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https://leetcode.com/problems/find-the-difference/discuss/1969523/Python3-runtime%3A-48ms-52.48-memory%3A-13.9mb-33.89 | class Solution:
def findTheDifference(self, string: str, target: str) -> str:
if not string and not target:
return True
if not string:
return target
if not target:
return False
visited = [0] * 26
for i in range(len(string)):
idx = ord(string[i]) - ord('a')
visited[idx] += 1
for i in range(len(target)):
idx = ord(target[i]) - ord('a')
if visited[idx] == 0:
return target[i]
visited[idx] -= 1
return True | find-the-difference | Python3 runtime: 48ms 52.48% memory: 13.9mb 33.89% | arshergon | 1 | 47 | find the difference | 389 | 0.603 | Easy | 6,700 |
https://leetcode.com/problems/find-the-difference/discuss/1920036/Easy-simple-python-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
d= defaultdict(int)
for i in s:
d[i]+=1
for j in t:
d[j]+=1
for key in d:
if d[key] %2==1:
return key | find-the-difference | Easy simple python solution | Buyanjargal | 1 | 84 | find the difference | 389 | 0.603 | Easy | 6,701 |
https://leetcode.com/problems/find-the-difference/discuss/1911593/Simple-Dictionary-oror-No-XOR-oror-No-Sorting-oror-Full-Explanation-with-Example | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
d = Counter(t) #stores the frequency of t
for i in s:
d[i] -= 1 #every time we decrement the frequency of dictionary which stores the frequency of t
if d[i] == 0:
del d[i] #if the frequency becomes zero delete that key from dictionary
a = str(d.keys())
#print(a) --->"dict_keys(['e'])"
return a[12] | find-the-difference | ✅Simple Dictionary || No XOR || No Sorting || Full Explanation with Example | Dev_Kesarwani | 1 | 57 | find the difference | 389 | 0.603 | Easy | 6,702 |
https://leetcode.com/problems/find-the-difference/discuss/1752397/Python-Curated-List-of-cool-oneliners | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return chr(reduce(xor, map(ord, chain(s, t)))) | find-the-difference | [Python] Curated List of cool oneliners | sidheshwar_s | 1 | 39 | find the difference | 389 | 0.603 | Easy | 6,703 |
https://leetcode.com/problems/find-the-difference/discuss/1752397/Python-Curated-List-of-cool-oneliners | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return chr(reduce(lambda x, y: x ^ y, map(ord, s + t))) | find-the-difference | [Python] Curated List of cool oneliners | sidheshwar_s | 1 | 39 | find the difference | 389 | 0.603 | Easy | 6,704 |
https://leetcode.com/problems/find-the-difference/discuss/1752397/Python-Curated-List-of-cool-oneliners | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return list(Counter(t) - Counter(s))[0] | find-the-difference | [Python] Curated List of cool oneliners | sidheshwar_s | 1 | 39 | find the difference | 389 | 0.603 | Easy | 6,705 |
https://leetcode.com/problems/find-the-difference/discuss/1752397/Python-Curated-List-of-cool-oneliners | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return next(iter(Counter(t) - Counter(s))) | find-the-difference | [Python] Curated List of cool oneliners | sidheshwar_s | 1 | 39 | find the difference | 389 | 0.603 | Easy | 6,706 |
https://leetcode.com/problems/find-the-difference/discuss/1752397/Python-Curated-List-of-cool-oneliners | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return "".join(Counter(t) - Counter(s)) | find-the-difference | [Python] Curated List of cool oneliners | sidheshwar_s | 1 | 39 | find the difference | 389 | 0.603 | Easy | 6,707 |
https://leetcode.com/problems/find-the-difference/discuss/1751858/Python-Easy-1-line-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return chr(sum([ord(c) for c in t]) - sum([ord(c) for c in s])) | find-the-difference | [Python] Easy 1 line solution | nomofika | 1 | 155 | find the difference | 389 | 0.603 | Easy | 6,708 |
https://leetcode.com/problems/find-the-difference/discuss/1751694/Python-oror-Easy-Solution-oror-96-Faster-and-96-Space-Efficient | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
for i in set(t):
if s.count(i) != t.count(i):
return i | find-the-difference | Python || Easy Solution || 96% Faster and 96% Space Efficient | cherrysri1997 | 1 | 19 | find the difference | 389 | 0.603 | Easy | 6,709 |
https://leetcode.com/problems/find-the-difference/discuss/1751193/python-easy-to-understand | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
a = []
for i in range(len(t)):
a.append(t[i])
for i in range(len(s)):
a.remove(s[i])
ans = a[0]
return ans | find-the-difference | python easy to understand | ggeeoorrggee | 1 | 25 | find the difference | 389 | 0.603 | Easy | 6,710 |
https://leetcode.com/problems/find-the-difference/discuss/1591194/Python-faster-than-87.94 | class Solution(object):
def findTheDifference(self, s, t):
"""
:type s: str
:type t: str
:rtype: str
"""
for c in t:
if s.count(c) != t.count(c):
return c | find-the-difference | Python faster than 87.94% | ckayfok | 1 | 92 | find the difference | 389 | 0.603 | Easy | 6,711 |
https://leetcode.com/problems/find-the-difference/discuss/1260511/Python3-dollarolution-(95-Faster) | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
s = Counter(s)
t = Counter(t)
for i in (t-s):
return i | find-the-difference | Python3 $olution (95% Faster) | AakRay | 1 | 316 | find the difference | 389 | 0.603 | Easy | 6,712 |
https://leetcode.com/problems/find-the-difference/discuss/1062566/Python-very-simple-solution-with-Counter | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
