaRakanAlMraikhat | 0 | 17 | power of four | 342 | 0.458 | Easy | 5,915 |
https://leetcode.com/problems/power-of-four/discuss/2461452/Simple-Python-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n>1:
n = n/4
if n == 1:
return True
return False | power-of-four | Simple Python Solution | saiavunoori4187 | 0 | 15 | power of four | 342 | 0.458 | Easy | 5,916 |
https://leetcode.com/problems/power-of-four/discuss/2461394/Python-Simple-and-Easy-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1: return True
while n > 1:
if n%4 == 0:
n = n // 4
else:
return False
if n == 1:
return True
return False | power-of-four | Python - Simple and Easy Solution | dayaniravi123 | 0 | 7 | power of four | 342 | 0.458 | Easy | 5,917 |
https://leetcode.com/problems/power-of-four/discuss/2461372/Easy-Python-Solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<1:
return False
elif n==1:
return True
else:
while n!=1:
if n%4!=0:
return False
else:
n=n//4
return True | power-of-four | Easy Python Solution | a_dityamishra | 0 | 13 | power of four | 342 | 0.458 | Easy | 5,918 |
https://leetcode.com/problems/power-of-four/discuss/2461309/Python-O(1)-oneliner-easy-to-understand-(no-loops-or-bitwise-math) | class Solution:
def isPowerOfFour(self, num):
return num == 4**(num.bit_length()//2) | power-of-four | [Python] O(1) oneliner - easy to understand (no loops or bitwise math) | user5285r | 0 | 7 | power of four | 342 | 0.458 | Easy | 5,919 |
https://leetcode.com/problems/power-of-four/discuss/2461196/Python-bit-check | class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n > 0 and n & (n - 1) == 0 and n & 0x55555555 | power-of-four | Python, bit check | blue_sky5 | 0 | 20 | power of four | 342 | 0.458 | Easy | 5,920 |
https://leetcode.com/problems/power-of-four/discuss/2446455/python3-using-a-while-loop-faster-than-86 | class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n%4==0<n:
n/=4
return n==1 | power-of-four | [python3] using a while loop, faster than 86% | hhlinwork | 0 | 8 | power of four | 342 | 0.458 | Easy | 5,921 |
https://leetcode.com/problems/power-of-four/discuss/2366691/Simple-fast-Python-solution-or-Easy-to-Understand-or-73.70-faster | class Solution:
def isPowerOfFour(self, n: int) -> bool:
x = 0
y = pow(4, x)
while y<n:
x+=1
y = pow(4, x)
if (n - y) == 0:
return True
else:
return False | power-of-four | Simple fast Python solution | Easy to Understand | 73.70% faster | harishmanjunatheswaran | 0 | 40 | power of four | 342 | 0.458 | Easy | 5,922 |
https://leetcode.com/problems/power-of-four/discuss/2120598/Python3-easy-understanding-solution | class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 0:
return False
pow_value = 1
while pow_value < n:
pow_value *=4
return pow_value == n | power-of-four | Python3 - easy understanding solution | thanhinterpol | 0 | 23 | power of four | 342 | 0.458 | Easy | 5,923 |
https://leetcode.com/problems/power-of-four/discuss/1936644/Python-One-Liner-or-Log-or-Math | class Solution:
def isPowerOfFour(self, n):
return n>0 and log(n,4).is_integer() | power-of-four | Python - One-Liner | Log | Math | domthedeveloper | 0 | 44 | power of four | 342 | 0.458 | Easy | 5,924 |
https://leetcode.com/problems/power-of-four/discuss/1815199/Easy-to-understand-trick | class Solution:
def isPowerOfFour(self, n: int) -> bool:
arr=[]
for i in range(100):
arr.append(4**i)
if n in arr:
return True
else:
return False | power-of-four | Easy to understand trick | prashantdahiya711 | 0 | 76 | power of four | 342 | 0.458 | Easy | 5,925 |
https://leetcode.com/problems/integer-break/discuss/2830343/O(1)-oror-TC1-10-line-code | class Solution:
def integerBreak(self, n: int) -> int:
if(n<=3):
return n-1
n3=n//3
r3=n%3
if(r3==0):
return 3**n3
if(r3==1):
r3=4
n3-=1
return r3*(3**n3) | integer-break | O(1) || TC=1 10 line code | droj | 7 | 71 | integer break | 343 | 0.554 | Medium | 5,926 |
https://leetcode.com/problems/integer-break/discuss/2180665/Python3-DP-top-down-approach-faster-than-96 | class Solution:
result = {1:1,2:1,3:2,4:4,5:6,6:9,7:12,8:18,9:27,10:36}
def integerBreak(self, n: int) -> int:
try:
return self.result[n]
except:
x = float("-inf")
for i in range(1,n):
j = n-1
while j>0:
if i+j==n:
k = self.integerBreak(i)*self.integerBreak(j)
x = max(x,k)
j-=1
self.result[n] = x
return self.result[n] | integer-break | 📌 Python3 DP top down approach faster than 96% | Dark_wolf_jss | 4 | 76 | integer break | 343 | 0.554 | Medium | 5,927 |
https://leetcode.com/problems/integer-break/discuss/1699371/Python-solution-O(1)-method-Easy-to-understand-with-explanation | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2 or n == 3:
return n - 1 # must break into 1 & (n-1)
num_3 = n // 3 # break the number into (3+3+3+...) + remainder
if n%3 == 0:
return 3**num_3 # (3+3+3+...)
