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https://leetcode.com/problems/reverse-string/discuss/2797640/Simple-Python3-Solution | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-1] | reverse-string | Simple Python3 Solution | vivekrajyaguru | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,000 |
https://leetcode.com/problems/reverse-string/discuss/2797550/Python-2-Approaches | class Solution:
def reverseString(self, s: List[str]) -> None:
# method 1: better speed
s[::]=s[::-1]
# method 2: better space
# l,r=0,len(s)-1
# while l<r:
# s[l],s[r]=s[r],s[l]
# l+=1
# r-=1 | reverse-string | Python 2 Approaches | sbhupender68 | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,001 |
https://leetcode.com/problems/reverse-string/discuss/2788957/Best-Easy-Approach-oror-99.63-Acceptance-oror-TC-O(N) | class Solution:
def reverseString(self, s: List[str]) -> None:
i=0
j=len(s)-1
while(i<j):
s[i],s[j]=s[j],s[i]
i+=1
j-=1
"""
Do not return anything, modify s in-place instead.
""" | reverse-string | Best Easy Approach || 99.63% Acceptance || TC O(N) | Kaustubhmishra | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,002 |
https://leetcode.com/problems/reverse-string/discuss/2782891/Two-pointers-beats-86 | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
l, r = 0,len(s)-1
while l < r:
s[l], s[r] = s[r], s[l]
l += 1
r -=1 | reverse-string | Two pointers, beats 86% | haniyeka | 0 | 3 | reverse string | 344 | 0.762 | Easy | 6,003 |
https://leetcode.com/problems/reverse-string/discuss/2782792/python-Easiest-one-line-solution-as-everyone-knows | class Solution:
def reverseString(self, s: List[str]) -> None:
s.reverse()
# Runtime: 633 ms, faster than 5.25% of Python3 online submissions for Reverse String.
# Memory Usage: 18.4 MB, less than 83.12% of Python3 online submissions for Reverse String. | reverse-string | python Easiest one-line solution as everyone knows | yutoun | 0 | 3 | reverse string | 344 | 0.762 | Easy | 6,004 |
https://leetcode.com/problems/reverse-string/discuss/2781302/Python-2-Simple-Solutions-oror-List-comprehension-oror-Swapping | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | reverse-string | [Python] 2 Simple Solutions || List comprehension || Swapping | shahbaz95ansari | 0 | 1 | reverse string | 344 | 0.762 | Easy | 6,005 |
https://leetcode.com/problems/reverse-string/discuss/2781302/Python-2-Simple-Solutions-oror-List-comprehension-oror-Swapping | class Solution:
def reverseString(self, s: List[str]) -> None:
l, r = 0, len(s) - 1
while l < r:
s[l], s[r] = s[r], s[l]
l += 1
r -= 1 | reverse-string | [Python] 2 Simple Solutions || List comprehension || Swapping | shahbaz95ansari | 0 | 1 | reverse string | 344 | 0.762 | Easy | 6,006 |
https://leetcode.com/problems/reverse-string/discuss/2779873/Easy-4-liner-Python-Solutions-Beats-96-of-Submissions | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
n = len(s)
for i in range(n // 2):
tmp = s[i]
s[i] = s[n - i - 1]
s[n - i - 1] = tmp | reverse-string | Easy 4 liner Python Solutions Beats 96% of Submissions | gpersonnat | 0 | 1 | reverse string | 344 | 0.762 | Easy | 6,007 |
https://leetcode.com/problems/reverse-string/discuss/2778176/easy-python-solutionbeats-95-!!! | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
first=0
last=len(s)-1
for i in range(len(s)//2):
s[first],s[last]=s[last],s[first]
first+=1
last-=1
return s | reverse-string | easy python solution,beats 95% !!! | Harshit-chaudhary_01 | 0 | 5 | reverse string | 344 | 0.762 | Easy | 6,008 |
https://leetcode.com/problems/reverse-string/discuss/2775298/faster-than-99.66-less-than-83.23 | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
first = 0
last = len(s) - 1
if last % 2 == 0:
mid = first + last // 2
else:
mid = first + last // 2 + 1
for i in range(0, mid):
s[i], s[last - i] = s[last - i], s[i] | reverse-string | faster than 99.66%, less than 83.23% | eckyrie0921 | 0 | 3 | reverse string | 344 | 0.762 | Easy | 6,009 |
https://leetcode.com/problems/reverse-string/discuss/2772061/Reverse-String-oror-Python-Simple-and-Crisp-Two-Pointer-Approach | class Solution:
def reverseString(self, s: List[str]) -> None:
# Keep 2 Pointers, 1 at the start of the string and the other at the end
i,j=0,len(s)-1
while i<j:
s[i],s[j]=s[j],s[i]
i+=1
j-=1 | reverse-string | Reverse String || Python Simple and Crisp Two Pointer Approach | vedanthvbaliga | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,010 |
https://leetcode.com/problems/reverse-string/discuss/2769725/Python-Super-Simple-beat-98.9 | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(0,int(len(s)/2)):
temp = s[i]
s[i] = s[-(1+i)]
s[-(1+i)] = temp | reverse-string | Python Super Simple, beat 98.9% | paddyveith987 | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,011 |
https://leetcode.com/problems/reverse-string/discuss/2769086/Reverse-string | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
res = s[::-1]
for i in range(0,len(s)):
s[i] = res[i] | reverse-string | Reverse string | keerthikrishnakumar93 | 0 | 4 | reverse string | 344 | 0.762 | Easy | 6,012 |
https://leetcode.com/problems/reverse-string/discuss/2765025/Two-pointers-solution. | class Solution:
def reverseString(self, s: List[str]) -> None:
L, R = 0, len(s)-1
while L < R:
s[L], s[R] = s[R], s[L]
L += 1
R -= 1
# or just simply s.reverse() | reverse-string | Two pointers solution. | woora3 | 0 | 2 | reverse string | 344 | 0.762 | Easy | 6,013 |
https://leetcode.com/problems/reverse-string/discuss/2746165/python-two-line-solution-memory-beats-99 | class Solution:
def reverseString(self, s: List[str]) -> None:
l = len(s)
for i in range(l//2):
s[i], s[l-1-i] = s[l-1-i],s[i] | reverse-string | python two line solution memory beats 99% | muge_zhang | 0 | 3 | reverse string | 344 | 0.762 | Easy | 6,014 |
https://leetcode.com/problems/reverse-string/discuss/2744560/Reverse-String-3-line-solution-in-python | class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
n = len(s)-1
for i in range(n,-1,-1):
s.append(s[i])
s.pop(i) | reverse-string | Reverse String 3 line solution in python | jashii96 | 0 | 4 | reverse string | 344 | 0.762 | Easy | 6,015 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1164745/Python-Solution-oror-99.58-faster-oror-86.96-less-memory | class Solution:
def reverseVowels(self, s: str) -> str:
s = list(s)
left = 0
right = len(s) - 1
m = 'aeiouAEIOU'
while left < right:
if s[left] in m and s[right] in m:
s[left], s[right] = s[right], s[left]
left += 1; right -= 1
elif s[left] not in m:
left += 1
elif s[right] not in m:
right -= 1
return ''.join(s) | reverse-vowels-of-a-string | Python Solution || 99.58% faster || 86.96% less memory | KiranUpase | 22 | 1,100 | reverse vowels of a string | 345 | 0.498 | Easy | 6,016 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2775366/python-iterator-simple-2-lines | class Solution:
def reverseVowels(self, s: str) -> str:
it = (ch for ch in s[::-1] if ch.lower() in 'aeiou')
return ''.join(next(it) if ch.lower() in 'aeiou' else ch for ch in s) | reverse-vowels-of-a-string | python iterator simple 2 lines | alvin-777 | 12 | 711 | reverse vowels of a string | 345 | 0.498 | Easy | 6,017 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2775704/easy-simple-bruteforce | class Solution:
def reverseVowels(self, s: str) -> str:
loc=[]
s=list(s)
for i in range(len(s)):
if(s[i] in "aeiouAEIOU"):
loc.append(i)
for i in range(len(loc)//2):
s[loc[i]],s[loc[-i-1]]=s[loc[-i-1]],s[loc[i]]
return "".join(s) | reverse-vowels-of-a-string | easy simple bruteforce | droj | 4 | 141 | reverse vowels of a string | 345 | 0.498 | Easy | 6,018 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2779022/Python-Simple-Python-Solution-97-ms | class Solution:
def reverseVowels(self, s: str) -> str:
l="aeiouAEIOU"
s=list(s)
i,j=0,len(s)-1
while(i<j):
if s[i] in l and s[j] in l:
s[i],s[j]=s[j],s[i]
i+=1
j-=1
elif s[i] not in l:
i+=1
elif s[j] not in l:
j-=1
return ''.join(s) | reverse-vowels-of-a-string | [ Python ] ππ Simple Python Solution β
β
97 ms | sourav638 | 2 | 11 | reverse vowels of a string | 345 | 0.498 | Easy | 6,019 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2022664/Python-2-Pointers | class Solution:
def reverseVowels(self, s: str) -> str:
i, j = 0, len(s)-1
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
s = list(s)
while i < j:
while i < j and s[i] not in vowels:
i+=1
while i < j and s[j] not in vowels:
j-=1
if i<j:
s[j], s[i] = s[i], s[j]
i+=1
j-=1
return ''.join(s) | reverse-vowels-of-a-string | Python 2 Pointers | constantine786 | 2 | 155 | reverse vowels of a string | 345 | 0.498 | Easy | 6,020 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1264637/Easy-Python-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
left=0
s=list(s)
x=['a','e','i','o','u']
right=len(s)-1
while left<right:
if(s[left].lower() in x and s[right].lower() in x):
s[left],s[right]=s[right],s[left]
left+=1
right-=1
if(s[left].lower() not in x):
left+=1
if(s[right].lower() not in x):
right-=1
s="".join(s)
return s | reverse-vowels-of-a-string | Easy Python Solution | Sneh17029 | 2 | 487 | reverse vowels of a string | 345 | 0.498 | Easy | 6,021 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/476707/Reverse-vowels-of-a-string | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = ["a","e","i","o","u","A","E","I","O","U"]
li = list(s)
i = 0
j = len(li)-1
while(i<j):
if(li[i] not in vowels and li[j] not in vowels):
i+=1
j-=1
if(li[i] not in vowels and li[j] in vowels):
i+=1
if(li[j] not in vowels and li[i] in vowels):
j-=1
if(li[i] in vowels and li[j] in vowels):
li[i],li[j] = li[j],li[i]
i+=1
j-=1
return "".join(li) | reverse-vowels-of-a-string | Reverse vowels of a string | qwe121 | 2 | 186 | reverse vowels of a string | 345 | 0.498 | Easy | 6,022 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2802052/Two-Pointer-or-Python | class Solution:
def reverseVowels(self, s: str) -> str:
v = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
l, r = 0, len(s) - 1
out = list(s)
while l < r:
if s[l] in v and s[r] in v:
out[l] = s[r]
out[r] = s[l]
l += 1
r -= 1
elif s[l] not in v:
l += 1
elif s[r] not in v:
r -= 1
return("".join(out)) | reverse-vowels-of-a-string | Two Pointer | Python | jaisalShah | 1 | 8 | reverse vowels of a string | 345 | 0.498 | Easy | 6,023 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777389/FASTEST-AND-EASIEST-oror-BEATS-99-SUBMISSIONS | class Solution:
def reverseVowels(self, s: str) -> str:
v="aeiouAEIOU"
s=list(s)
l=0
r=len(s)-1
while l<r :
if s[l] in v and s[r] in v :
s[l],s[r]=s[r],s[l]
l +=1
r -=1
elif s[l] not in v :
l +=1
else :
r -=1
return ''.join(s) | reverse-vowels-of-a-string | FASTEST AND EASIEST || BEATS 99% SUBMISSIONS | Pritz10 | 1 | 7 | reverse vowels of a string | 345 | 0.498 | Easy | 6,024 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2776981/Python-Two-Pointer-Approach-O(n) | class Solution:
def reverseVowels(self, s: str) -> str:
s_list = list(s)
vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]
start = 0
end = len(s_list) - 1
while start < end:
if (s_list[start] not in vowels):
start +=1
if (s_list[end] not in vowels):
end -= 1
if s_list[start] in vowels and s_list[end] in vowels:
s_list[start], s_list[end] = s_list[end], s_list[start]
start +=1
end -= 1
return "".join(s_list) | reverse-vowels-of-a-string | Python Two Pointer Approach O(n) | brains_Up | 1 | 24 | reverse vowels of a string | 345 | 0.498 | Easy | 6,025 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2776111/Python-2-Simple-and-Easy-Way-to-Solve-or-97-Faster | class Solution:
def reverseVowels(self, s: str) -> str:
s = list(s)
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
vowels_in_s = []
for i, c in enumerate(s):
if c in vowels:
vowels_in_s.append(c)
s[i] = None
p = 0
vowels_in_s = vowels_in_s[::-1]
for i, c in enumerate(s):
if c == None:
s[i] = vowels_in_s[p]
p+=1
return ''.join(s) | reverse-vowels-of-a-string | βοΈ Python 2 Simple and Easy Way to Solve | 97% Faster π₯ | pniraj657 | 1 | 87 | reverse vowels of a string | 345 | 0.498 | Easy | 6,026 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2776111/Python-2-Simple-and-Easy-Way-to-Solve-or-97-Faster | class Solution:
def reverseVowels(self, s: str) -> str:
v="aeiouAEIOU"
l=list(s)
i=0
j=(len(s)-1)
while i<j:
while i<j and l[i] not in v:
i+=1
while j>i and l[j] not in v:
j-=1
l[i],l[j]=l[j],l[i]
i+=1
j-=1
return "".join(l) | reverse-vowels-of-a-string | βοΈ Python 2 Simple and Easy Way to Solve | 97% Faster π₯ | pniraj657 | 1 | 87 | reverse vowels of a string | 345 | 0.498 | Easy | 6,027 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2775641/Python-easy-to-read-linear-solution | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = ['a', 'e', 'i', 'o', 'u']
s = list(s)
left, right = 0, len(s)-1
while right > left:
leftS = s[left].lower()
rightS = s[right].lower()
if leftS not in vowels:
left += 1
continue
if rightS not in vowels:
right -= 1
continue
s[left], s[right] = s[right], s[left]
left, right = left+1, right-1
return ''.join(s) | reverse-vowels-of-a-string | Python easy to read linear solution | really_cool_person | 1 | 45 | reverse vowels of a string | 345 | 0.498 | Easy | 6,028 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2775401/Simple-Python-Two-pointers-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = set('aeiouAEIOU')
s = list(s)
l, r = 0, len(s)-1
while l < r:
while s[l] not in vowels and l < r:
l+=1
while s[r] not in vowels and r > l:
r-=1
s[l], s[r] = s[r], s[l]
l+=1
r-=1