diff = Counter(t) - Counter(s)
return list(diff.keys())[0] | find-the-difference | Python - very simple solution with Counter | angelique_ | 1 | 71 | find the difference | 389 | 0.603 | Easy | 6,713 |
https://leetcode.com/problems/find-the-difference/discuss/249886/Solution-using-Counter | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
from collections import Counter
s_counter = Counter(s)
t_counter = Counter(t)
return list(t_counter - s_counter)[0] | find-the-difference | Solution using Counter | compbuzz09 | 1 | 73 | find the difference | 389 | 0.603 | Easy | 6,714 |
https://leetcode.com/problems/find-the-difference/discuss/2848549/python3 | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
# for each unique char in t
for c in set(t):
# if # of char in t > # of char in s, it was added to t
if t.count(c) > s.count(c):
return c | find-the-difference | python3 | wduf | 0 | 1 | find the difference | 389 | 0.603 | Easy | 6,715 |
https://leetcode.com/problems/find-the-difference/discuss/2846864/Easist-Solution-in-Python | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
set_t = set(t)
for char in set_t:
if t.count(char) != s.count(char):
return char | find-the-difference | Easist Solution in Python | namashin | 0 | 1 | find the difference | 389 | 0.603 | Easy | 6,716 |
https://leetcode.com/problems/find-the-difference/discuss/2846432/bit-manip-python3-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
res = 0
for i in s+t:
res^=ord(i)
return chr(res) | find-the-difference | bit manip python3 solution | Cosmodude | 0 | 1 | find the difference | 389 | 0.603 | Easy | 6,717 |
https://leetcode.com/problems/find-the-difference/discuss/2834982/Python-simple-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
list1 = [x for x in t]
list2 = [x for x in s]
list1.sort()
list2.sort()
list2.append(0)
print(list1)
print(list2)
reslist = [list1[i] for i in range(len(list1)) if list1[i] != list2[i] ]
res = ''.join(reslist)
print (res)
return reslist[0] | find-the-difference | Python simple solution | user1079z | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,718 |
https://leetcode.com/problems/find-the-difference/discuss/2834528/Python-solution-oror-O(n)-time-oror-O(n)-space | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
d = {}
for i in s:
try:
d[i] += 1
except:
d[i] = 1
for i in t:
try:
d[i] -= 1
if d[i] < 0:
return i
except:
return i | find-the-difference | Python solution || O(n) time || O(n) space | Pavel_Kos | 0 | 2 | find the difference | 389 | 0.603 | Easy | 6,719 |
https://leetcode.com/problems/find-the-difference/discuss/2832365/dictionary-solution-python | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
dict1={}
dict2 = {}
for x in s:
if x not in dict1:
dict1[x]=1
else:
dict1[x]=dict1[x]+1
for x in t:
if x not in dict2:
dict2[x]=1
else:
dict2[x]=dict2[x]+1
print(dict1)
print(dict2)
for x in dict2:
if x not in dict1:
return x
else:
if dict2[x]!=dict1[x]:
return x | find-the-difference | dictionary solution python | sahityasetu1996 | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,720 |
https://leetcode.com/problems/find-the-difference/discuss/2831104/Better-one-line-python-solution-O(n) | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return chr(sum([ord(l) for l in t]) - sum([ord(l) for l in s])) | find-the-difference | Better one line python solution O(n) | lornedot | 0 | 1 | find the difference | 389 | 0.603 | Easy | 6,721 |
https://leetcode.com/problems/find-the-difference/discuss/2831092/One-line-python-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return chr(abs(sum([ord(l) for l in s]) + sum([-1* ord(l) for l in t]))) | find-the-difference | One line python solution | lornedot | 0 | 2 | find the difference | 389 | 0.603 | Easy | 6,722 |
https://leetcode.com/problems/find-the-difference/discuss/2829324/Easy-python-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
for i in set(t):
if s.count(i) != t.count(i):
return i | find-the-difference | Easy python solution | _debanjan_10 | 0 | 1 | find the difference | 389 | 0.603 | Easy | 6,723 |
https://leetcode.com/problems/find-the-difference/discuss/2826203/Easy-Python-Solution-Using-HashMap-or-dict | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
final_word = s + t
hashMap = {}
for i in final_word:
hashMap[i] = final_word.count(i)
print(hashMap)
for i in hashMap:
if hashMap[i] % 2 != 0:
return i | find-the-difference | Easy Python Solution - Using HashMap or dict | danishs | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,724 |
https://leetcode.com/problems/find-the-difference/discuss/2820287/simple-python-using-dictionary-beats-75 | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
string = "abcdefghijklmnopqrstuvwxyz"
hashmap = {string[k]:0 for k in range(len(string))}
for i in range(len(s)):
hashmap[s[i]]+=1
for i in range(len(t)):
hashmap[t[i]]+=1
for k,v in hashmap.items():
if v%2:
return k | find-the-difference | simple python using dictionary beats 75% | sudharsan1000m | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,725 |
https://leetcode.com/problems/find-the-difference/discuss/2795178/Python-Solution-(With-basics-concepts) | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
s1 = []
t1 = []
alpha = string.ascii_lowercase
for i in alpha:
s1.append(s.count(i))
t1.append(t.count(i))
for i in range(0,26):
if (s1[i]-t1[i])!=0:
return alpha[i] | find-the-difference | Python Solution (With basics concepts) | VINAY_KUMAR_V_C | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,726 |
https://leetcode.com/problems/find-the-difference/discuss/2788536/Simple-Python3-Solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
if s == t:
return ""
for i in t:
if s.count(i) != t.count(i):
return i | find-the-difference | Simple Python3 Solution | vivekrajyaguru | 0 | 4 | find the difference | 389 | 0.603 | Easy | 6,727 |
https://leetcode.com/problems/find-the-difference/discuss/2778442/python-one-line | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
return[i for i in t if s.count(i) != t.count(i)][0] | find-the-difference | python one line | seifsoliman | 0 | 4 | find the difference | 389 | 0.603 | Easy | 6,728 |
https://leetcode.com/problems/find-the-difference/discuss/2770114/Python3-Simple-beginners-approach.-Please-upvote-if-it-helped | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
temp = {}
for letter in s:
if letter in temp:
temp[letter] += 1
else:
temp[letter] = 1
letters = {}
for letter in t:
if letter not in temp:
return letter
else:
if letter in letters:
letters[letter] += 1
else:
letters[letter] = 1
if letters[letter] > temp[letter]:
return letter | find-the-difference | Python3 Simple beginners approach. Please upvote if it helped | OGLearns | 0 | 4 | find the difference | 389 | 0.603 | Easy | 6,729 |
https://leetcode.com/problems/find-the-difference/discuss/2738653/Simple-Python-Solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
arr = list(t)
for i in range(len(s)):
arr.remove(s[i])
return arr[0] | find-the-difference | Simple Python Solution | dnvavinash | 0 | 7 | find the difference | 389 | 0.603 | Easy | 6,730 |
https://leetcode.com/problems/find-the-difference/discuss/2736496/Python-Solution-93.86-faster-TC-O(n)-Space-O(1) | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
ans = 0
#iterating over the list 1 and performing XOR
for i in s:
# converting the element to its ascii value for XOR
ans^=ord(i)
for j in t:
ans^=ord(j)
# converting the extra element back to its character from ascii value
return chr(ans) | find-the-difference | Python Solution 93.86% faster TC O(n) Space O(1) | nidhi_nishad26 | 0 | 6 | find the difference | 389 | 0.603 | Easy | 6,731 |
https://leetcode.com/problems/find-the-difference/discuss/2735628/Python-solution-Faster-than-9818-using-Counters | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
Cs=Counter(s)
Ct=Counter(t)
return list((Ct-Cs).keys())[0] | find-the-difference | Python solution Faster than 98,18%, using Counters | irouis | 0 | 4 | find the difference | 389 | 0.603 | Easy | 6,732 |
https://leetcode.com/problems/find-the-difference/discuss/2732694/easy-approach-using-python | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
s = sorted(list(s))
t = sorted(list(t))
for i in range(len(s)):
if s[i]!=t[i]:
return t[i]
return t[-1] | find-the-difference | easy approach using python | sindhu_300 | 0 | 2 | find the difference | 389 | 0.603 | Easy | 6,733 |
https://leetcode.com/problems/find-the-difference/discuss/2732362/Python-(2-line-Code) | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
for i in t:
if s.count(i)!=t.count(i):
return i | find-the-difference | Python (2 line Code) | durgaraopolamarasetti | 0 | 6 | find the difference | 389 | 0.603 | Easy | 6,734 |
https://leetcode.com/problems/find-the-difference/discuss/2721715/Python-Solution-Using-hashmap | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
sHash = {}
tHash = {}
for char in s:
if char in sHash:
sHash[char] += 1
else:
sHash[char] = 1
for char in t:
if char in tHash:
tHash[char] += 1
else:
tHash[char] = 1
for key in tHash:
if key in sHash and sHash[key] == tHash[key]:
continue
return key | find-the-difference | Python Solution Using hashmap | Furat | 0 | 9 | find the difference | 389 | 0.603 | Easy | 6,735 |
https://leetcode.com/problems/find-the-difference/discuss/2699340/Python-easy-solution | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
s_arr = list(s)
s_arr.sort()
t_arr = list(t)
t_arr.sort()
for i in range(len(t_arr)-1):
if t_arr[i]!=s_arr[i]:
return t_arr[i]
return t_arr[len(t_arr)-1] | find-the-difference | Python easy solution | dyussenovaanel | 0 | 7 | find the difference | 389 | 0.603 | Easy | 6,736 |
https://leetcode.com/problems/find-the-difference/discuss/2693125/easy-solytion-using-XOR | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
KeyNo = 0
for i in s:
KeyNo ^= ord(i)
for i in t:
KeyNo ^= ord(i)
return chr(KeyNo) | find-the-difference | easy solytion using XOR | algoon2002 | 0 | 3 | find the difference | 389 | 0.603 | Easy | 6,737 |
https://leetcode.com/problems/find-the-difference/discuss/2690706/PYTHON3-FASTEST | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
for i in t:
if s.count(i)!=t.count(i):
return i | find-the-difference | PYTHON3 FASTEST | Gurugubelli_Anil | 0 | 2 | find the difference | 389 | 0.603 | Easy | 6,738 |
https://leetcode.com/problems/elimination-game/discuss/824126/Python3-3-line-O(logN) | class Solution:
def lastRemaining(self, n: int) -> int:
if n == 1: return 1
if n&1: n -= 1
return n + 2 - 2*self.lastRemaining(n//2) | elimination-game | [Python3] 3-line O(logN) | ye15 | 5 | 588 | elimination game | 390 | 0.465 | Medium | 6,739 |
https://leetcode.com/problems/elimination-game/discuss/824126/Python3-3-line-O(logN) | class Solution:
def lastRemaining(self, n: int) -> int:
def fn(n):
"""Return the final number of a list of length n."""