if n%3 == 2:
return (3**num_3) * 2 # (3+3+3+...) + 2
if n%3 == 1:
return 3**(num_3-1) * 4 # (3+3+3+...) + 1 --> (3+3+...) + 3 + 1 -- > (3+3+...) + 4 | integer-break | Python solution -- O(1) method Easy to understand with explanation | byroncharly3 | 4 | 130 | integer break | 343 | 0.554 | Medium | 5,928 |
https://leetcode.com/problems/integer-break/discuss/479923/Python-O(n)-Dynamic-Programming-with-Memoization | class Solution:
def integerBreak(self, n: int) -> int:
memo = {}
def intBreak(n):
if n in memo:
return memo[n]
if n<2:
memo[n] = 0
return 0
if n==2:
memo[2] = 2
return 1
maxval = 0
for i in range(1,n//2+1):
maxval = max(maxval,i*intBreak(n-i),i*(n-i))
memo[n] = maxval
return maxval
ans = intBreak(n)
return ans | integer-break | Python O(n) Dynamic Programming with Memoization | user8307e | 3 | 692 | integer break | 343 | 0.554 | Medium | 5,929 |
https://leetcode.com/problems/integer-break/discuss/2407006/Python3-or-Solved-using-Bottom-Up-DP-%2B-Tabulation | class Solution:
#Time-Complexity: O(n^2)
#Space-Complexity: O(n)
def integerBreak(self, n: int) -> int:
#we know we can reduce n as
# n
# / \
# 1 n-1
# / \
# 1 n-2
# ...
#Basically, we can keep reducing n like this in this tree structure above!
#This is the pattern I recognized! I recognized for given n, there are
#potential sums of (1, n-1), (2, n-2), (3, n-3), ..., (n//2, n//2)!
#For each pair, I can compare the direct number with the max product decomposition
#and take the max of two!
#Reason for comparison: for each of the sum factor of given n, either leave it
#undecomposed or decompose it into further sum factors if the product of sum
#factors produce ultimately a number that exceeds the orignal sum factor! This way
#I am maximing product contribution for each and every sum factor!
#For example, for 5, we decompose it into 2 and 3, since 2*3 > 5, so it will
#maximize our product further!
#However, for 3, we don't decompose since we can maximally decompose to
#1 and 2 but 1*2 < 3!
#Do that for both numbers of each pair and take the product!
#Whatever is largest across the pairs will be answer for given input n!
dp = [-1] * (n+1)
#add dp-base!
dp[1] = 1
#this problem has only one state parameter: the given number to start decomposing #from!
#iterate through each subproblem or state!
#Bottom-Up
for i in range(2, n+1, 1):
upper_bound = (i // 2) + 1
#iterate through all possible pairs!
for j in range(1, upper_bound, 1):
#current pair (j, i-j), which we probably already solved its subproblems!
first = max(j, dp[j])
second = max(i-j, dp[i-j])
#get product for current pair!
sub_ans = first * second
#compare current pair's product against built up answer maximum!
dp[i] = max(dp[i], sub_ans)
#then, once we are done, we can return dp[n]!
return dp[n] | integer-break | Python3 | Solved using Bottom-Up DP + Tabulation | JOON1234 | 2 | 29 | integer break | 343 | 0.554 | Medium | 5,930 |
https://leetcode.com/problems/integer-break/discuss/1136369/O(n)-DP-Python3-bites-97 | class Solution:
def integerBreak(self, n: int) -> int:
dp = [None, None, 1, 2, 4, 6, 9]
for i in range(7, n+1):
dp.append(dp[i-3] * 3)
return dp[n] | integer-break | O(n) - DP - Python3 bites 97% | ShSaktagan | 2 | 167 | integer break | 343 | 0.554 | Medium | 5,931 |
https://leetcode.com/problems/integer-break/discuss/1778956/Python-3-Solution-Using-power-of-3 | class Solution:
def integerBreak(self, n: int) -> int:
if n==2:
return 1
if n==3:
return 2
if n==4:
return 4
x=n%3
a=n//3
if x==0:
return 3**a
if x==1:
return 3**(a-1)*4
return 3**a*2 | integer-break | [Python 3 Solution Using power of 3] | jiteshbhansali | 1 | 81 | integer break | 343 | 0.554 | Medium | 5,932 |
https://leetcode.com/problems/integer-break/discuss/1148577/Python-math-solutiion-O(1) | class Solution:
def integerBreak(self, n: int) -> int:
q, r = divmod(n, 3)
if r == 0:
return 2 if n == 3 else 3**q
elif r == 1:
return 1 if n == 1 else 4*3**(q-1)
else: # r == 2
return 1 if n == 2 else 2*3**q | integer-break | Python, math solutiion O(1) | blue_sky5 | 1 | 136 | integer break | 343 | 0.554 | Medium | 5,933 |
https://leetcode.com/problems/integer-break/discuss/1007487/Python3Golang-Solution-with-using-dp | class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n + 1)
dp[1] = 1
dp[2] = 1
for i in range(3, n + 1):
for j in range(1, i // 2 + 1):
dp[i] = max(dp[i], max((i - j), dp[i - j]) * j)
return dp[-1] | integer-break | [Python3/Golang] Solution with using dp | maosipov11 | 1 | 55 | integer break | 343 | 0.554 | Medium | 5,934 |
https://leetcode.com/problems/integer-break/discuss/2842149/Python-Solution-with-O(1)-time-and-space-complexity | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
remainder = n%3
numberOfThree = n//3
if remainder == 0:
return 3 ** numberOfThree
if remainder == 1:
return (3 ** (numberOfThree-1)) * 4
else:
return (3 ** numberOfThree) * 2 | integer-break | Python Solution with O(1) time and space complexity | jxlau | 0 | 3 | integer break | 343 | 0.554 | Medium | 5,935 |
https://leetcode.com/problems/integer-break/discuss/2817001/intuitive-sub-optimal-solution-from-someone-who-is-terrible-at-math | class Solution:
def integerBreak(self, n: int) -> int:
# to max product, try to break into k similar-sized pieces
def build_pieces( k ):
if k < 2:
return None
# k >= 2
div, mod = divmod(n,k)
pieces = [div] * k
while mod >0:
pieces[mod] += 1 # spread the residule evenly
mod -= 1
return pieces
def product( nums ):
ans = 1
for num in nums:
ans *= num
return ans
ans = 0
for k in range(2, n+1):
pieces = build_pieces( k ) # for debug purpose
ans = max(ans, product( pieces ) )
return ans | integer-break | intuitive, sub-optimal solution from someone who is terrible at math | SkinheadBob | 0 | 2 | integer break | 343 | 0.554 | Medium | 5,936 |
https://leetcode.com/problems/integer-break/discuss/2776999/Simple-Python-Solution-or-Math | class Solution:
def integerBreak(self, n: int) -> int:
three = 0
if n<4: return n-1
while n>4:
n-=3
three+=1
return 3**three*n | integer-break | Simple Python Solution | Math | RajatGanguly | 0 | 6 | integer break | 343 | 0.554 | Medium | 5,937 |
https://leetcode.com/problems/integer-break/discuss/2740476/Python-solution | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
if n % 3 == 0:
return pow(3, n//3)
else:
r3 = 0
while n > 4:
r3 += 1
n -= 3
return pow(3, r3) * n | integer-break | Python solution | maomao1010 | 0 | 5 | integer break | 343 | 0.554 | Medium | 5,938 |
https://leetcode.com/problems/integer-break/discuss/2733624/Very-short-dynamic-programming-solution-beat-99 | class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] + [1] * n
for i in range(3,n+1):
dp[i] = max(max(i-2,dp[i-2]) * 2, max(i-3,dp[i-3]) * 3)
return dp[n] | integer-break | Very short dynamic programming solution, beat 99% | zhaoqiang | 0 | 7 | integer break | 343 | 0.554 | Medium | 5,939 |
https://leetcode.com/problems/integer-break/discuss/2691100/python-working-solution-or-dp | class Solution:
def integerBreak(self, n: int) -> int:
dp = {}
def dfs(n):
if n == 1:
return 1
if n in dp:
return dp[n]
ans = float('-inf')
for l in range(1,n):
r = n-l
left,right = dfs(l),dfs(r)
ans = max(ans ,left * r , l * right ,r*l)
dp[n] = ans
return ans
return dfs(n) | integer-break | python working solution | dp | Sayyad-Abdul-Latif | 0 | 10 | integer break | 343 | 0.554 | Medium | 5,940 |
https://leetcode.com/problems/integer-break/discuss/2682681/Easy-to-Understand.-Not-the-best-solution-I-feel.-Faster-than-50 | class Solution:
def integerBreak(self, n: int) -> int:
def dfs(a):
if a == 1:
return 1
max_res = 1
