post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/longest-palindromic-substring/discuss/2288241/simple-python-dp | class Solution:
def longestPalindrome(self, s):
longest_palindrome = ''
dp = [[0] * len(s) for _ in range(len(s))]
for i in range(len(s)):
dp[i][i] = True
longest_palindrome = s[i]
for i in range(len(s) - 1, -1, -1):
for j in range(i + 1, len(s)):... | longest-palindromic-substring | simple python dp | gasohel336 | 1 | 448 | longest palindromic substring | 5 | 0.324 | Medium | 200 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2251262/python3-oror-easy-oror-solution | class Solution:
def longestPalindrome(self, s: str) -> str:
res=""
resLen=0
finalLeftIndex=0
finalRightIndex=0
for i in range(len(s)):
#for odd length of palindromes
l,r=i,i
while l>=0 and r<len(s) and s[l]==s[r]:
... | longest-palindromic-substring | python3 || easy || solution | _soninirav | 1 | 231 | longest palindromic substring | 5 | 0.324 | Medium | 201 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2156912/Python3-or-Very-Easy-to-Understand-or-simple-approach | class Solution:
def longestPalindrome(self, s: str) -> str:
res = ''
n = len(s)
mxlen = 0
for i in range(n):
for j in range(max(i+1, i+mxlen), n+1):
if s[i]==s[j-1]:
sub = s[i: j]
if sub == sub[::-1] and len(sub)>len... | longest-palindromic-substring | Python3 | Very Easy to Understand | simple approach | H-R-S | 1 | 62 | longest palindromic substring | 5 | 0.324 | Medium | 202 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/1861817/Simple-Python-Solution-oror-85-Faster-oror-Memory-less-than-99 | class Solution:
def longestPalindrome(self, s: str) -> str:
def get_palindrome(s,l,r):
while l>=0 and r<len(s) and s[l]==s[r]: l-=1 ; r+=1
return s[l+1:r]
ans=''
for i in range(len(s)):
ans1=get_palindrome(s,i,i) ; ans2=get_palindrome(s,i,i+1)
... | longest-palindromic-substring | Simple Python Solution || 85% Faster || Memory less than 99% | Taha-C | 1 | 348 | longest palindromic substring | 5 | 0.324 | Medium | 203 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/1761378/Python-3-Expand-from-center | class Solution:
def longestPalindrome(self, s: str) -> str:
longest = 0
sub = ''
def expand(left, right):
nonlocal longest, sub
while left >= 0 and right < len(s) and s[left] == s[right]:
length = right-left+1
if length > longe... | longest-palindromic-substring | [Python 3] Expand from center | leet4ever | 1 | 230 | longest palindromic substring | 5 | 0.324 | Medium | 204 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/1471768/PyPy3-Simple-solution-w-comments | class Solution:
def longestPalindrome(self, s: str) -> str:
# Check if a string is palindrome
# within the given index i to j
def isPal(s,i,j):
while i < j:
if s[i] != s[j]:
break
i += 1
j -= 1
... | longest-palindromic-substring | [Py/Py3] Simple solution w/ comments | ssshukla26 | 1 | 665 | longest palindromic substring | 5 | 0.324 | Medium | 205 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/756544/Simple-Python-code-oror-Faster-than-70.29-of-Python3 | class Solution:
def longestPalindrome(self, s: str) -> str:
answer = ""
for i in range(len(s)):
odd = self.palindrome(s, i, i)
even = self.palindrome(s, i, i+1)
large = odd if len(odd) > len(even) else even
answer = l... | longest-palindromic-substring | Simple Python code || Faster than 70.29% of Python3 | dharmakshii | 1 | 439 | longest palindromic substring | 5 | 0.324 | Medium | 206 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/249433/My-simple-and-easy-understanding-DP-Python-solution | class Solution:
def longestPalindrome(self, s: str) -> str:
length = len(s)
dp = {}
max_len = 1
start = 0