if n == 1: return 1
if n&1: n -= 1
return n + 2*(1 - fn(n//2))
return fn(n) | elimination-game | [Python3] 3-line O(logN) | ye15 | 5 | 588 | elimination game | 390 | 0.465 | Medium | 6,740 |
https://leetcode.com/problems/elimination-game/discuss/824126/Python3-3-line-O(logN) | class Solution:
def lastRemaining(self, n: int) -> int:
i, di = 0, 2
for _ in range(n-1):
if not 0 <= i+di < n:
i += di//2 if 0 <= i + di//2 < n else -di//2
di *= -2
else: i += di
return i+1 | elimination-game | [Python3] 3-line O(logN) | ye15 | 5 | 588 | elimination game | 390 | 0.465 | Medium | 6,741 |
https://leetcode.com/problems/elimination-game/discuss/1016735/Python3-O(logN)-and-constant-space | class Solution:
def lastRemaining(self, n: int) -> int:
first=1
distance=1
step=0
while n>1:
if step%2==0 or n%2!=0:
first+=distance
# Multiply distance by 2
distance=distance<<1
step+=1
# Divide n by 2
n=n>>1
return first | elimination-game | Python3 O(logN) and constant space | bindhushreehr | 1 | 492 | elimination game | 390 | 0.465 | Medium | 6,742 |
https://leetcode.com/problems/elimination-game/discuss/2733788/Detail-1-line-python-explanations | class Solution:
def lastRemaining(self, n: int) -> int:
# explanation
# F algorithem is take the even index element from the list and reverse the list
# lastRemaining(9) = F([1, 2, 3, 4, 5, 6, 7, 8, 9])
# first round: F([2, 4, 6, 8]) = F(2 * [1, 2, 3, 4]) = 2 * F([4, 3, 2, 1])
# convert F([4, 3, 2, 1]) = F([3, 1])
# now we need to convert F[(3, 1)] to F([1, 3]), which is the original order then do F algorithm
# 4 - 4 + 1 = 1
# 4 - 2 + 1 = 3
# n//2 - F(n//2) + 1
# base case is 1
return 2 * (n // 2 - self.lastRemaining(n//2) + 1) if n != 1 else 1 | elimination-game | Detail 1-line python explanations | jackson-cmd | 0 | 24 | elimination game | 390 | 0.465 | Medium | 6,743 |
https://leetcode.com/problems/elimination-game/discuss/2723369/Simple-recursive-solution-(beat-95-TC-beat-97-SC) | class Solution:
def lastRemaining(self, n: int) -> int:
if n < 3:
return n
if n % 2 == 1:
return self.lastRemaining(n-1)
elif n % 4 == 2:
return 4 * self.lastRemaining((n-2)//4)
else:
return 4 * self.lastRemaining(n//4) - 2 | elimination-game | Simple recursive solution (beat 95% TC, beat 97% SC) | zhaoqiang | 0 | 22 | elimination game | 390 | 0.465 | Medium | 6,744 |
https://leetcode.com/problems/elimination-game/discuss/2642590/python3 | class Solution:
def lastRemaining(self, n: int) -> int:
ans = 1
k, cnt, step = 0, n, 1
while cnt > 1:
if k % 2 == 0:
ans += step
else:
if cnt % 2:
ans += step
k += 1
cnt >>= 1
step <<= 1
return ans | elimination-game | python3 | xy01 | 0 | 134 | elimination game | 390 | 0.465 | Medium | 6,745 |
https://leetcode.com/problems/elimination-game/discuss/2316855/C%2B%2B-solution | class Solution:
def lastRemaining(self, n: int) -> int:
beg = 1
len = n
d = 1
fromleft = True
while len > 1:
if(fromleft or len%2 == 1):
beg += d
d <<= 1
len >>= 1
fromleft = not fromleft
return beg | elimination-game | C++ solution | 487o | 0 | 192 | elimination game | 390 | 0.465 | Medium | 6,746 |
https://leetcode.com/problems/elimination-game/discuss/514388/Python3-different-simple-solutions | class Solution:
def lastRemaining(self, n: int) -> int:
return 1 if n==1 else 2*(1+n//2-self.lastRemaining(n//2))
def lastRemaining1(self, n: int) -> int:
""""
removing from left to right [1 2 3 4 5 6 7 8 9]==[2 4 6 8]==2*[1 2 3 4]
"""
def helper(n,is_left):
if n == 1: return 1
if is_left: return 2*helper(n//2,False)
if n%2==1: return 2*helper(n//2,True)
return 2*helper(n//2,2)-1
return helper(n,True)
def lastRemaining2(self, n: int) -> int:
if n==0: return
st1,st2,left=list(range(1,n+1)),[],True
while len(st1)>1:
if not left:
st1=st1[::-1]
for i in range(1,len(st1),2):
st2.append(st1[i])
st1=list(st2) if left else list(st2[::-1])
st2=[]
left = not left
return st1[0] | elimination-game | Python3 different simple solutions | jb07 | 0 | 379 | elimination game | 390 | 0.465 | Medium | 6,747 |
https://leetcode.com/problems/perfect-rectangle/discuss/968076/Python-Fast-and-clear-solution-with-explanation | class Solution:
def isRectangleCover(self, rectangles: List[List[int]]) -> bool:
area = 0
corners = set()
a = lambda: (Y-y) * (X-x)
for x, y, X, Y in rectangles:
area += a()
corners ^= {(x,y), (x,Y), (X,y), (X,Y)}
if len(corners) != 4: return False
x, y = min(corners, key=lambda x: x[0] + x[1])
X, Y = max(corners, key=lambda x: x[0] + x[1])
return a() == area | perfect-rectangle | [Python] Fast and clear solution with explanation | modusV | 40 | 1,200 | perfect rectangle | 391 | 0.325 | Hard | 6,748 |
https://leetcode.com/problems/perfect-rectangle/discuss/2666778/Python-soluton | class Solution:
def isRectangleCover(self, rectangles: List[List[int]]) -> bool:
xs=[]
ys=[]
for sx,sy,ex,ey in rectangles:
xs.append(sx)
xs.append(ex)
ys.append(sy)
ys.append(ey)
xlookup={x:i for i,x in enumerate(sorted(set(xs)))}
ylookup={y:i for i,y in enumerate(sorted(set(ys)))}
N=len(xlookup)-1
M=len(ylookup)-1
grid=[[0]*M for _ in range(N)]
for sx,sy,ex,ey in rectangles:
for cx in range(xlookup[sx],xlookup[ex]):
for cy in range(ylookup[sy],ylookup[ey]):
grid[cx][cy]+=1
for row in grid:
for cell in row:
if cell!=1:
return False
return True | perfect-rectangle | Python soluton | Motaharozzaman1996 | 0 | 8 | perfect rectangle | 391 | 0.325 | Hard | 6,749 |
https://leetcode.com/problems/perfect-rectangle/discuss/2629231/Straightforward-Python-Solution | class Solution:
def isRectangleCover(self, rectangles: List[List[int]]) -> bool:
X1, Y1 = float('inf'), float('inf')
X2, Y2 = float('-inf'), float('-inf')
# keeps track of the vertices that appear 1 or 3 times and eliminates those appearing 2 or 4 times
points = set()
actual_area = 0
for x1, y1, x2, y2 in rectangles:
# find 4 vertices of the perfect rectangle
X1, Y1 = min(X1, x1), min(Y1, y1)
X2, Y2 = max(X2, x2), max(Y2, y2)
actual_area += (x2 - x1) * (y2 - y1)
vertices = [(x1, y1), (x1, y2), (x2, y1), (x2, y2)]
for v in vertices:
if v in points:
points.remove(v)
else:
points.add(v)
# check if actual_area is equal to expected_area
expected_area = (X2 - X1) * (Y2 - Y1)
if (actual_area != expected_area):
return False
# checks if there are four vertices left in the points set
if (len(points) != 4): return False
# checks if four vertices correspond to the actual perfect rectangle
if (X1, Y1) not in points or (X2, Y1) not in points or (X1, Y2) not in points or (X2, Y2) not in points:
return False
return True | perfect-rectangle | Straightforward Python Solution | leqinancy | 0 | 14 | perfect rectangle | 391 | 0.325 | Hard | 6,750 |
https://leetcode.com/problems/perfect-rectangle/discuss/1102039/Python3-locate-the-corners | class Solution:
def isRectangleCover(self, rectangles: List[List[int]]) -> bool:
area = 0
corner = set()
X0 = Y0 = inf
X1 = Y1 = -inf
for x0, y0, x1, y1 in rectangles:
area += (x1-x0)*(y1-y0)
X0 = min(x0, X0)
Y0 = min(y0, Y0)
X1 = max(x1, X1)
Y1 = max(y1, Y1)
corner ^= {(x0, y0), (x0, y1), (x1, y0), (x1, y1)}
return area == (X1-X0)*(Y1-Y0) and corner == {(X0, Y0), (X0, Y1), (X1, Y0), (X1, Y1)} | perfect-rectangle | [Python3] locate the corners | ye15 | 0 | 280 | perfect rectangle | 391 | 0.325 | Hard | 6,751 |
https://leetcode.com/problems/is-subsequence/discuss/2473010/Very-Easy-oror-100-oror-Fully-Explained-oror-Java-C%2B%2B-Python-JS-C-Python3-(Two-Pointers-Approach) | class Solution(object):
def isSubsequence(self, s, t):
# Base case
if not s:
return True
i = 0
# Traverse elements of t string
for j in t:
# If this index matches to the index of s string, increment i pointer...
if j == s[i]:
i += 1
# If the pointer is equal to the size of s...
if i == len(s):
break
return i == len(s) | is-subsequence | Very Easy || 100% || Fully Explained || Java, C++, Python, JS, C, Python3 (Two-Pointers Approach) | PratikSen07 | 16 | 998 | is subsequence | 392 | 0.49 | Easy | 6,752 |
https://leetcode.com/problems/is-subsequence/discuss/1811624/Simple-2-Pointer-Python-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i, j, n, m = 0, 0, len(s), len(t)
while i < n and j < m: # Loop till any of the strings is fully traversed
if s[i] == t[j]: # If char at i and j are equal then update i
i += 1
# Could also write i += s[i]==t[j]
j += 1 # Update j always.
return i == n | is-subsequence | Simple 2 Pointer Python Solution | anCoderr | 10 | 831 | is subsequence | 392 | 0.49 | Easy | 6,753 |
https://leetcode.com/problems/is-subsequence/discuss/2679763/Python-Easy-Solution-or-99-Faster-or-Is-Subsequence | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
sl, tl = 0, 0
while sl<len(s) and tl<len(t):
if s[sl] == t[tl]:
sl+=1
tl+=1
else:
tl+=1
if sl==len(s):
return True
else:
return False | is-subsequence | ✔️ Python Easy Solution | 99% Faster | Is Subsequence | pniraj657 | 9 | 745 | is subsequence | 392 | 0.49 | Easy | 6,754 |
https://leetcode.com/problems/is-subsequence/discuss/1077869/Python.-faster-than-99.45.-Easy-understanding-solution. | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i, j, m, n = 0, 0, len(s), len(t)
while i < m and j < n and n - j >= m - i:
if s[i] == t[j]:
i += 1
j += 1
return i == m | is-subsequence | Python. faster than 99.45%. Easy-understanding solution. | m-d-f | 5 | 736 | is subsequence | 392 | 0.49 | Easy | 6,755 |
https://leetcode.com/problems/is-subsequence/discuss/423566/Python3-4-solutions-(quickest-99.98-shortest-two-lines) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
it = iter(t)
return all(ch in it for ch in s) | is-subsequence | [Python3] 4 solutions (quickest 99.98%, shortest two lines) | ye15 | 5 | 422 | is subsequence | 392 | 0.49 | Easy | 6,756 |
https://leetcode.com/problems/is-subsequence/discuss/423566/Python3-4-solutions-(quickest-99.98-shortest-two-lines) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
k = 0
for c in s:
k = t.find(c, k) + 1
if k == 0: return False
return True | is-subsequence | [Python3] 4 solutions (quickest 99.98%, shortest two lines) | ye15 | 5 | 422 | is subsequence | 392 | 0.49 | Easy | 6,757 |
https://leetcode.com/problems/is-subsequence/discuss/423566/Python3-4-solutions-(quickest-99.98-shortest-two-lines) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
k = 0
for c in s:
try: k = t.index(c, k) + 1
except: return False
return True | is-subsequence | [Python3] 4 solutions (quickest 99.98%, shortest two lines) | ye15 | 5 | 422 | is subsequence | 392 | 0.49 | Easy | 6,758 |
https://leetcode.com/problems/is-subsequence/discuss/423566/Python3-4-solutions-(quickest-99.98-shortest-two-lines) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i = 0
for c in t:
if i < len(s) and s[i] == c: i += 1
return i == len(s) | is-subsequence | [Python3] 4 solutions (quickest 99.98%, shortest two lines) | ye15 | 5 | 422 | is subsequence | 392 | 0.49 | Easy | 6,759 |
https://leetcode.com/problems/is-subsequence/discuss/2126747/Python-or-One-of-the-most-elegant-solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
for char in t:
if len(s) > 0 and char == s[0]:
s = s[1:]
return len(s) == 0 | is-subsequence | Python | One of the most elegant solution | __Asrar | 3 | 96 | is subsequence | 392 | 0.49 | Easy | 6,760 |
https://leetcode.com/problems/is-subsequence/discuss/1811501/Pythonic-O(n)-Iterative-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i,j=0,0
while(j<len(s) and i<len(t)):
if(s[j]==t[i]):
j+=1
i+=1
return j==len(s) | is-subsequence | Pythonic O(n) Iterative Solution | js5809 | 3 | 134 | is subsequence | 392 | 0.49 | Easy | 6,761 |
https://leetcode.com/problems/is-subsequence/discuss/1197645/Python-DP-Standard-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
n = len(s)
m = len(t)
if n==0:
return True
if m==0:
return False
dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
for i in range(1,m+1):
for j in range(1,n+1):
if t[i-1] == s[j-1]:
dp[i][j] = dp[i-1][j-1] + 1
else:
dp[i][j] = max(dp[i-1][j],dp[i][j-1])
if dp[i][j] == len(s):
return True
return False | is-subsequence | Python DP Standard Solution | iamkshitij77 | 3 | 136 | is subsequence | 392 | 0.49 | Easy | 6,762 |
https://leetcode.com/problems/is-subsequence/discuss/380060/Solution-in-Python-3-(beats-~100)-(five-lines) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
I = -1
for i in s:
I = t.find(i,I+1)
if I == -1: return False
return True
- Junaid Mansuri
(LeetCode ID)@hotmail.com | is-subsequence | Solution in Python 3 (beats ~100%) (five lines) | junaidmansuri | 3 | 672 | is subsequence | 392 | 0.49 | Easy | 6,763 |
https://leetcode.com/problems/is-subsequence/discuss/2770032/The-most-readable-solution-in-Python-for-%22392.-Is-subsequence%22 | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
n = len(s)
j = 0
for letter in t:
if j == n:
return True
if letter == s[j]:
j += 1
return j == n | is-subsequence | The most readable solution in Python for "392. Is subsequence" | bekbull | 2 | 169 | is subsequence | 392 | 0.49 | Easy | 6,764 |