for right in range(1, a):
left = a - right
max_res = max(max_res, left * dfs(right), left * right)
return max_res
return dfs(n) | integer-break | Easy to Understand. Not the best solution I feel. Faster than 50% | devanshsolani30 | 0 | 18 | integer break | 343 | 0.554 | Medium | 5,941 |
https://leetcode.com/problems/integer-break/discuss/2662453/python-easy-dp-solution | class Solution:
def integerBreak(self, n: int) -> int:
dp = [0]*(n+1)
dp[1] = 1
for i in range(2,n+1):
for j in range(1,i):
dp[i] =max(dp[i],j*(i-j),j*dp[i-j])
return dp[n] | integer-break | python easy dp solution | Akash_chavan | 0 | 13 | integer break | 343 | 0.554 | Medium | 5,942 |
https://leetcode.com/problems/integer-break/discuss/2657315/Simple-1D-Bottom-Up-DP | class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n+1)
for i in range(1, n+1):
for j in range(1, i):
dp[i] = max(dp[i], j * (i-j), j * dp[i-j])
return dp[-1] | integer-break | Simple 1D Bottom-Up DP | jonathanbrophy47 | 0 | 6 | integer break | 343 | 0.554 | Medium | 5,943 |
https://leetcode.com/problems/integer-break/discuss/2296772/Python3-or-Memoization-or-Recursion | class Solution:
def integerBreak(self, n: int) -> int:
@lru_cache(None)
def recursive(n):
if n==1:
return 1
max_prod=0
for index in range(1,n):
max_prod=max(max_prod,index*recursive(n-index))
max_prod=max(max_prod,index*(n-index))
return max_prod
return recursive(n) | integer-break | Python3 | Memoization | Recursion | Sefinehtesfa34 | 0 | 19 | integer break | 343 | 0.554 | Medium | 5,944 |
https://leetcode.com/problems/integer-break/discuss/1642913/Liner-solution-90-speed | class Solution:
def integerBreak(self, n: int) -> int:
max_prod = 1
for k in range(2, n // 2 + 2):
num, rem = divmod(n, k)
max_prod = max(max_prod, pow(num, k - rem) * pow(num + 1, rem))
return max_prod | integer-break | Liner solution, 90% speed | EvgenySH | 0 | 92 | integer break | 343 | 0.554 | Medium | 5,945 |
https://leetcode.com/problems/integer-break/discuss/1593305/Python-Straightforward-Solution-or-Pure-Easy-Mathematics | class Solution:
def integerBreak(self, n: int) -> int:
if n <= 3:
return n-1
three = n//3
rem = n % 3
if rem == 0:
return 3**three
elif rem == 1:
return (3**(three-1))*4
else:
return (3**three)*2
# 6 = 3x3 = 9
# 7 = 3x4 = 12
# 8 = 3x3x2 = 18
# 9 = 3x3x3 = 27
# 10 = 3x3x4 = 36
# 54 = 3 ^ 18 = 387420489
# 55 = (3 ^ 17)*4 = 516560652
# 56 = (3 ^ 18)*2 = 774840978 | integer-break | Python Straightforward Solution | Pure Easy Mathematics | leet_satyam | 0 | 94 | integer break | 343 | 0.554 | Medium | 5,946 |
https://leetcode.com/problems/integer-break/discuss/1498409/we-only-need-2-and-3. | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2: return 1
if n == 3: return 2
num3 = n//3
maxv = 1
for i in range(num3+1):
res = n - 3*i
if res %2 == 0:
num2 = res//2
maxv = max(maxv, 2**num2*3**i)
return maxv | integer-break | we only need 2 and 3. | byuns9334 | 0 | 103 | integer break | 343 | 0.554 | Medium | 5,947 |
https://leetcode.com/problems/integer-break/discuss/1492277/Python-solution-using-DP | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
# 1
dp = [sys.maxsize] * (n+1)
# 2
dp[0] = 0
dp[1] = 1
dp[2] = 2
# 3
for i in range(3,n+1):
# 4
# Generate all combos that add up to i
# Figure out what is bigger: i, or max((a x dp[b]) for (a,b) in combos)
max_combo = max([(j * dp[i-j]) for j in range(1,i)])
# If i is bigger than the max_combo at i, just use i (this only applies to 3 and 4)
dp[i] = max(i, max_combo)
return dp[n] | integer-break | Python solution using DP | mikewillard87 | 0 | 46 | integer break | 343 | 0.554 | Medium | 5,948 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math | class Solution:
def integerBreak(self, n: int) -> int:
@lru_cache(None)
def fn(n):
"""Return the max product by splitting n."""