for _len in range(2, length + 1):
i = 0
while i <= length - _len:
j = i + _len - 1
if _len <= 3:
... | longest-palindromic-substring | My simple and easy understanding DP Python solution | mazheng | 1 | 802 | longest palindromic substring | 5 | 0.324 | Medium | 207 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/249433/My-simple-and-easy-understanding-DP-Python-solution | class Solution:
def longestPalindrome(self, s: str) -> str:
length = len(s)
dp = {}
max_len = 1
start = 0
for _len in range(2, length + 1):
i = 0
while i <= length - _len:
j = i + _len - 1
if _len <= 3 and s[i] == s[j]:... | longest-palindromic-substring | My simple and easy understanding DP Python solution | mazheng | 1 | 802 | longest palindromic substring | 5 | 0.324 | Medium | 208 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/249433/My-simple-and-easy-understanding-DP-Python-solution | class Solution:
def longestPalindrome(self, s: str) -> str:
length = len(s)
dp = {}
max_len = 1
start = 0
for _len in range(2, length + 1):
i = 0
while i <= length - _len:
j = i + _len - 1
if (_len <= 3 and s[i] == s[j]... | longest-palindromic-substring | My simple and easy understanding DP Python solution | mazheng | 1 | 802 | longest palindromic substring | 5 | 0.324 | Medium | 209 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2845651/Python | class Solution:
def longestPalindrome(self, s: str) -> str:
if not s:
return ''
def extend(s:str, i:int, j:int) -> Tuple[int, int]:
while i>=0 and j < len(s):
if s[i] != s[j]:
break
i -= 1
j += 1
... | longest-palindromic-substring | Python | Kishore1K | 0 | 2 | longest palindromic substring | 5 | 0.324 | Medium | 210 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2840979/Manacher's-Algorithm-oror-Python-oror-Beats-98-oror-O(n)-time | class Solution:
def longestPalindrome(self, text: str) -> str:
N = len(text)
if N == 0:
return
if N == 1:
return text
N = 2*N+1 # Position count
L = [0] * N
L[0] = 0
L[1] = 1
C = 1 # centerPosition
R = 2 # centerRig... | longest-palindromic-substring | Manacher’s Algorithm || Python || Beats 98% || O(n) time | joetrankang | 0 | 5 | longest palindromic substring | 5 | 0.324 | Medium | 211 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2838180/Python-On2-solution-very-simple. | class Solution:
def longestPalindrome(self, s: str) -> str:
n = len(s)
res = ''
def find(i, j):
nonlocal res
while 0<=i<n and 0<=j<n:
ss = s[i: j+1]
if ss == ss[::-1]:
if len(ss) > len(res):
r... | longest-palindromic-substring | Python On2 solution - very simple. | florinbasca | 0 | 7 | longest palindromic substring | 5 | 0.324 | Medium | 212 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2827771/Python-(Simple-Maths) | class Solution:
def is_palindrome(self,s,left,right):
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return s[left+1:right]
def longestPalindrome(self, s):
ans = []
for i in range(len(s)):
ans.append(self.is... | longest-palindromic-substring | Python (Simple Maths) | rnotappl | 0 | 6 | longest palindromic substring | 5 | 0.324 | Medium | 213 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2825598/Python%3A-tracking-character-positions | class Solution:
def longestPalindrome(self, s: str) -> str:
positions = dict()
palindrome = s[0]
for j, char in enumerate(s):
if char in positions:
for i in positions[char]:
if (j - i + 1) <= len(palindrome):
break
... | longest-palindromic-substring | Python: tracking character positions | kwabenantim | 0 | 6 | longest palindromic substring | 5 | 0.324 | Medium | 214 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2825598/Python%3A-tracking-character-positions | class Solution:
def longestPalindrome(self, s: str) -> str:
n = len(s)
isPalindrome = [[False] * n for _ in range(n)]
for i in range(n-1):
isPalindrome[i][i] = True
if s[i] == s[i+1]:
isPalindrome[i+1][i] = True
isPalindrome[n-1][n-1] = True
... | longest-palindromic-substring | Python: tracking character positions | kwabenantim | 0 | 6 | longest palindromic substring | 5 | 0.324 | Medium | 215 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2823984/Manacher-Algorithm-faster-than-97-O(n) | class Solution:
def longestPalindrome(self, s: str) -> str:
# as you see we will use a string called t
# t will be our old s string but delimited by ^# and #$
# Between each character of s there will be #
t = "^#" + "#".join(s) + "#$"
size_t = len(t)
# p will contain ... | longest-palindromic-substring | Manacher Algorithm faster than 97% O(n) | diavollo | 0 | 12 | longest palindromic substring | 5 | 0.324 | Medium | 216 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2819045/Python-Slow-Solution-Beginner | class Solution:
def longestPalindrome(self, s: str) -> str:
if s == s[::-1]:
return s
check = []
l = 0
while l != len(s):
count = 0
for i in range(l,len(s)):
sub = s[l:len(s)- count]
if sub == sub[::-1]:
... | longest-palindromic-substring | Python Slow Solution Beginner | Jashan6 | 0 | 2 | longest palindromic substring | 5 | 0.324 | Medium | 217 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2811057/Two-Pointer-Approach-oror-Python | class Solution:
def longestPalindrome(self, s: str) -> str:
res = ''
length = 0
for i in range(len(s)):
# checking for odd length
left, right = i, i
while left >= 0 and right < len(s) and s[left] == s[right]:
if (right - left + 1) > length... | longest-palindromic-substring | Two Pointer Approach || Python | darsigangothri0698 | 0 | 10 | longest palindromic substring | 5 | 0.324 | Medium | 218 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2808156/Counting-Ripples-in-a-Pond | class Solution:
def longestPalindrome(self, s: str) -> str:
start, extent = 0, 1
for i in range(len(s) - 1):
possLen = 2 * min(i + 1, len(s) - i)
if i >= len(s) // 2 and extent > possLen:
break
if possLen > extent:
j, jEnd = 0, min(... | longest-palindromic-substring | Counting Ripples in a Pond | constantstranger | 0 | 5 | longest palindromic substring | 5 | 0.324 | Medium | 219 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2806129/Very-easy-python-solution-(ACCEPTED) | class Solution:
def longestPalindrome(self, s: str) -> str:
left = right = 0
palindrome = ""
while left < len(s) and (len(s)-left) > len(palindrome):
right = left
while right < len(s):
if s[left: right + 1] == s[left: right + 1][::-1]:
... | longest-palindromic-substring | Very easy python solution (ACCEPTED) | nehavari | 0 | 23 | longest palindromic substring | 5 | 0.324 | Medium | 220 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2804789/hehe | class Solution:
def longestPalindrome(self, s: str) -> str:
pal = ''
temp = ''
f = 0
t = s[:-1]
if s == s[::-1]:
return s
for i in range(len(s)):
for j in range(f, len(s)+1):
if s[i:j] == (s[i:j])[::-1] and len(s[i:j]) > f:
... | longest-palindromic-substring | hehe | user0355Kw | 0 | 3 | longest palindromic substring | 5 | 0.324 | Medium | 221 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2803633/Palindrome-substring-Python-Solution-Time-complexity-O(n*n) | class Solution:
def longestPalindrome(self, s: str) -> str:
res=''
reslen=0
for i in range(len(s)):
l,r=i,i
while l>=0 and r<len(s) and s[l]==s[r]:
if (r-l+1) > reslen:
res=s[l:r+1]
reslen=r-l+1