return "".join(s) | reverse-vowels-of-a-string | Simple Python Two pointers Solution | tragob | 1 | 9 | reverse vowels of a string | 345 | 0.498 | Easy | 6,029 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1871260/Python-easy-to-read-and-understand-or-stack | class Solution:
def reverseVowels(self, s: str) -> str:
vow = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
t = []
for i in s:
if i in vow:
t.append(i)
ans = ''
for i in s:
if i in vow:
ans += t.pop()
else:
ans += i
return ans | reverse-vowels-of-a-string | Python easy to read and understand | stack | sanial2001 | 1 | 162 | reverse vowels of a string | 345 | 0.498 | Easy | 6,030 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1579170/Runtime%3A-48-ms-faster-than-88.96-of-Python3 | class Solution:
def reverseVowels(self, s: str) -> str:
vowel={'a','e','i','o','u','A','E','I','O','U'}
i=0
j=len(s)-1
s=list(s)
while i<j:
if s[i] in vowel and s[j] in vowel:
s[i],s[j]=s[j],s[i]
i+=1
j-=1
elif s[i] in vowel and s[j] not in vowel:
j-=1
elif s[i] not in vowel and s[j] in vowel:
i+=1
else:
i+=1
j-=1
return "".join(s) | reverse-vowels-of-a-string | Runtime: 48 ms, faster than 88.96% of Python3 | topensite | 1 | 110 | reverse vowels of a string | 345 | 0.498 | Easy | 6,031 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1500820/Build-Approach-from-Scratch-greater-Thought-Process | class Solution:
def reverseVowels(self, s: str) -> str:
arr = list(s) #arr = ["h", "e", "l", "l", "o"]
n = len(arr)
vowels = {"a", "e", "i", "o", "u", "A", "E", "I", "O", "U"} #checking an element across a set is O(1)
#initialising Two Pointers - one to move from front and other to move from back
i = 0 #this will move from 0th index - will take care from front
j = n - 1 #this will move from (n-1)th index - will take care from back
#Termination Condition is i < j because as soon as i >= j, it means we have covered all the elements
#from front as well as back
while i < j :
#if we do not find a vowel while moving from front, lets move ahead
if arr[i] not in vowels:
i += 1
#if we do not find a vowel while moving from back, lets move ahead
if arr[j] not in vowels:
j -= 1
#if we find a vowel from front as well as from behind, we have to swap them
#and then, move ahead from front and move back from behind by 1 unit respectively
#Note - The termination condition will take care of any eventuality
if arr[i] in vowels and arr[j] in vowels:
arr[i],arr[j] = arr[j],arr[i] #swapping them
i += 1
j -= 1
answer = ""
#we will iterate over the array and keep adding all elements to a string
#as output is required in the form of a string
for i in arr:
answer += i
return answer | reverse-vowels-of-a-string | Build Approach from Scratch --> Thought Process | aarushsharmaa | 1 | 97 | reverse vowels of a string | 345 | 0.498 | Easy | 6,032 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1392280/WEEB-DOES-PYTHON-(BEATS-99.71) | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = set(list("AEIOUaeiou"))
high, s = len(s)-1, list(s)
for low in range(len(s)):
if s[low] in vowels:
while s[high] not in vowels:
high-=1
if low == high:
break
# print("b4: ", s[low], s[high])
s[low], s[high] = s[high], s[low]
# print("after: ", s[low], s[high])
high-=1
if low == high:
break
return "".join(s) | reverse-vowels-of-a-string | WEEB DOES PYTHON (BEATS 99.71%) | Skywalker5423 | 1 | 199 | reverse vowels of a string | 345 | 0.498 | Easy | 6,033 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1239737/Python3-simple-solution-using-two-pointer-approach | class Solution:
def reverseVowels(self, s: str) -> str:
i,j=0,len(s)-1
l = list(s)
vowels = ['a','e','i','o','u']
while i < j:
if l[i].lower() not in vowels:
i += 1
if l[j].lower() not in vowels:
j -= 1
if l[i].lower() in vowels and l[j].lower() in vowels:
l[i],l[j] = l[j],l[i]
i += 1
j -= 1
return ''.join(l) | reverse-vowels-of-a-string | Python3 simple solution using two pointer approach | EklavyaJoshi | 1 | 69 | reverse vowels of a string | 345 | 0.498 | Easy | 6,034 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/807615/simple-of-simple | class Solution:
def reverseVowels(self, s: str) -> str:
words = [x for x in s]
vowels = [x for x in 'aeiou']
t = []
for i in range(len(words)):
if words[i].lower() in vowels:
t.append(words[i])
words[i] = None
for i in range(len(words)):
if words[i] is None:
words[i] = t.pop()
return ''.join(words) | reverse-vowels-of-a-string | simple of simple | seunggabi | 1 | 57 | reverse vowels of a string | 345 | 0.498 | Easy | 6,035 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/737069/Python-Two-Pointers | class Solution:
def reverseVowels(self, s: str) -> str:
# Two pointers solution
s = list(s)
i, j = 0, len(s) -1
vowels = set("aeiouAEIOU")
while i < j:
# Swap
if s[i] in vowels and s[j] in vowels:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1
# Decrement by 1
elif s[j] not in vowels:
j -= 1
# Increment by 1
elif s[i] not in vowels:
i += 1
return "".join(s) | reverse-vowels-of-a-string | Python Two Pointers | adhishthite | 1 | 202 | reverse vowels of a string | 345 | 0.498 | Easy | 6,036 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2842645/Python-2-Pointer-Easy-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
s=list(s)
ot=""
l=[]
vowel=["a","e","i","o","u","A","E","I","O","U"]
for i in range(len(s)):
if s[i] in vowel :
l.append(i)
for i in range(len(l)//2) :
s[l[i]],s[l[len(l)-i-1]]=s[l[len(l)-1-i]],s[l[i]]
het=""
return(het.join(s)) | reverse-vowels-of-a-string | Python 2 Pointer Easy Solution | patelhet050603 | 0 | 1 | reverse vowels of a string | 345 | 0.498 | Easy | 6,037 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2838706/Python3-Clean-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
i = 0
j = len(s)-1
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
s = list(s)
while i <= j:
ci = s[i]
cj = s[j]
if s[i] in vowels and s[j] in vowels:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1
elif s[i] in vowels and s[j] not in vowels:
j -= 1
elif s[i] not in vowels and s[j] in vowels:
i += 1
else:
j -= 1
i += 1
return "".join(s) | reverse-vowels-of-a-string | β
Python3 Clean Solution | Cecilia_GuoChen | 0 | 1 | reverse vowels of a string | 345 | 0.498 | Easy | 6,038 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2810694/easy-solution-in-python | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = ['a', 'e', 'i', 'o','u', 'A', 'E', 'I', 'O', 'U']
vs = ''
vc = 0
res = ''
for i in range(0, len(s)):
if s[i] in vowels:
vs+=s[i]
vs = vs[::-1]
for i in range(0, len(s)):
if s[i] in vowels:
res+=vs[vc]
vc+=1
continue
res+=s[i]
return res | reverse-vowels-of-a-string | easy solution in python | user1366SF | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,039 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2791882/PYTHON91.54-FASTEREXPLAINED. | class Solution:
def reverseVowels(self, s: str) -> str:
chars = list(s)
N = len(chars)
left = 0
right = len(s)-1
vowels = "aeiou"
while left < right:
while left < N and chars[left].lower() not in vowels:
left += 1
while right >=0 and chars[right].lower() not in vowels:
right -= 1
if left < right:
chars[left], chars[right] = chars[right], chars[left]
left += 1
right -= 1
return "".join(chars) | reverse-vowels-of-a-string | PYTHONπ91.54% FASTERβ
EXPLAINEDπ₯. | shubhamdraj | 0 | 5 | reverse vowels of a string | 345 | 0.498 | Easy | 6,040 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2788241/Python-Iterative-solution-(Easy-to-understand) | class Solution:
def reverseVowels(self, s: str) -> str:
w = ['a', 'e', 'i', 'o','u',"A","E","I","O","U"]
wl = ""
tw = ''
for i in range(len(s)):
if(s[i] in w):
tw+="="
wl+=s[i]
else:
tw+=s[i]
wl = wl[::-1]
out =""
cnt=0
for i in range(len(tw)):
if(tw[i]=="="):
out+=wl[cnt]
cnt+=1
else:
out+=tw[i]
return out | reverse-vowels-of-a-string | [Python] Iterative solution (Easy to understand) | thesaderror | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,041 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2785560/Two-Pointers-Simple-or-Python | class Solution:
def reverseVowels(self, s: str) -> str:
s = list(s)
l, r = 0, len(s)-1
vows = 'aeiouAEIOU'
while l <r:
while l< r and s[l] not in vows:
l +=1
while l < r and s[r] not in vows:
r -= 1
if l > r:
break
s[l],s[r] = s[r],s[l]
l += 1
r -= 1
return ''.join(s) | reverse-vowels-of-a-string | Two Pointers Simple | Python | Abhi_-_- | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,042 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2782558/Python-2-pointer-approach-w.-intuition-and-approach-explanation | class Solution:
def reverseVowels(self, s: str) -> str:
i = 0
j = len(s) - 1
s = list(s)
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
while i < j:
if s[i] in vowels and s[j] in vowels:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1
if s[i] not in vowels:
i += 1
if s[j] not in vowels:
j -= 1
return "".join(s) | reverse-vowels-of-a-string | Python - 2 pointer approach w. intuition & approach explanation | abhandar | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,043 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2781403/Python3-Two-Pointers-or-Reverse-and-Replace-(2-approaches)-O(n) | class Solution:
def reverseVowels(self, s: str) :
ss_list = []
for item in s:
ss_list.append(item)
data_vowel = "aeiouAEIOU"
pL,pR = 0,len(ss_list)-1
while(pL<pR):
if(ss_list[pL] in data_vowel and ss_list[pR] in data_vowel):
temp = ss_list[pR]
ss_list[pR]=ss_list[pL]
ss_list[pL]=temp
pL+=1
pR-=1
continue
if(ss_list[pL] not in data_vowel):
pL+=1
if(ss_list[pR] not in data_vowel):
pR-=1
return "".join(ss_list) | reverse-vowels-of-a-string | [Python3] Two Pointers | Reverse and Replace (2 approaches) O(n) | worachote659 | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,044 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2779272/Two-Pointers-Solution-or-Python3 | class Solution:
def reverseVowels(self, s: str) -> str:
l, r = 0, len(s) - 1
vowels = {'a', 'e', 'o', 'i', 'u', 'A', 'E', 'O', 'U', 'I'}
s = list(s)
if r == 0:
return ''.join(s)
while l < r:
if s[l] in vowels and s[r] in vowels:
s[l], s[r] = s[r], s[l]
l += 1
r -=1
elif s[l] in vowels and s[r] not in vowels:
r -= 1
elif s[l] not in vowels and s[r] in vowels:
l += 1
else:
l += 1
r -=1
return ''.join(s) | reverse-vowels-of-a-string | Two Pointers Solution | Python3 | omniaosman4 | 0 | 6 | reverse vowels of a string | 345 | 0.498 | Easy | 6,045 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2779189/Python-or-LeetCode-or-345.-Reverse-Vowels-of-a-String | class Solution:
def reverseVowels(self, s: str) -> str:
# vowel_letters = v
v = "aeiouAEIOU"
left = 0
right = len(s) - 1
s = list(s)
while left < right:
if (s[left] in v) and (s[right] in v):
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
elif (s[left] in v) and (not (s[right] in v)):
right -= 1
elif (not (s[left] in v)) and (s[right] in v):
left += 1
else:
left += 1
right -= 1
return "".join(s) | reverse-vowels-of-a-string | Python | LeetCode | 345. Reverse Vowels of a String | UzbekDasturchisiman | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,046 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778900/Python-O(n)-with-stack | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = ['a', 'e', 'i', 'o', 'u']
stack = []
[stack.append(x) for x in s if x.lower() in vowels]
print(stack)
result = ""
for letter in s:
if letter.lower() in vowels:
letter = stack.pop()
result += letter
return result | reverse-vowels-of-a-string | Python O(n) with stack | suhika | 0 | 5 | reverse vowels of a string | 345 | 0.498 | Easy | 6,047 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778885/PYTHON-159-ms | class Solution:
def reverseVowels(self, s: str) -> str:
i = 0
j = len(s) - 1
vowels = list("aeiouAEIOU")
s = list(s)
while i<=j:
if s[i] in vowels:
while s[j] not in vowels and j>=i:
j-=1
s[i],s[j] = s[j],s[i]
j-=1
i+=1
return "".join(s) | reverse-vowels-of-a-string | β
PYTHON 159 ms | logolica99 | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,048 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778681/Simple-Two-Pointer-Approach-Using-List | class Solution:
def reverseVowels(self, s: str) -> str:
left, right = 0, len(s)-1
vowels = {'a','e','i','o','u','A', 'E', 'I', 'O', 'U'}
s = list(s)
while left < right:
# get vowel from left
while left < right and s[left] not in vowels:
left += 1
# print('[left] found vowel: ', s[left])
# get vowel from right
while left < right and s[right] not in vowels:
right -= 1
# print('[right] found vowel: ', s[right])
if left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
return ''.join(s) | reverse-vowels-of-a-string | π Simple Two Pointer Approach Using List | pruthvi_hingu | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,049 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778641/Easy-Python-Approach | class Solution:
def reverseVowels(self, s: str) -> str:
vo = "aeiouAEIOU"
sv = []
for i in s:
if i in vo:
sv.append(i)
sv = sv[::-1]
zi = 0
ns = ""
for i in range(0,len(s)):
if s[i] in vo:
ns += sv[zi]
zi+= 1
else:
ns+=s[i]
return ns | reverse-vowels-of-a-string | Easy Python Approach | Shagun_Mittal | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,050 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778477/Python-3-Two-Index-Solution%3A-O(n) | class Solution:
def reverseVowels(self, s: str) -> str:
result = [c for c in s]
left = 0
right = len(s) - 1
vowels = set([v for v in "aeiouAEIOU"])
while left < right:
if result[left] not in vowels:
left += 1
continue
if result[right] not in vowels:
right -= 1
continue
tmp = result[left]
result[left] = result[right]
result[right] = tmp