if n == 1: return 1
return max(max(i, fn(i))*max(n-i, fn(n-i)) for i in range(1, n//2+1))
return fn(n) | integer-break | [Python3] dp & math | ye15 | 0 | 91 | integer break | 343 | 0.554 | Medium | 5,949 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math | class Solution:
def integerBreak(self, n: int) -> int:
ans = [None, 1] + [0]*(n-1)
for k in range(2, n+1):
for i in range(1, k//2+1):
ans[k] = max(ans[k], max(i, ans[i])*max(k-i, ans[k-i]))
return ans[-1] | integer-break | [Python3] dp & math | ye15 | 0 | 91 | integer break | 343 | 0.554 | Medium | 5,950 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math | class Solution:
def integerBreak(self, n: int) -> int:
if n == 2: return 1
if n == 3: return 2
ans = 1
while n > 4:
ans *= 3
n -= 3
return ans*n | integer-break | [Python3] dp & math | ye15 | 0 | 91 | integer break | 343 | 0.554 | Medium | 5,951 |
https://leetcode.com/problems/integer-break/discuss/457138/python-DP-solution-with-explanation-80-faster | class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n+3) # extra buffer
dp[0] = 0
dp[1] = 1
dp[2] = 1
if n <= 2: return dp[n]
for i in range(3, n+1):
for j in range(1, i // 2 + 1):
dp[i] = max(
dp[i],
j * dp[i-j],
j * (i-j)
)
return dp[n] | integer-break | python DP solution with explanation 80% faster | sokketsu | 0 | 141 | integer break | 343 | 0.554 | Medium | 5,952 |
https://leetcode.com/problems/integer-break/discuss/362181/Solution-in-Python-3-(beats-100)-(one-line)-(direct-solution-using-Calculus) | class Solution:
def integerBreak(self, n: int) -> int:
return 3**(n//3-((n-3*(n//3))<=1))*((n-3*(n//3))+3*((n-3*(n//3))<=1)) if n>3 else n-1 | integer-break | Solution in Python 3 (beats 100%) (one line) (direct solution using Calculus) | junaidmansuri | 0 | 241 | integer break | 343 | 0.554 | Medium | 5,953 |
https://leetcode.com/problems/reverse-string/discuss/670137/Python-3-~actually~-easiest-solution | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | reverse-string | Python 3 ~actually~ easiest solution | drblessing | 77 | 6,800 | reverse string | 344 | 0.762 | Easy | 5,954 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
s.reverse() | reverse-string | Python and Java - One Line, Iterative, Recursive Solutions | Pootato | 8 | 554 | reverse string | 344 | 0.762 | Easy | 5,955 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
N = len(s)
for i in range(N // 2):
s[i], s[-i-1] == s[-i-1], s[i] | reverse-string | Python and Java - One Line, Iterative, Recursive Solutions | Pootato | 8 | 554 | reverse string | 344 | 0.762 | Easy | 5,956 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions | class Solution:
def reverse(self, l, r, s):
if l >= r:
return
s[l], s[r] = s[r], s[l]
self.reverse(l+1, r-1, s)
def reverseString(self, s: List[str]) -> None:
self.reverse(0, len(s)-1, s) | reverse-string | Python and Java - One Line, Iterative, Recursive Solutions | Pootato | 8 | 554 | reverse string | 344 | 0.762 | Easy | 5,957 |
https://leetcode.com/problems/reverse-string/discuss/946287/Python-Easy-Solution | class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(len(s)//2):
s[i],s[-i-1]=s[-i-1],s[i]
return s | reverse-string | Python Easy Solution | lokeshsenthilkumar | 8 | 1,400 | reverse string | 344 | 0.762 | Easy | 5,958 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner | class Solution:
def reverseString(self, s: List[str]) -> None:
lo, hi = 0, len(s) - 1
while lo < hi:
s[lo], s[hi] = s[hi], s[lo]
lo += 1
hi -= 1 | reverse-string | ✅ Python Easy Solutions | One Liner | dhananjay79 | 6 | 952 | reverse string | 344 | 0.762 | Easy | 5,959 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner | class Solution:
def reverseString(self, s: List[str]) -> None:
def reverse(lo,hi):
if lo > hi: return
s[lo], s[hi] = s[hi], s[lo]
reverse(lo+1, hi-1)
reverse(0, len(s)-1) | reverse-string | ✅ Python Easy Solutions | One Liner | dhananjay79 | 6 | 952 | reverse string | 344 | 0.762 | Easy | 5,960 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner | class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(1,len(s) // 2 + 1): s[i-1], s[-i] = s[-i], s[i-1] | reverse-string | ✅ Python Easy Solutions | One Liner | dhananjay79 | 6 | 952 | reverse string | 344 | 0.762 | Easy | 5,961 |
https://leetcode.com/problems/reverse-string/discuss/2403620/Pythonoror4-methodoror-beginner-friendlyororeasyto-understandororvery-simple | class Solution:
def reverseString(self, s: List[str]) -> None:
i=0
j=len(s)-1
def rev(s,i,j):
if i>=j:
return
s[i],s[j]=s[j],s[i]
rev(s,i+1,j-1)
rev(s,i,j) | reverse-string | Python||4 method|| beginner-friendly||easyto understand||very simple | shikhar_srivastava391 | 4 | 184 | reverse string | 344 | 0.762 | Easy | 5,962 |
https://leetcode.com/problems/reverse-string/discuss/2403620/Pythonoror4-methodoror-beginner-friendlyororeasyto-understandororvery-simple | class Solution:
def reverseString(self, s: List[str]) -> None:
size = len(s)
for i in range(size//2):