l... | longest-palindromic-substring | Palindrome substring Python Solution Time complexity O(n*n) | Jai_Trivedi | 0 | 6 | longest palindromic substring | 5 | 0.324 | Medium | 222 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2798602/Longest-Palindrome-using-python3 | class Solution:
def longestPalindrome(self, s: str) -> str:
n=len(s)
def expand_pallindrome(i,j):
while 0<=i<=j<n and s[i]==s[j]:
i-=1
j+=1
return (i+1, j)
res=(0,0)
for i in range... | longest-palindromic-substring | Longest Palindrome using python3 | user1885uW | 0 | 6 | longest palindromic substring | 5 | 0.324 | Medium | 223 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2790471/dynammic-programming-O(n2) | class Solution:
def longestPalindrome(self, s: str) -> str:
dp = [[False]*len(s) for _ in range(len(s))]
for i in range(len(s)):
dp[i][i] = True
ans=s[0]
for j in range(len(s)):
for i in range(j):
if s[i]==s[j] and (dp[i+1][j-1] or j == i+1):
... | longest-palindromic-substring | dynammic programming O(n^2) | haniyeka | 0 | 14 | longest palindromic substring | 5 | 0.324 | Medium | 224 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2790446/brute-force-python-O(n2) | class Solution:
def longestPalindrome(self, s: str) -> str:
res = ''
for i in range(len(s)):
pos = ''
neg = ''
for j in range(i, len(s)):
pos = pos + s[j]
neg = s[j] + neg
# print('pos is ', pos)
#... | longest-palindromic-substring | brute force python O(n^2) | ben_wei | 0 | 3 | longest palindromic substring | 5 | 0.324 | Medium | 225 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2784909/Python-O(n)-95-faster-O(n)-Solution | class Solution:
def longestPalindrome(self, s: str) -> str:
# edge cases: for a char or less return str
# 2 pointers, from beginning and end
# if they == append to another list.
N = len(s)
if N <= 1:
return s
N = 2*N+1 # Position count
... | longest-palindromic-substring | Python O(n) 95% faster O(n) Solution | mahsumkcby | 0 | 11 | longest palindromic substring | 5 | 0.324 | Medium | 226 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2777807/Python3-or-O(N2)-or-Easy-To-Understand | class Solution:
def longestPalindrome(self, A: str) -> str:
n = len(A)
res = "" # for storing longest palindrome
resLen = 0 # for storing length of longest palindrome
for i in range(0 , n):
# for odd length palindrome check
l , r = i,... | longest-palindromic-substring | Python3 | O(N2) | Easy To Understand | vishal7085 | 0 | 2 | longest palindromic substring | 5 | 0.324 | Medium | 227 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2755339/Simple-Python-Solution | class Solution:
def longestPalindrome(self, s: str) -> str:
res = ""
resLen = 0
for i in range(len(s)):
# odd cases
l, r = i, i
while l >= 0 and r < len(s) and s[l] == s[r]:
if (r-l+1)>resLen:
res = s[l:r+1]
... | longest-palindromic-substring | Simple Python Solution | ekomboy012 | 0 | 18 | longest palindromic substring | 5 | 0.324 | Medium | 228 |
https://leetcode.com/problems/longest-palindromic-substring/discuss/2752656/Palindromic-Subtring-(Best-90-by-Memory) | class Solution:
def longestPalindrome(self, s: str) -> str:
self.res = "";
self.resLen = 0;
for i in range(len(s)):
self.getPalindrome(s,i,i);
self.getPalindrome(s,i,i+1);
return self.res;
def getPalindrome(self,s,l,r):
while l >= 0 and r < len(s) ... | longest-palindromic-substring | Palindromic Subtring (Best 90% by Memory) | mohidk02 | 0 | 13 | longest palindromic substring | 5 | 0.324 | Medium | 229 |
https://leetcode.com/problems/zigzag-conversion/discuss/817306/Very-simple-and-intuitive-O(n)-python-solution-with-explanation | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
row_arr = [""] * numRows
row_idx = 1
going_up = True
for ch in s:
row_arr[row_idx-1] += ch
if row_idx == numRows:
going_... | zigzag-conversion | Very simple and intuitive O(n) python solution with explanation | wmv3317 | 96 | 3,000 | zigzag conversion | 6 | 0.432 | Medium | 230 |
https://leetcode.com/problems/zigzag-conversion/discuss/791453/90-faster-and-90-less-space-%2B-explanation | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows < 2:
return s
i = 0
res = [""]*numRows # We will fill in each line in the zigzag
for letter in s:
if i == numRows-1: # If this is the last line in the zigzag we go up
... | zigzag-conversion | 90% faster and 90% less space + explanation | MariaMozgunova | 27 | 1,600 | zigzag conversion | 6 | 0.432 | Medium | 231 |
https://leetcode.com/problems/zigzag-conversion/discuss/2709075/97-BEATED! | class Solution:
def convert(self, s: str, numRows: int) -> str:
list_of_items = [[] for i in range(numRows)]
DOWN = -1
i = 0
if numRows == 1:
return s
for char in s:
list_of_items[i].append(char)
if i == 0 or i == numRows - 1:
... | zigzag-conversion | 97% BEATED! | lil_timofey | 2 | 435 | zigzag conversion | 6 | 0.432 | Medium | 232 |
https://leetcode.com/problems/zigzag-conversion/discuss/1645784/Python3-8-Lines-or-Clean-%2B-Simple-Solution-or-Time-O(n)-or-Space-O(1) | class Solution:
def convert(self, s: str, numRows: int) -> str:
rows, direction, i = [[] for _ in range(numRows)], 1, 0
for ch in s:
rows[i].append(ch)
i = min(numRows - 1, max(0, i + direction))
if i == 0 or i == numRows - 1: direction *= -1
return ''.joi... | zigzag-conversion | [Python3] 8 Lines | Clean + Simple Solution | Time O(n) | Space O(1) | PatrickOweijane | 2 | 410 | zigzag conversion | 6 | 0.432 | Medium | 233 |
https://leetcode.com/problems/zigzag-conversion/discuss/2216401/O(n)-O(n)-A-simple-Python3-solution-to-an-annoyingly-confusing-problem | class Solution:
def convert(self, s: str, numRows: int) -> str:
# rules for columns that have a diagonal
diagcols = max(numRows-2,1) if numRows>2 else 0
# return diagcols
grid = [""]*numRows
# while the string isn't empty
while s:
# insert characters 1 by ... | zigzag-conversion | O(n), O(n) A simple Python3 solution to an annoyingly confusing problem | StackingFrog | 1 | 115 | zigzag conversion | 6 | 0.432 | Medium | 234 |
https://leetcode.com/problems/zigzag-conversion/discuss/727686/Python3Simple-implementation-less-than-97.89 | class Solution:
def convert(self, s: str, numRows: int) -> str:
length = len(s)
if numRows == 1 or numRows >= length:
return s
step = 2*numRows - 2
ret = ""
for i in range(0, numRows):
j = i
step_one = step - 2*i
while j < len... | zigzag-conversion | [Python3]Simple implementation less than 97.89% | abclzm1989 | 1 | 225 | zigzag conversion | 6 | 0.432 | Medium | 235 |
https://leetcode.com/problems/zigzag-conversion/discuss/2846241/Python-3-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
actual_row = 0
actual_col = 0
s = [letter for letter in s]
matriz = [[0]*(len(s)) for _ in range(numRows)]
while s:
if actual_row == 0:
for row in range(numRows):
i... | zigzag-conversion | Python 3 solution | user0106Ez | 0 | 2 | zigzag conversion | 6 | 0.432 | Medium | 236 |
https://leetcode.com/problems/zigzag-conversion/discuss/2827899/Python-Solution-Using-Dictionnary | class Solution:
def convert(self, s: str, numRows: int) -> str:
numCols = (numRows * 2) - 1
d = {}
i = 0
col = 0
row = 0
zig = numRows
while i < len(s):
r = abs(row)
... | zigzag-conversion | Python Solution Using Dictionnary | chiboubys | 0 | 2 | zigzag conversion | 6 | 0.432 | Medium | 237 |
https://leetcode.com/problems/zigzag-conversion/discuss/2813961/6.-Zigzag-Conversion-Python-solution-with-explanation-or-Iterate-array-in-zigzag-order | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
skip = (2 * numRows) - 2