https://leetcode.com/problems/is-subsequence/discuss/1813856/Python3-oror-Linear-Solution-oror-Almost-100-Faster | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) > len(t): return False
j = 0
for i in t:
if j < len(s):
if i == s[j]:
j += 1
return True if j == len(s) else False | is-subsequence | Python3 || Linear Solution || Almost 100% Faster | cherrysri1997 | 2 | 83 | is subsequence | 392 | 0.49 | Easy | 6,765 |
https://leetcode.com/problems/is-subsequence/discuss/1811282/Python-Simple-Python-Solution-With-Two-Pointers | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i=0
j=0
while i<len(s) and j<len(t):
if s[i]==t[j]:
i=i+1
j=j+1
else:
j=j+1
if i==len(s):
return True
else:
return False | is-subsequence | [ Python ] ✔✔ Simple Python Solution With Two Pointers 🔥✌ | ASHOK_KUMAR_MEGHVANSHI | 2 | 183 | is subsequence | 392 | 0.49 | Easy | 6,766 |
https://leetcode.com/problems/is-subsequence/discuss/1496681/Difference-between-Subarray-or-Subsequence-or-Subset-oror-Beginner-Approach | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
# A very intuitive way to check if one is a subsequence of other or not
# Here, We will check if s is a subsequence of t or not
i=0 #to keep a check of elements across s
j=0 #to keep a check of elements across t
while(j < len(t) and i < len(s)):
if (s[i] == t[j]):
i += 1
j += 1
else:
j+=1
if(i == len(s)): #it means s is a subsequence of t as i covered all elements of s across t
return True
else:
return False | is-subsequence | Difference between Subarray | Subsequence | Subset || Beginner Approach | aarushsharmaa | 2 | 126 | is subsequence | 392 | 0.49 | Easy | 6,767 |
https://leetcode.com/problems/is-subsequence/discuss/2573596/SIMPLE-PYTHON3-SOLUTION-O(t) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s)==0:return True
count = 0
for i in range(len(t)):
if count <= len(s) - 1:
if s[count] == t[i]:
count += 1
return count == len(s) | is-subsequence | ✅✔ SIMPLE PYTHON3 SOLUTION ✅✔O(t) | rajukommula | 1 | 67 | is subsequence | 392 | 0.49 | Easy | 6,768 |
https://leetcode.com/problems/is-subsequence/discuss/2483813/Runtime%3A-56-ms-faster-than-38.48-of-Python3-Memory-Usage%3A-13.9-MB-less-than-43.43-of-Python3 | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
for i in range(0,len(t)):
if len(s)==0:
return True
elif t[i]==s[0]:
s=s[1:]
return len(s)==0 | is-subsequence | Runtime: 56 ms, faster than 38.48% of Python3 ,Memory Usage: 13.9 MB, less than 43.43% of Python3 | DG-Problemsolver | 1 | 6 | is subsequence | 392 | 0.49 | Easy | 6,769 |
https://leetcode.com/problems/is-subsequence/discuss/2327519/Python3-O(n%2Bm)-oror-O(1)-Runtime%3A-38ms-83.94-oror-memory%3A-13.8mb-82.20 | class Solution:
# O(n+m) || O(1)
# Runtime: 38ms 83.94% || memory: 13.8mb 82.20%
def isSubsequence(self, string: str, target: str) -> bool:
i = 0
for char in target:
if i == len(string):
return True
if char == string[i]:
i += 1
return i == len(string) | is-subsequence | Python3 O(n+m) || O(1) # Runtime: 38ms 83.94% || memory: 13.8mb 82.20% | arshergon | 1 | 36 | is subsequence | 392 | 0.49 | Easy | 6,770 |
https://leetcode.com/problems/is-subsequence/discuss/2279444/Python3-two-pointers-approach | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
first = 0
second = 0
while first<len(s) and second<len(t):
if s[first]==t[second]:
first+=1
second+=1
return first>=len(s) | is-subsequence | 📌 Python3 two pointers approach | Dark_wolf_jss | 1 | 32 | is subsequence | 392 | 0.49 | Easy | 6,771 |
https://leetcode.com/problems/is-subsequence/discuss/1930728/Two-Pointer-Approach | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
# initialize pointers
pointer_1, pointer_2 = 0, 0
# to ensure pointers traverse till the end of both strings
while pointer_1 < len(s) and pointer_2 <len(t):
# if both the pointers are pointing to same element, just shift the pointers to next position
if s[pointer_1] == t[pointer_2]:
pointer_1 += 1
pointer_2 += 1
# otherwise, just shift the right pointer by one position keeping left intact
else:
pointer_2 += 1
return pointer_1 == len(s) | is-subsequence | Two Pointer Approach | Jayesh_Suryavanshi | 1 | 43 | is subsequence | 392 | 0.49 | Easy | 6,772 |
https://leetcode.com/problems/is-subsequence/discuss/1847339/Python-easy-solution-beats-96 | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
left, right = 0, 0
while left != len(s) and right != len(t) :
if s[left] != t[right]:
right += 1
elif s[left] == t[right]:
left += 1
right += 1
if left == len(s):
return True
return False | is-subsequence | Python easy solution beats 96% | yusianglin11010 | 1 | 101 | is subsequence | 392 | 0.49 | Easy | 6,773 |
https://leetcode.com/problems/is-subsequence/discuss/1813489/Python-Queue-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) == 0: return True
q = [c for c in s]
for c in t:
if q[0] == c:
q.pop(0)
if len(q) == 0:
return True
return False | is-subsequence | Python Queue Solution | Pootato | 1 | 20 | is subsequence | 392 | 0.49 | Easy | 6,774 |
https://leetcode.com/problems/is-subsequence/discuss/1811595/Self-understandable-Python-%3A | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s)>len(t):
return False
if len(s)==0:
return True
j=0
p=""
for i in t:
if i==s[j]:
p+=i
j+=1
if j>=len(s):
break
return p==s | is-subsequence | Self understandable Python : | goxy_coder | 1 | 81 | is subsequence | 392 | 0.49 | Easy | 6,775 |
https://leetcode.com/problems/is-subsequence/discuss/1370625/2-pointer-solution-or-Python | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i,j = 0,0