s[i], s[-i-1] = s[-i-1], s[i] | reverse-string | Python||4 method|| beginner-friendly||easyto understand||very simple | shikhar_srivastava391 | 4 | 184 | reverse string | 344 | 0.762 | Easy | 5,963 |
https://leetcode.com/problems/reverse-string/discuss/1764369/Python-3-(200ms)-or-Two-Pointers-Approach-or-2-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
st,e=0,len(s)-1
while st<=e:
s[st],s[e]=s[e],s[st]
st+=1
e-=1 | reverse-string | Python 3 (200ms) | Two Pointers Approach | 2 Solutions | MrShobhit | 4 | 338 | reverse string | 344 | 0.762 | Easy | 5,964 |
https://leetcode.com/problems/reverse-string/discuss/1764369/Python-3-(200ms)-or-Two-Pointers-Approach-or-2-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(len(s)//2): s[i], s[-i-1] = s[-i-1], s[i] | reverse-string | Python 3 (200ms) | Two Pointers Approach | 2 Solutions | MrShobhit | 4 | 338 | reverse string | 344 | 0.762 | Easy | 5,965 |
https://leetcode.com/problems/reverse-string/discuss/1264496/Easy-Python-Solution | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(int(len(s)/2)):
s[i],s[len(s)-i-1]=s[len(s)-i-1],s[i] | reverse-string | Easy Python Solution | Sneh17029 | 3 | 1,000 | reverse string | 344 | 0.762 | Easy | 5,966 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods) | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse() | reverse-string | Simple python solutions (3 different methods) | tylerpruitt | 2 | 143 | reverse string | 344 | 0.762 | Easy | 5,967 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods) | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[-1::-1] | reverse-string | Simple python solutions (3 different methods) | tylerpruitt | 2 | 143 | reverse string | 344 | 0.762 | Easy | 5,968 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods) | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# First pointer (start of array)
i = 0
# Second pointer (end of array)
j = len(s) - 1
while i < j:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1 | reverse-string | Simple python solutions (3 different methods) | tylerpruitt | 2 | 143 | reverse string | 344 | 0.762 | Easy | 5,969 |
https://leetcode.com/problems/reverse-string/discuss/2346952/3-SOLUTIONS-oror-SINGLE-LINE-PYTHON-oror-oror-Beginner-friendly | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | reverse-string | 3 SOLUTIONS || SINGLE LINE PYTHON || ✔✔|| Beginner friendly | adithya_s_k | 1 | 100 | reverse string | 344 | 0.762 | Easy | 5,970 |
https://leetcode.com/problems/reverse-string/discuss/2346952/3-SOLUTIONS-oror-SINGLE-LINE-PYTHON-oror-oror-Beginner-friendly | class Solution(object):
def reverseString(self, s):
l = len(s)
if l < 2:
return s
return self.reverseString(s[l/2:]) + self.reverseString(s[:l/2])
**other two general solutions**
class SolutionClassic(object):
def reverseString(self, s):
r = list(s)
i, j = 0, len(r) - 1
while i < j:
r[i], r[j] = r[j], r[i]
i += 1
j -= 1
return "".join(r) | reverse-string | 3 SOLUTIONS || SINGLE LINE PYTHON || ✔✔|| Beginner friendly | adithya_s_k | 1 | 100 | reverse string | 344 | 0.762 | Easy | 5,971 |
https://leetcode.com/problems/reverse-string/discuss/2308497/Python-2-Easy-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
left = 0
right = len(s) - 1
while left <= right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
return s | reverse-string | ✅Python 2 Easy Solutions | Skiper228 | 1 | 121 | reverse string | 344 | 0.762 | Easy | 5,972 |
https://leetcode.com/problems/reverse-string/discuss/2308497/Python-2-Easy-Solutions | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | reverse-string | ✅Python 2 Easy Solutions | Skiper228 | 1 | 121 | reverse string | 344 | 0.762 | Easy | 5,973 |
https://leetcode.com/problems/reverse-string/discuss/2253500/O(n)-Runtime%3A-185-ms-faster-than-99.86-of-Python3-online-submissions-for-Reverse-String. | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
start=0
end = len(s)-1
while start<end:
s[start],s[end] = s[end], s[start]
start = start+1
end = end-1 | reverse-string | O(n) Runtime: 185 ms, faster than 99.86% of Python3 online submissions for Reverse String. | grishptl | 1 | 105 | reverse string | 344 | 0.762 | Easy | 5,974 |
https://leetcode.com/problems/reverse-string/discuss/2134039/2-Approaches.-Easy-to-Understand-in-Python. | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:]=s[::-1] | reverse-string | 2 Approaches. Easy to Understand in Python. | AY_ | 1 | 103 | reverse string | 344 | 0.762 | Easy | 5,975 |
https://leetcode.com/problems/reverse-string/discuss/2134039/2-Approaches.-Easy-to-Understand-in-Python. | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
sLen = len(s)
j = sLen-1
for i in range(sLen//2):
s[i],s[j] = s[j],s[i]
j -= 1 | reverse-string | 2 Approaches. Easy to Understand in Python. | AY_ | 1 | 103 | reverse string | 344 | 0.762 | Easy | 5,976 |
https://leetcode.com/problems/reverse-string/discuss/2131067/Python3-Two-pointer-Solution-faster-than-90.31-easily-explained. | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# so we use two pointers; left at 0 index and right at last index (len(numbers)-1)