res = []
for i in range(numRows): # use i to iterate vertically
j = 0 # use j to iterate horizontally
while i + j < len(s):
... | zigzag-conversion | 6. Zigzag Conversion - Python solution with explanation | Iterate array in zigzag order | jonjon98 | 0 | 3 | zigzag conversion | 6 | 0.432 | Medium | 238 |
https://leetcode.com/problems/zigzag-conversion/discuss/2809942/Python-slick-solution-using-generator | class Solution:
def convert(self, s: str, numRows: int) -> str:
# I will use numRows = 4 to provide examples here
# There is nothing to do if numRows == 1
if numRows == 1:
return s
# Example: for n=4, generator yields numbers in pattern:
# 0 1 2 3 2 1 ... | zigzag-conversion | Python slick solution using generator | nbashkir | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 239 |
https://leetcode.com/problems/zigzag-conversion/discuss/2804719/Clean-Python-code | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
res = [""] * numRows
cnt, period = 0, 2*numRows-2
for i in range(len(s)):
if cnt < numRows:
res[cnt] += s[i]
else:
res[perio... | zigzag-conversion | Clean Python code | codebreakblock | 0 | 7 | zigzag conversion | 6 | 0.432 | Medium | 240 |
https://leetcode.com/problems/zigzag-conversion/discuss/2802918/Simple-Python-Solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
row_map = {row:"" for row in range(1,numRows+1)}
row = 1
down = True
for c in s:
row_map[row] += c
if row == 1 or ((row < numRows) and down):
... | zigzag-conversion | Simple Python Solution | ekomboy012 | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 241 |
https://leetcode.com/problems/zigzag-conversion/discuss/2801123/Easy-to-Understand-O(n)-space-and-time-complexity-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
direction, row = "positive", -1
rows = [""] * numRows
for char in s:
row += 1 if direction is "positive" else -1
rows[row] += char
if row < 1:
direction = "positive"
... | zigzag-conversion | Easy to Understand O(n) space, and time complexity solution | Henok2011 | 0 | 8 | zigzag conversion | 6 | 0.432 | Medium | 242 |
https://leetcode.com/problems/zigzag-conversion/discuss/2800084/Python-3-easy-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
zigzag = [""] * numRows
row_num = 0
dirn = 1
for c in list(s):
zigzag[row_num] += c
if(row_num >= numRows-1):
dirn = -1
elif row_num <= 0:
dirn = 1
... | zigzag-conversion | Python 3 easy solution | pranjal51193 | 0 | 5 | zigzag conversion | 6 | 0.432 | Medium | 243 |
https://leetcode.com/problems/zigzag-conversion/discuss/2790733/Simple-solution-in-python-for-any-interested | class Solution:
def convert(self, s: str, numRows: int) -> str:
###### Empty/Trivial case########
if len(s) <2 or s == "" or s == None or numRows == 1:
return s
#################################
ret = ""
rows = list()
for i in range(0,numRows):
... | zigzag-conversion | Simple solution in python for any interested | yqd5143 | 0 | 3 | zigzag conversion | 6 | 0.432 | Medium | 244 |
https://leetcode.com/problems/zigzag-conversion/discuss/2790431/Very-Easy-level-mapping-solution-or-95%2B | class Solution:
def convert(self, s: str, numRows: int) -> str:
if len(s)<=numRows or numRows==1:
return s
premap=defaultdict(str)
i=0
while i<len(s):
k=1
while k<numRows and i<len(s):
premap[k]+=s[i]
k+=1
... | zigzag-conversion | Very Easy level mapping solution | 95%+ | AbhayKadapa007 | 0 | 7 | zigzag conversion | 6 | 0.432 | Medium | 245 |
https://leetcode.com/problems/zigzag-conversion/discuss/2786880/Python-implementation-with-Slice | class Solution:
def convert(self, s: str, numRows: int) -> str:
n = 2 * numRows - 2
if n == 0:
return s
ans = s[::n]
for i in range(1, n//2):