if(len(s)>len(t)):
return False
while( i<len(s) and j<len(t) ):
if(s[i] == t[j]):
i+=1
j+=1
else:
j+=1
if(i == len(s)):
return True
return False | is-subsequence | 2 pointer solution | Python | RishithaRamesh | 1 | 74 | is subsequence | 392 | 0.49 | Easy | 6,776 |
https://leetcode.com/problems/is-subsequence/discuss/1079112/Python-solution-using-while-and-for-loop. | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) > len(t): return False
i, j = 0, 0
while i < len(s) and j < len(t):
if s[i] == t[j]:
i += 1
j += 1
if len(s) == i:
return True
return False | is-subsequence | Python solution using while and for loop. | vsahoo | 1 | 80 | is subsequence | 392 | 0.49 | Easy | 6,777 |
https://leetcode.com/problems/is-subsequence/discuss/679493/python-two-pointers-and-NlogN-follow-up-by-binary-search | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i = j = 0
while i < len(s) and j < len(t):
if s[i] == t[j]:
i+=1
j+=1
else:
j+=1
return i == len(s) | is-subsequence | python two pointers and NlogN follow-up by binary search | nightybear | 1 | 86 | is subsequence | 392 | 0.49 | Easy | 6,778 |
https://leetcode.com/problems/is-subsequence/discuss/679493/python-two-pointers-and-NlogN-follow-up-by-binary-search | class Solution:
'''
Time complexity : O(nlogn)
Space complexity : O(n)
'''
def isSubsequence(self, s: str, t: str) -> bool:
indexs = collections.defaultdict(list)
for idx, x in enumerate(t):
indexs[x].append(idx)
prev = 0
for x in s:
i = bisect.bisect_left(indexs[x], prev)
if i == len(indexs[x]):
return False
prev = indexs[x][i]+1
return True | is-subsequence | python two pointers and NlogN follow-up by binary search | nightybear | 1 | 86 | is subsequence | 392 | 0.49 | Easy | 6,779 |
https://leetcode.com/problems/is-subsequence/discuss/2848574/python3 | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
idx = 0
for c in s:
idx = t.find(c, idx)
if idx == -1:
return False
idx += 1
return True | is-subsequence | python3 | wduf | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,780 |
https://leetcode.com/problems/is-subsequence/discuss/2846740/Easy-understand-python-solution-(beginer-friendly) | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
count=0
if not s :
return True
if not t:
return False
for ele in t:
if ele==s[count]:
count+=1
if count==len(s):
return True
return False | is-subsequence | Easy understand python solution (beginer friendly) | amangarg0599 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,781 |
https://leetcode.com/problems/is-subsequence/discuss/2838020/Python3-solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
position = -1
for char in s:
position = t.find(char, position + 1)
if position == -1:
return False
return True | is-subsequence | Python3 solution | arthur54342 | 0 | 3 | is subsequence | 392 | 0.49 | Easy | 6,782 |
https://leetcode.com/problems/is-subsequence/discuss/2835462/Simple-One-Pass-O(1)-space-with-two-pointers | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
p1 = p2 = 0
while p1 < len(s) and p2 < len(t):
if s[p1] == t[p2]:
p1 += 1
p2 += 1
return p1 == len(s) | is-subsequence | Simple One Pass O(1) space with two pointers | WhyYouCryingMama | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,783 |
https://leetcode.com/problems/is-subsequence/discuss/2835461/Simple-One-Pass-O(1)-space-with-two-pointers | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
p1 = p2 = 0
while p1 < len(s) and p2 < len(t):
if s[p1] == t[p2]:
p1 += 1
p2 += 1
return p1 == len(s) | is-subsequence | Simple One Pass O(1) space with two pointers | WhyYouCryingMama | 0 | 0 | is subsequence | 392 | 0.49 | Easy | 6,784 |
https://leetcode.com/problems/is-subsequence/discuss/2835150/Python3-Simple-Clean-Code | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
for i in range(len(s)):
findIndex = t.find(s[i]) if s[i] in t else -1
if findIndex == -1: return False
t = t[findIndex+1:]
return True | is-subsequence | ✅ Python3 Simple Clean Code | Cecilia_GuoChen | 0 | 3 | is subsequence | 392 | 0.49 | Easy | 6,785 |
https://leetcode.com/problems/is-subsequence/discuss/2835046/My-python-solution-or-beats-90 | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
for e in s:
if e not in t:
return False
i = 0
j = 0
while j < len(t):
if i <len(s) and s[i] ==t[j]:
i+=1
j+=1
return i == len(s) | is-subsequence | My python solution | beats 90% | sahil193101 | 0 | 4 | is subsequence | 392 | 0.49 | Easy | 6,786 |
https://leetcode.com/problems/is-subsequence/discuss/2829435/Python-or-Two-Pointer-or-Merge-two-array-way | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if not s:
return True
if len(s) > len(t):
return False
end = len(t)-1
first_end = len(s)-1
t_start = 0
s_start = 0
while t_start <= end and s_start <= first_end:
if s[s_start] == t[t_start]:
s_start += 1
t_start += 1
else:
t_start += 1
if s_start < len(s):
return False
return True | is-subsequence | Python | Two Pointer | Merge two array way | ajay_gc | 0 | 5 | is subsequence | 392 | 0.49 | Easy | 6,787 |
https://leetcode.com/problems/is-subsequence/discuss/2828195/Two-Pointers-easy | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i=0
j=0
if len(s)>len(t) or (len(t)==0 and len(s)!=0):
return False
if len(s)==0:
return True
for i in range(len(t)):
if s[j]==t[i]:
j+=1
if j==len(s):
return True
return False | is-subsequence | Two Pointers easy | rhi_1 | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,788 |