left += 1
right -= 1
return ''.join(result) | reverse-vowels-of-a-string | Python 3 Two-Index Solution: O(n) | theleastinterestingman | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,051 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778364/Python-(Faster-than-greater90)-or-Two-pointers-or-Stack | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = 'aeiouAEIOU'
string = list(s)
l, r = 0, len(s) - 1
while l < r:
while l < len(s) and string[l] not in vowels:
l += 1
while r >= 0 and string[r] not in vowels:
r -= 1
if l < r:
string[l], string[r] = string[r], string[l]
l += 1
r -= 1
return ''.join(string) | reverse-vowels-of-a-string | Python (Faster than >90%) | Two-pointers | Stack | KevinJM17 | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,052 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778364/Python-(Faster-than-greater90)-or-Two-pointers-or-Stack | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = 'aeiouAEIOU'
string = list(s)
stack = []
for c in s:
if c in vowels:
stack.append(c)
for i in range(len(string)):
if string[i] in vowels:
string[i] = stack.pop()
return ''.join(string) | reverse-vowels-of-a-string | Python (Faster than >90%) | Two-pointers | Stack | KevinJM17 | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,053 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778328/python3-2-line-solution | class Solution:
def reverseVowels(self, s: str, vwls = list("aeiouAEIOU")) -> str:
stack = [v for v in s if v in vwls]
return "".join(c if c not in vwls else stack.pop() for c in s) | reverse-vowels-of-a-string | python3 2 line solution | avs-abhishek123 | 0 | 6 | reverse vowels of a string | 345 | 0.498 | Easy | 6,054 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778315/Simple-stack-solution-or-Python3 | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = set('aeiouAEIOU')
s = list(s)
indices = []
stack = []
for i in range(len(s)):
if s[i] in vowels:
indices.append(i)
stack.append(s[i])
for index in indices:
s[index] = stack.pop()
return ''.join(s) | reverse-vowels-of-a-string | Simple stack solution | Python3 | saamenerve | 0 | 5 | reverse vowels of a string | 345 | 0.498 | Easy | 6,055 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778302/easy-thinking | class Solution:
def reverseVowels(self, s: str) -> str:
l, r = 0, len(s)-1
vowel = set(('a', 'e', 'i', 'o', 'u'))
s = list(s)
while l <= r:
if s[r].lower() in vowel:
if s[l].lower() in vowel:
s[l], s[r] = s[r], s[l]
l +=1
r-=1
else:
l+=1
elif s[l].lower() in vowel:
if s[r].lower() in vowel:
s[l], s[r] = s[r], s[l]
l +=1
r-=1
else:
r-=1
else:
r-=1
l+=1
return "".join(s) | reverse-vowels-of-a-string | easy thinking | fsubhani | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,056 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778268/Clean-solution-or-Easy-understanding | class Solution:
def reverseVowels(self, s: str) -> str:
volwels={'a','e','i','o','u','A','E','I','O','U'}
temp=[]
res=""
for i in range(len(s)):
if s[i] in volwels:
temp.append(s[i])
for i in range(len(s)):
if s[i] in volwels:
res+=temp.pop()
else:
res+=s[i]
return res | reverse-vowels-of-a-string | Clean solution | Easy understanding | sundram_somnath | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,057 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778200/Python3-Simple-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
vowels = "aeiouAEIOU"
vowel_letters = [l for l in s if l in vowels]
new_word = ""
idx = -1
for l in s:
if l in vowels:
new_word += vowel_letters[idx]
idx += -1
else:
new_word += l
return new_word | reverse-vowels-of-a-string | Python3 Simple Solution | Nadhir_Hasan | 0 | 1 | reverse vowels of a string | 345 | 0.498 | Easy | 6,058 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778159/LeetCode-345-%3A-Reverse-Vowels-of-a-String | class Solution:
def reverseVowels(self, s: str) -> str:
s=list(s)
vowel=set(list("aeiouAEIOU"))
left=0
right=len(s)-1
while left<right:
while left<right and s[left] not in vowel:
left+=1
while left<right and s[right] not in vowel:
right-=1
s[left],s[right]=s[right],s[left]
left+=1
right-=1
return ''.join(s) | reverse-vowels-of-a-string | LeetCode 345 : Reverse Vowels of a String | Nakshu_Python | 0 | 7 | reverse vowels of a string | 345 | 0.498 | Easy | 6,059 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778120/Python-oror-Two-Pointers-oror-HashSet | class Solution:
def reverseVowels(self, s: str) -> str:
arr = [v for v in s]
vowels = set(["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"])
l = 0
r = len(arr) - 1
while l < r:
if arr[l] in vowels and arr[r] in vowels:
arr[l], arr[r] = arr[r], arr[l]
l += 1
r -= 1
elif arr[l] in vowels and arr[r] not in vowels:
r -= 1
elif arr[l] not in vowels and arr[r] in vowels:
l += 1
else:
l += 1
r -= 1
return ''.join(arr) | reverse-vowels-of-a-string | Python || Two Pointers || HashSet | Vamsi995 | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,060 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778115/Simple-Solution | class Solution:
def reverseVowels(self, s: str) -> str:
vow=['a','e','i','o','u','A','E','I','O','U']
l=list(s)
l1=[]
i1=[]
for i in range(len(s)):
if l[i] in vow:
l1.append(s[i])
i1.append(i)
l1=l1[::-1]
lr=l
for i in range(len(i1)):
lr[i1[i]]=l1[i]
s1=''
for i in lr:
s1=s1+i
return s1 | reverse-vowels-of-a-string | Simple Solution | SnehaGanesh | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,061 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778067/Python%3A-Single-Pass-Two-Pointers | class Solution:
def reverseVowels(self, s: str) -> str:
# One time set conversion of string for o(1) look-ups
vowels = set('aeiouAEIOU')
# Conversion of string to list for ability to assign
s = list(s)
# Our two pointers
left, right = 0, len(s)-1
# We exit once the two pointers are at the same point
while left < right:
if s[left] in vowels and s[right] in vowels:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
elif s[left] in vowels:
right -= 1
continue
else:
left += 1
continue
return ''.join(s) | reverse-vowels-of-a-string | Python: Single-Pass, Two-Pointers | jessewalker2010 | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,062 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778059/Python | class Solution:
def reverseVowels(self, s: str) -> str:
vow = {"a","e","i","o","u","A","E","I","O","U"}
arr=[i for i in s]
lptr = 0
rptr = len(arr)-1
while lptr<=rptr:
while arr[lptr] not in vow:
lptr+=1
if lptr>rptr or lptr>len(arr):
return "".join(arr)
while arr[rptr] not in vow:
rptr-=1
if lptr>rptr or rptr<0 :
return "".join(arr)
temp = arr[lptr]
arr[lptr]=arr[rptr]
arr[rptr]=temp
lptr+=1
rptr-=1
return "".join(arr) | reverse-vowels-of-a-string | Python | shzaheer514 | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,063 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2778015/Very-Bad-But-Easy-to-Understand | class Solution:
def reverseVowels(self, s: str) -> str:
found = ""
vowels = "EOAUIeaiou"
for i in s :
if i in vowels :
found += i
s = s.replace(i, "*")
found = found[::-1]
j = 0
for i in s :
if i == "*" :
s = s.replace(i, found[j] , 1)
j+=1
return s | reverse-vowels-of-a-string | Very Bad But Easy to Understand | nedaladham8 | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,064 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777946/Slow-but-simple-Python-O(2n) | class Solution:
def reverseVowels(self, s: str) -> str:
_vowels = ["a","e","i","o","u"]
_v = []
for i in range(len(s)):
if s[i].lower() in _vowels:
_v.append(s[i])
s = s[:i]+"*"+s[i+1:]
_v = _v[::-1]
for i in range(len(s)):
if s[i] == "*":
s= s[:i] + _v[0] + s[i+1:]
_v = _v[1:]
return s | reverse-vowels-of-a-string | Slow but simple Python O(2n) | ATHBuys | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,065 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777931/Py-Best-solution-Easy-Approch | class Solution:
def reverseVowels(self, s: str) -> str:
s=list(s)
i=0
j=len(s)-1
v=["a","e","i","o","u","A","E","I","O","U"]
while(i<j):
if s[i] in v and s[j] in v:
s[i],s[j]=s[j],s[i]
i+=1
j-=1
if s[i] not in v:
i+=1
if s[j] not in v:
j-=1
return "".join(s) | reverse-vowels-of-a-string | Py Best solution Easy Approch π―β
β
| adarshg04 | 0 | 4 | reverse vowels of a string | 345 | 0.498 | Easy | 6,066 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777758/Two-pointers-solution | class Solution:
def reverseVowels(self, s: str) -> str:
left = 0
right = len(s)-1
vowels = set(['a','e','i','o','u', 'A','E','I','O','U'])
s = list(s)
while left<right:
lv = s[left] in vowels
rv = s[right] in vowels
if (lv and rv):
s[left], s[right] = s[right], s[left]
left+=1
right-=1
else:
left+=int(not lv)
right-=int(not rv)
return "".join(s) | reverse-vowels-of-a-string | Two pointers solution | user9698Ra | 0 | 12 | reverse vowels of a string | 345 | 0.498 | Easy | 6,067 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777669/Python3-Solution-with-using-two-pointers | class Solution:
def reverseVowels(self, s: str) -> str:
it1, it2 = 0, len(s) - 1
vowels = set(['a', 'A', 'e', 'E', 'I', 'i', 'o', 'O', 'u', 'U'])
s = list(s)
while it1 < it2:
while it1 < it2 and s[it1] not in vowels:
it1 += 1
while it1 < it2 and s[it2] not in vowels:
it2 -= 1
s[it1], s[it2] = s[it2], s[it1]
it1 += 1
it2 -= 1
return ''.join(s) | reverse-vowels-of-a-string | [Python3] Solution with using two-pointers | maosipov11 | 0 | 3 | reverse vowels of a string | 345 | 0.498 | Easy | 6,068 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777640/Two-pointers-approach | class Solution:
def reverseVowels(self, s: str) -> str:
v = ['a', 'e' , 'i' , 'o' , 'u' , 'A' , 'E' , 'I' , 'O' , 'U']
i = 0
j = len(s) - 1
while (i < j) :
while (i < j) :
if s[i] in v : break
else :
i += 1
while (i < j) :
if s[j] in v :
break
else :
j -= 1
if (i < j) :
s = s[:i]+s[j]+s[i+1:j]+s[i] + s[j+1:]
i += 1
j -= 1
else : break
return s | reverse-vowels-of-a-string | Two pointers approach | bhavithas153 | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,069 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777587/fast | class Solution:
def reverseVowels(self, s: str) -> str:
yuan = 'aeiouAEIOU'
s = list(s)
s_yuan = []
index_list=[]
index = 0
for i in s:
if i in yuan:
s_yuan.append(i)
index_list.append(index)
index += 1
s_yuan.reverse()
index = 0
for c in s_yuan:
s[index_list[index]]=c
index+=1
return ''.join(s) | reverse-vowels-of-a-string | fast | hlwin | 0 | 2 | reverse vowels of a string | 345 | 0.498 | Easy | 6,070 |
https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/2777449/simple-python-solution-Beats-86.15-Memory-14.9-MB-Beats-89.63 | class Solution:
def reverseVowels(self, s: str) -> str:
s = list(s)
vowel = list('AEIOU')
left_counter, right_counter = 0, len(s) - 1
l_val, r_val = None, None
while left_counter<right_counter:
if s[left_counter].upper() in vowel:
l_val = left_counter
elif l_val is None:
left_counter += 1
if s[right_counter].upper() in vowel:
r_val = right_counter
elif r_val is None:
right_counter -= 1
if l_val is not None and r_val is not None:
s[l_val], s[r_val] = s[r_val], s[l_val]
l_val = r_val = None
left_counter +=1
right_counter -=1
if left_counter > right_counter:
break
return "".join(s) | reverse-vowels-of-a-string | simple python solution Beats 86.15% Memory 14.9 MB Beats 89.63% | lakhannawalani911 | 0 | 1 | reverse vowels of a string | 345 | 0.498 | Easy | 6,071 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1928198/Python-Simple-Python-Solution-Using-Dictionary-(-HashMap-) | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
frequency = {}
for num in nums:
if num not in frequency:
frequency[num] = 1
else:
frequency[num] = frequency[num] + 1
frequency = dict(sorted(frequency.items(), key=lambda x: x[1], reverse=True))
result = list(frequency.keys())[:k]
return result | top-k-frequent-elements | [ Python ] β
β
Simple Python Solution Using Dictionary ( HashMap ) βπ | ASHOK_KUMAR_MEGHVANSHI | 24 | 2,800 | top k frequent elements | 347 | 0.648 | Medium | 6,072 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1927383/Python-one-liner-beats-98 | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return [x[0] for x in Counter(nums).most_common(k)] | top-k-frequent-elements | Python one-liner beats 98% | eduardocereto | 13 | 2,100 | top k frequent elements | 347 | 0.648 | Medium | 6,073 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1705495/Python-or-4-Ways-of-doing-same-simple-thing | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
freq_table = Counter(nums)
ans_table = freq_table.most_common()
ans = []
for key, _ in ans_table:
if k <= 0:
break
k -= 1
ans.append(key)
return ans | top-k-frequent-elements | Python | 4 Ways of doing same simple thing | anCoderr | 11 | 939 | top k frequent elements | 347 | 0.648 | Medium | 6,074 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1705495/Python-or-4-Ways-of-doing-same-simple-thing | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
freq_table = Counter(nums)
heap = []
for i in freq_table.keys():
heappush(heap, (freq_table[i], i))
freq_table = nlargest(k,heap)
ans = []
for i, j in freq_table:
ans.append(j)
return ans | top-k-frequent-elements | Python | 4 Ways of doing same simple thing | anCoderr | 11 | 939 | top k frequent elements | 347 | 0.648 | Medium | 6,075 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1705495/Python-or-4-Ways-of-doing-same-simple-thing | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
freq_table = Counter(nums)
heap = []
for i in freq_table.keys():
heappush(heap, (-freq_table[i], i))
ans = []
while k > 0:
k -= 1