# then in a while loop we swap our left index value to the right index value
# in one liner. That's it.
left, right = 0, len(s)-1
while left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
# return s | reverse-string | Python3 Two-pointer Solution; faster than 90.31%, easily explained. | shubhamdraj | 1 | 56 | reverse string | 344 | 0.762 | Easy | 5,977 |
https://leetcode.com/problems/reverse-string/discuss/2098031/Python3-1-optimal-solution-1-brute-force-2-builtIn | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
return self.reverseStringOptimal(s)
# O(n) || O(1)
# runtime: 342 19.09%
def reverseStringOptimal(self, string):
if not string: return string
left, right = 0, len(string) - 1
while left < right:
string[left], string[right] = string[right], string[left]
left += 1
right -= 1
return string
# O(n) || O(n)
# brute force
def reverseStringWithNewList(self, string):
if not string:
return string
newList = [0] * len(string)
j = 0
for i in reversed(range(len(string))):
newList[i] = string[j]
j += 1
return newList
# below are just 'some' python built in
def reverseStringWithListCompression(self, string):
if not string: return string
return [string[i] for i in reversed(range(len(string)))]
def reversedStringWithReverse(self, string):
string.reverse()
return string or string[::-1] | reverse-string | Python3 1 optimal solution, 1 brute force, 2 builtIn | arshergon | 1 | 69 | reverse string | 344 | 0.762 | Easy | 5,978 |
https://leetcode.com/problems/reverse-string/discuss/1787196/Dumb-python-solution-oror-faster-than-93-with-space-less-than-99 | class Solution:
def reverseString(self, s: List[str]) -> None:
for i, v in enumerate(s[::-1]):
s[i] = v | reverse-string | Dumb python solution || faster than 93% with space less than 99% | shafzgamer | 1 | 239 | reverse string | 344 | 0.762 | Easy | 5,979 |
https://leetcode.com/problems/reverse-string/discuss/1751782/Python-Simple-and-Clean-Python-Solution-By-Swapping-First-and-Last-Number | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
length = len(s)
for i in range(length//2):
s[i],s[-(i+1)]=s[-(i+1)],s[i] | reverse-string | [ Python ] ✔✔ Simple and Clean Python Solution By Swapping First and Last Number | ASHOK_KUMAR_MEGHVANSHI | 1 | 103 | reverse string | 344 | 0.762 | Easy | 5,980 |
https://leetcode.com/problems/reverse-string/discuss/1736515/Python-93.24-faster-ororSimplest-solution-with-explanationoror-Beg-to-Advoror-Two-Pointer | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse()
return s | reverse-string | Python 93.24% faster ||Simplest solution with explanation|| Beg to Adv|| Two Pointer | rlakshay14 | 1 | 152 | reverse string | 344 | 0.762 | Easy | 5,981 |
https://leetcode.com/problems/reverse-string/discuss/1736515/Python-93.24-faster-ororSimplest-solution-with-explanationoror-Beg-to-Advoror-Two-Pointer | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
left, right = 0, len(s) - 1
while left < right:
#swaping the positions of the elements.
s[left], s[right] = s[right], s[left]
# incrementing both the pointers.
right -= 1
left += 1 | reverse-string | Python 93.24% faster ||Simplest solution with explanation|| Beg to Adv|| Two Pointer | rlakshay14 | 1 | 152 | reverse string | 344 | 0.762 | Easy | 5,982 |
https://leetcode.com/problems/reverse-string/discuss/1621867/faster-than-82.25-of-Python3-online-submissions-for-Reverse-String. | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
i, j = 0, len(s) - 1
while i != j and i < j:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1 | reverse-string | faster than 82.25% of Python3 online submissions for Reverse String. | sagarhasan273 | 1 | 46 | reverse string | 344 | 0.762 | Easy | 5,983 |
https://leetcode.com/problems/reverse-string/discuss/1559714/Python-two-pointers-O(n)-time-O(1)-space-beats-85-explained | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# two pointers, one at the beginning, the other at the end
# swap, unti you meet at center
# O(N) time, O(1) space
left_pointer, right_pointer = 0, len(s) - 1
while left_pointer < right_pointer:
s[left_pointer], s[right_pointer] = s[right_pointer], s[left_pointer]
left_pointer += 1
right_pointer -= 1
class Solution2:
def reverseString(self, s:List[str]) -> None:
# for loop + bitwise inversion might be faser, but while loop makes it more readable
for l in range(len(s) // 2):
s[l], s[~l] = s[~l], s[l] | reverse-string | [Python] two-pointers O(n) time, O(1) space, beats 85%, explained | mateoruiz5171 | 1 | 135 | reverse string | 344 | 0.762 | Easy | 5,984 |
https://leetcode.com/problems/reverse-string/discuss/1388002/Python3-faster-than-95.34 | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
l = len(s)
for i in range(l//2):
s[i], s[l-i-1] = s[l-i-1],s[i]
return | reverse-string | Python3 faster than 95.34% | saumyasuvarna | 1 | 258 | reverse string | 344 | 0.762 | Easy | 5,985 |
https://leetcode.com/problems/reverse-string/discuss/1259231/Python3-dollarolution | class Solution:
def reverseString(self, s: List[str]) -> None:
i, j = 0, len(s)-1
while i < j:
temp = s[i]
s[i] = s[j]
s[j] = temp
i += 1
j -= 1 | reverse-string | Python3 $olution | AakRay | 1 | 296 | reverse string | 344 | 0.762 | Easy | 5,986 |
https://leetcode.com/problems/reverse-string/discuss/670009/Python-one-liner-solution | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s=s.reverse() | reverse-string | Python one liner solution | rajesh_26 | 1 | 80 | reverse string | 344 | 0.762 | Easy | 5,987 |
https://leetcode.com/problems/reverse-string/discuss/669559/simplest-solution-for-reverse-string-python3 | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | reverse-string | simplest solution for reverse string [python3] | user4410vu | 1 | 118 | reverse string | 344 | 0.762 | Easy | 5,988 |
https://leetcode.com/problems/reverse-string/discuss/291391/Python-basic-recursive-and-iterative-formula | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(len(s) // 2):
s[i], s[len(s) - i - 1] = s[len(s) - i - 1], s[i] | reverse-string | Python -- basic recursive and iterative formula | CorvusEtiam | 1 | 293 | reverse string | 344 | 0.762 | Easy | 5,989 |
https://leetcode.com/problems/reverse-string/discuss/291391/Python-basic-recursive-and-iterative-formula | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
return self.loop(s, 0, len(s) // 2)
def loop(self, s, i, final):
if i == final:
return
else:
s[i], s[len(s) - i - 1] = s[len(s) - i - 1], s[i]
return self.loop(s, i + 1, final) | reverse-string | Python -- basic recursive and iterative formula | CorvusEtiam | 1 | 293 | reverse string | 344 | 0.762 | Easy | 5,990 |
https://leetcode.com/problems/reverse-string/discuss/2847979/Quick-Python-solution-to-reverse-string | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(0,len(s)):
s.insert(i,s.pop())
print(s)
sol=Solution()
sol.reverseString(["h","e","l","l","o"]) | reverse-string | Quick Python solution to reverse string | dassdipanwita | 0 | 1 | reverse string | 344 | 0.762 | Easy | 5,991 |
https://leetcode.com/problems/reverse-string/discuss/2846086/python3-oror-in-place-o(1)-memory | class Solution:
def reverseString(self, s: List[str]) -> None:
l, r = 0, len(s) - 1 # left and right pointer
while(l < r):
# switch chars at l and r
s[l], s[r] = s[r], s[l]
l += 1
r -= 1 | reverse-string | python3 || in-place, o(1) memory | wduf | 0 | 2 | reverse string | 344 | 0.762 | Easy | 5,992 |
https://leetcode.com/problems/reverse-string/discuss/2842646/Python-solution. | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
l = 0
h = len(s) - 1
for i in range(len(s) // 2):
s[i], s[h] = s[h], s[l]
l += 1
h -= 1 | reverse-string | Python solution. | ahti1405 | 0 | 2 | reverse string | 344 | 0.762 | Easy | 5,993 |
https://leetcode.com/problems/reverse-string/discuss/2840063/Reverse-String-or-Python-3.x-or-Two-pointers | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
head = -1
tail = len(s) - 1
while head != tail:
tmp = s.pop(tail)
tail -= 1
s.append(tmp) | reverse-string | Reverse String | Python 3.x | Two pointers | SnLn | 0 | 1 | reverse string | 344 | 0.762 | Easy | 5,994 |
https://leetcode.com/problems/reverse-string/discuss/2830609/Python-Stack-Two-Pointers | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
stack = []
l, r = 0, len(s) - 1
stack.append((l, r))
while stack and l < r:
if stack:
l, r = stack.pop()
s[l], s[r] = s[r], s[l]
l += 1
r -= 1
stack.append((l , r)) | reverse-string | Python Stack Two Pointers | user6168bh | 0 | 1 | reverse string | 344 | 0.762 | Easy | 5,995 |
https://leetcode.com/problems/reverse-string/discuss/2819025/Simple-and-Fast-Python-Solution-One-Liner | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse() | reverse-string | Simple and Fast Python Solution - One-Liner | PranavBhatt | 0 | 5 | reverse string | 344 | 0.762 | Easy | 5,996 |
https://leetcode.com/problems/reverse-string/discuss/2816647/list-reverse | class Solution:
def reverseString(self, s: List[str]) -> None:
s2 = s.copy()
lens = len(s)
for i in range(lens):
s[i] = s2.pop() | reverse-string | list-reverse | dexck7770 | 0 | 1 | reverse string | 344 | 0.762 | Easy | 5,997 |
https://leetcode.com/problems/reverse-string/discuss/2815904/Simple-Python-using-2-Pointers | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
first, last = 0, len(s)-1
while first < last:
s[first],s[last] = s[last],s[first]
first +=1
last -= 1
return s | reverse-string | Simple Python using 2 Pointers | BhavyaBusireddy | 0 | 2 | reverse string | 344 | 0.762 | Easy | 5,998 |
https://leetcode.com/problems/reverse-string/discuss/2813967/TC-%3A-95.40-Simple-solution | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-1] | reverse-string | 😎 TC : 95.40 % Simple solution | Pragadeeshwaran_Pasupathi | 0 | 3 | reverse string | 344 | 0.762 | Easy | 5,999 |
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