https://leetcode.com/problems/is-subsequence/discuss/2819037/Python-My-O(n)-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if not s:
return True
if not t:
return False
s_pointer = 0
for t_char in t:
if t_char == s[s_pointer]:
s_pointer += 1
if s_pointer == len(s):
return True
return False | is-subsequence | [Python] My O(n) Solution | manytenks | 0 | 3 | is subsequence | 392 | 0.49 | Easy | 6,789 |
https://leetcode.com/problems/is-subsequence/discuss/2818144/two-pointer-and-some-conclutions | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) < 1:return True
if len(s) > len(t):return False
s_pointer = 0
for t_pointer in range(0,len(t)):
if s_pointer <= len(s)-1:
if s[s_pointer]==t[t_pointer]:s_pointer += 1
return s_pointer == len(s) | is-subsequence | two pointer and some conclutions | roger880327 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,790 |
https://leetcode.com/problems/is-subsequence/discuss/2818113/using-two-pointers | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) < 1:return True
s_pointer = 0
for t_pointer in range(0,len(t)):
if s_pointer < len(s):
if s[s_pointer]==t[t_pointer]:s_pointer += 1
if s_pointer == 0:return False
return s_pointer == len(s) | is-subsequence | using two pointers | roger880327 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,791 |
https://leetcode.com/problems/is-subsequence/discuss/2816077/Python-o(n)-solution-simple-to-understand | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
i = 0
for c in s:
while i < len(t) and t[i] != c:
i+= 1
if i == len(t): return False
i+=1
return True | is-subsequence | Python o(n) solution, simple to understand | pradyumna04 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,792 |
https://leetcode.com/problems/is-subsequence/discuss/2807838/easy-understnding-python | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
counter = 0
for c in s:
if c in t:
t = t[t.index(c) + 1:]
counter += 1
if len(s) == counter:
return True
return False | is-subsequence | easy understnding python | don_masih | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,793 |
https://leetcode.com/problems/is-subsequence/discuss/2799242/Subsequence-My-Solution | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(s) == 0:
return True
si, ti = 0,0
for ti in range(len(t)):
if s[si] == t[ti]:
si += 1
if si == len(s):
return True
return False | is-subsequence | Subsequence - My Solution | marked01one | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,794 |
https://leetcode.com/problems/is-subsequence/discuss/2796564/Is-Substring-Python-easy-to-understand-beginner-friendly-comments | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
"""
Using a 2-pointer approach, compare two strings and determines if the first string is a subset of the second string.
:param s:
str: String of alpha characters
:param t:
str: String of alpha characters
:return
bool: True if first string is a contiguous subset of the second string
"""
# Set variables equal to the length of each input string
left_bound, right_bound = len(s), len(t)
# Initialize index for each pointer
p_left = p_right = 0
while p_left < left_bound and p_right < right_bound:
# Move both pointers if the equal each other
if s[p_left] == t[p_right]:
p_left += 1
p_right += 1
# Return when left pointer equals length of source string
return p_left == left_bound | is-subsequence | Is Substring - Python, easy-to-understand, beginner friendly, comments | mthomp89 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,795 |
https://leetcode.com/problems/is-subsequence/discuss/2790867/Simple-solution-using-exception-to-catch-completed-substring | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
# Make an iterator of the sub string
s_iter = iter(s)
try:
# Start by taking the first character in the substring
sub = next(s_iter)
for curr in t:
if curr == sub:
sub = next(s_iter)
# When next() is called, and s_iter is emptied,
# StopIteration will be raised
except StopIteration:
return True
return False | is-subsequence | Simple solution using exception to catch completed substring | stereo1 | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,796 |
https://leetcode.com/problems/is-subsequence/discuss/2786048/Easy-python-solution-or-Is-Subsequence | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(t) < len(s):
return False
for i in range(len(t)):
if len(s) > 0 and t[i] == s[0]:
s = s[1:]
if len(s) == 0:
return True
return False | is-subsequence | Easy python solution | Is Subsequence | nishanrahman1994 | 0 | 2 | is subsequence | 392 | 0.49 | Easy | 6,797 |
https://leetcode.com/problems/is-subsequence/discuss/2786047/Easy-python-solution-or-Is-Subsequence | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
if len(t) < len(s):
return False
for i in range(len(t)):
if len(s) > 0 and t[i] == s[0]:
s = s[1:]
if len(s) == 0:
return True
return False | is-subsequence | Easy python solution | Is Subsequence | nishanrahman1994 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,798 |
https://leetcode.com/problems/is-subsequence/discuss/2784933/Pretty-happy-i-managed-to-get-a-pretty-good-solution-on-my-own-for-once. | class Solution:
def isSubsequence(self, s: str, t: str) -> bool:
len_s = len(s)
if len_s == 0: return True
i: int = 0
for char in t:
if s[i] == char:
i += 1
if i == len_s: return True
return False | is-subsequence | Pretty happy i managed to get a pretty good solution on my own for once. | Seawolf159 | 0 | 1 | is subsequence | 392 | 0.49 | Easy | 6,799 |
Subsets and Splits