ans.append(heappop(heap)[1])
return ans | top-k-frequent-elements | Python | 4 Ways of doing same simple thing | anCoderr | 11 | 939 | top k frequent elements | 347 | 0.648 | Medium | 6,076 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1705495/Python-or-4-Ways-of-doing-same-simple-thing | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
freq_table = {}
for i in nums:
freq_table[i] = freq_table.get(i, 0) + 1
heap = []
for i in freq_table.keys():
if len(heap) == k: # If size is k then we dont want to increase the size further
heappushpop(heap, (freq_table[i], i))
else: # Size is not k then freely push values
heappush(heap, (freq_table[i], i))
# After this operation the heap contains only k largest values of all the values in nums
ans = []
while k > 0:
k -= 1
ans.append(heappop(heap)[1])
return ans | top-k-frequent-elements | Python | 4 Ways of doing same simple thing | anCoderr | 11 | 939 | top k frequent elements | 347 | 0.648 | Medium | 6,077 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1507137/96-faster-oror-All-three-Approaches-oror-Well-Coded | class Solution:
def topKFrequent(self, nums, k):
bucket = [[] for _ in range(len(nums) + 1)]
Count = Counter(nums)
for num, freq in Count.items():
bucket[freq].append(num)
flat_list = list(chain(*bucket))
return flat_list[::-1][:k] | top-k-frequent-elements | ππ 96% faster || All-three Approaches || Well-Coded π | abhi9Rai | 5 | 699 | top k frequent elements | 347 | 0.648 | Medium | 6,078 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1507137/96-faster-oror-All-three-Approaches-oror-Well-Coded | class Solution:
def topKFrequent(self, nums: List[int], size: int) -> List[int]:
dic = Counter(nums)
heap = []
heapq.heapify(heap)
for k in dic:
if len(heap)<size:
heapq.heappush(heap,[dic[k],k])
elif dic[k]>heap[0][0]:
heapq.heappop(heap)
heapq.heappush(heap,[dic[k],k])
res = []
while heap:
res.append(heapq.heappop(heap)[1])
return res | top-k-frequent-elements | ππ 96% faster || All-three Approaches || Well-Coded π | abhi9Rai | 5 | 699 | top k frequent elements | 347 | 0.648 | Medium | 6,079 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1507137/96-faster-oror-All-three-Approaches-oror-Well-Coded | class Solution:
def topKFrequent(self, nums: List[int], size: int) -> List[int]:
dic = Counter(nums)
res = []
dic = sorted(dic.items(),key = lambda x:x[1],reverse = True)
for k in dic:
res.append(k[0])
size-=1
if size<=0:
return res
return res | top-k-frequent-elements | ππ 96% faster || All-three Approaches || Well-Coded π | abhi9Rai | 5 | 699 | top k frequent elements | 347 | 0.648 | Medium | 6,080 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2133564/Python3-O(n)-Time-Optimized-Solution-Easy-to-understand | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
# nums = [1,1,1,2,2,3]; k = 2
numCountDict = {} # key = num; value = count
for num in nums:
if num not in numCountDict: numCountDict[num] = 1
else: numCountDict[num] += 1
# numCountDict = {1:3, 2:2, 3:1}
maxCount = max(numCountDict.values()) # 3
countNumDict = {i:[] for i in range(1, maxCount+1)} # key = count; value = [num,...]
for num in numCountDict:
countNumDict[numCountDict[num]].append(num)
# countNumDict = [3:[1], 2:[2], 1:[3]]
res = []
for count in range(maxCount, 0, -1): # count = 3, 2, 1
if len(res) >= k: break
if len(countNumDict[count]) > 0:
res += countNumDict[count]
return res
# Time: O(N)
# SPace: O(N) | top-k-frequent-elements | [Python3] O(n) Time Optimized Solution Easy to understand | samirpaul1 | 4 | 375 | top k frequent elements | 347 | 0.648 | Medium | 6,081 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1180163/Simple-Python-O(n)-Solution-Heaps-(Beats-99.97) | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
d = dict(collections.Counter(nums))
heap = []
for key, val in d.items():
if len(heap) == k:
heapq.heappushpop(heap, (val,key))
else:
heapq.heappush(heap, (val,key))
return [y for x,y in heap] | top-k-frequent-elements | Simple Python O(n) Solution - Heaps (Beats 99.97%) | zealorez | 4 | 545 | top k frequent elements | 347 | 0.648 | Medium | 6,082 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2416406/2-Python-solutions | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
stats = {}
for i in nums:
if i not in stats:
stats[i] = 1
else:
stats[i] += 1
buckets = [[] for i in range(len(nums)+1)]
for i, j in stats.items():
buckets[j].append(i)
buckets = [j for i in buckets for j in i]
return buckets[::-1][:k] | top-k-frequent-elements | π 2 Python solutions | croatoan | 2 | 85 | top k frequent elements | 347 | 0.648 | Medium | 6,083 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2416406/2-Python-solutions | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return [_[0] for _ in collections.Counter(nums).most_common(k)] | top-k-frequent-elements | π 2 Python solutions | croatoan | 2 | 85 | top k frequent elements | 347 | 0.648 | Medium | 6,084 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2310619/Two-Python-Solutions-or-Heap-or-Bucket-Sort | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
num_dict = {}
for num in nums:
if num not in num_dict:
num_dict[num] = 1
else:
num_dict[num] += 1
heap = []
for key,val in num_dict.items():
heap.append((-val, key))
heapq.heapify(heap)
res = []
for _ in range(k):
_,ans = heapq.heappop(heap)
res.append(ans)
return res | top-k-frequent-elements | Two Python Solutions | Heap | Bucket Sort | PythonicLava | 2 | 285 | top k frequent elements | 347 | 0.648 | Medium | 6,085 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2310619/Two-Python-Solutions-or-Heap-or-Bucket-Sort | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
buckets = [[] for _ in range(len(nums)+1)]
num_dict = {}
for num in nums:
if num not in num_dict:
num_dict[num] = 1
else:
num_dict[num] += 1
for key,value in num_dict.items():
buckets[value].append(key)
res = []
for i in range(len(buckets)-1, 0, -1):
for val in buckets[i]:
res.append(val)
if len(res) == k:
return res | top-k-frequent-elements | Two Python Solutions | Heap | Bucket Sort | PythonicLava | 2 | 285 | top k frequent elements | 347 | 0.648 | Medium | 6,086 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/740463/PythonPython3-Top-K-Frequent-Elements | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return [x for x,_ in sorted(Counter(nums).items(), key=lambda x:-x[1])[:k]] | top-k-frequent-elements | [Python/Python3] Top K Frequent Elements | newborncoder | 2 | 909 | top k frequent elements | 347 | 0.648 | Medium | 6,087 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2302645/Hashmap-easy-python | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
dict_count = {}
for i in nums:
if i in dict_count:
dict_count[i]+=1
else:
dict_count[i]=1
rev_count = {}
for i in dict_count:
if dict_count[i] in rev_count:
rev_count[dict_count[i]].append(i)
else:
rev_count[dict_count[i]] = [i]
ans = []
for c in range(len(nums),-1,-1):
if c in rev_count:
ans.extend(rev_count[c])
if len(ans)>=k:
return ans[:k]
return ans[:k] | top-k-frequent-elements | Hashmap easy python | sunakshi132 | 1 | 87 | top k frequent elements | 347 | 0.648 | Medium | 6,088 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2243515/Python-Solution | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
helper=collections.Counter(nums)
elements = helper.most_common(k)
result = []
for tup in elements:
result.append(tup[0])
return result | top-k-frequent-elements | Python Solution | rtyagi1 | 1 | 86 | top k frequent elements | 347 | 0.648 | Medium | 6,089 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/2081265/Python3-or-Heap-%2B-Counter-or-Easy-Understand | class Solution
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
count = Counter(nums)
heap = [[-cnt, value] for value, cnt in count.items()]
heapify(heap)
result = []
for _ in range(k):
_, num = heappop(heap)
result.append(num)
return result | top-k-frequent-elements | Python3 | Heap + Counter | Easy Understand | itachieve | 1 | 58 | top k frequent elements | 347 | 0.648 | Medium | 6,090 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1947831/Optimal-O(n)-Time-Solution | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
counts = defaultdict(int)
freq = defaultdict(list)
max_count = 0
result = []
for num in nums:
counts[num] += 1
max_count = max(max_count, counts[num])
for key in counts:
freq[counts[key]].append(key)
while k > 0:
values = freq[max_count]
result += values[:k]
k -= len(values)
max_count -= 1
return result | top-k-frequent-elements | Optimal O(n) Time Solution | EdwinJagger | 1 | 86 | top k frequent elements | 347 | 0.648 | Medium | 6,091 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1929249/Python-or-Buckets-or-Beats-96 | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
count = Counter(nums)
max_fq, min_fq = float('-inf'), float('inf')
for x in count:
max_fq = max(max_fq, count[x])
min_fq = min(min_fq, count[x])
listMap , ans = defaultdict(list), []
for x in count.items():
listMap[x[1]].append(x[0])
for i in range(max_fq, min_fq-1, -1):
itemList = listMap.get(i)
if not itemList:
continue
for item in itemList:
ans.append(item)
if len(ans) == k:
return ans
return ans | top-k-frequent-elements | Python | Buckets | Beats 96% | prajyotgurav | 1 | 73 | top k frequent elements | 347 | 0.648 | Medium | 6,092 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1929004/Single-line-python-solution | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return list(dict(sorted(collections.Counter(nums).items(),key=lambda x:x[1],reverse=True)).keys())[:k] | top-k-frequent-elements | Single line python solution | amannarayansingh10 | 1 | 47 | top k frequent elements | 347 | 0.648 | Medium | 6,093 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1928908/Python-solution-Explained-with-diagram-or-hashmap-or-Easy-understanding | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
dict={}
freq=[[] for i in range(len(nums)+1)]
for num in nums:
dict[num]= 1+dict.get(num,0)
for value,key in dict.items():
freq[key].append(value)
res =[]
for i in range(len(freq)-1,0,-1):
for value in freq[i]:
res.append(value)
if len(res)==k:
return(res) | top-k-frequent-elements | Python solution Explained with diagram | hashmap | Easy understanding | poojatripathi0_0 | 1 | 18 | top k frequent elements | 347 | 0.648 | Medium | 6,094 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1928411/Python3-Solution-with-using-heap | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
c = collections.Counter(nums)
heap = []
for key in c:
heapq.heappush(heap, (c[key], key))
if len(heap) > k:
heapq.heappop(heap)
return [v[1] for v in heap] | top-k-frequent-elements | [Python3] Solution with using heap | maosipov11 | 1 | 85 | top k frequent elements | 347 | 0.648 | Medium | 6,095 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1928325/Python-Solution-or-One-Liner-or-collections.Counter | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
return [element[0] for element in Counter(nums).most_common(k)] | top-k-frequent-elements | Python Solution | One Liner | collections.Counter | Gautam_ProMax | 1 | 73 | top k frequent elements | 347 | 0.648 | Medium | 6,096 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1891992/Python-3-or-Counter-or-Heap-or-Priority-Queue-or-Easy-to-Understand | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
c, temp, ans = collections.Counter(nums), [], []
for i, j in zip(c.keys(), c.values()):
temp.append((j, i))
heapq._heapify_max(temp)
for i in range(k):
t = heapq._heappop_max(temp)
ans.append(t[1])
return ans | top-k-frequent-elements | Python 3 | Counter | Heap | Priority Queue | Easy to Understand | milannzz | 1 | 68 | top k frequent elements | 347 | 0.648 | Medium | 6,097 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1708324/3-liner-in-Python-with-explanation | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
# (number, count)
# e.g. [(1,3), (2,2), (3,1)]
c = list(Counter(nums).items())
# sort by the 2nd element in each tuple
c.sort(key=lambda x: x[1], reverse=True)
# return the top k
return [c.pop(0)[0] for _ in range(k)] | top-k-frequent-elements | 3 liner in Python with explanation | tokamaka3 | 1 | 107 | top k frequent elements | 347 | 0.648 | Medium | 6,098 |
https://leetcode.com/problems/top-k-frequent-elements/discuss/1499963/Quickselect-beats-99.86-vs-minheap-beats-90.27 | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
def partit(nums, l, r)->int:
'''
partition and return pivot index. After execution, nums is partitial ordered, and pivot and final position.
Based on JeffE's Algorithms.wtf book.
'''
pi = randint(l,r)
pivot = nums[pi]
nums[pi], nums[r] = nums[r], nums[pi]
s = l-1 # [0..s] < pivot
for f in range(l,r):
if nums[f][1] < pivot[1]: # XXX: only change from default lomuto partition template
s += 1
nums[s], nums[f] = nums[f], nums[s]
# put pivot in correct place: s+1
s += 1
nums[s], nums[r] = nums[r], nums[s]
return s
def qselect(nums, l, r, k)->int:
'''
Based on https://en.wikipedia.org/wiki/Quickselect
'''
# base
if l == r:
return nums[l]
pi = partit(nums, l, r)
if pi == k:
return nums[pi]
elif pi > k:
return qselect(nums, l, pi-1, k)
else:
return qselect(nums, pi+1, r, k)
ctr = Counter(nums)
n2f = list(ctr.items())
qselect(n2f, 0, len(n2f)-1, len(n2f)-k)
res = []
for i in range(len(n2f)-1, len(n2f)-k-1, -1):
res.append(n2f[i][0])
return res | top-k-frequent-elements | [ηθ¨] Quickselect beats 99.86%, vs minheap beats 90.27% | hzhang413 | 1 | 80 | top k frequent elements | 347 | 0.648 | Medium | 6,099 |
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