x, y = s[i::n], s[n-i::n]
tmp = [None] *len(x+y)
tmp[::2] = x
tmp[1::2] = y
... | zigzag-conversion | Python implementation with Slice | derbuihan | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 246 |
https://leetcode.com/problems/zigzag-conversion/discuss/2782777/Solve-ZigZag-Conversion-in-linear-time | class Solution:
def convert(self, s: str, numRows: int) -> str:
ans=[]
for i in range(numRows):
ans.insert(len(ans),[])
i=0
while i<len(s):
j=0
while i<len(s) and j<(numRows):
ans[j].insert(len(ans[j]),... | zigzag-conversion | Solve ZigZag Conversion in linear time | sijan_stha016 | 0 | 1 | zigzag conversion | 6 | 0.432 | Medium | 247 |
https://leetcode.com/problems/zigzag-conversion/discuss/2765011/Quick-line-efficient-solution-in-Py3 | class Solution:
def convert(self, s: str, numRows: int) -> str:
solutionArray = [""] * numRows
row = 0
for ele in range(len(s)):
if (numRows == 1): return s
solutionArray[row] += s[ele]
if (row == 0): step = 1
elif (row == (numRo... | zigzag-conversion | Quick, line efficient solution in Py3 | mantone6 | 0 | 5 | zigzag conversion | 6 | 0.432 | Medium | 248 |
https://leetcode.com/problems/zigzag-conversion/discuss/2756149/not-simple-python3 | class Solution:
def convert(self, s: str, numRows: int) -> str:
a = [""]*numRows
k = 0
boo = True
if numRows==1:
return s
for i in range(len(s)):
a[k] += s[i]
if boo == True:
if k<numRows:
if k==(numRows-... | zigzag-conversion | not simple python3 | meetHadvani | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 249 |
https://leetcode.com/problems/zigzag-conversion/discuss/2755195/Simple-Step-Algorithm-for-zigzag-pattern-using-python | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows <= 1: return s
res = ""
for i in range(numRows):
increment = 2 * (numRows - 1)
for j in range(i,len(s),increment):
res += s[j]
if (i > 0 and i < numRows - 1
... | zigzag-conversion | Simple Step Algorithm for zigzag pattern using python | mohidk02 | 0 | 2 | zigzag conversion | 6 | 0.432 | Medium | 250 |
https://leetcode.com/problems/zigzag-conversion/discuss/2754172/Concise-Python-Solution-(Beats-95) | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
result = ["" for i in range(numRows)]
for i in range(len(s)):
if ((i // (numRows-1)) % 2) == 0:
result[i % (numRows-1)] += s[i]
else:
... | zigzag-conversion | Concise Python Solution (Beats 95%) | ivan_n | 0 | 1 | zigzag conversion | 6 | 0.432 | Medium | 251 |
https://leetcode.com/problems/zigzag-conversion/discuss/2749707/Simple-Python-Solution-with-Explanation-or-Beginner-Friendly-(Beats-90) | class Solution:
def convert(self, s: str, numRows: int) -> str:
# If there is only one row, save time and return the string itself
if numRows == 1:
return s
# Create an empty list containing lists, these represent each row
result = []
for x in ra... | zigzag-conversion | Simple Python Solution with Explanation | Beginner Friendly (Beats 90%) | vijaivir | 0 | 9 | zigzag conversion | 6 | 0.432 | Medium | 252 |
https://leetcode.com/problems/zigzag-conversion/discuss/2744027/O(N)-implementation-using-a-pattern | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
#distance between characters for the top row
d0 = 2*numRows - 2
s_res = ''
d = 0 #it will store a variable distance for subsequent rows, except top and bottom ones
#r... | zigzag-conversion | O(N) implementation using a pattern | nonchalant-enthusiast | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 253 |
https://leetcode.com/problems/zigzag-conversion/discuss/2720579/Easy-to-understand-Python-3-Solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
zigzag = [""]*numRows
goingUp = False
i = 0
for ch in s:
if not goingUp:
zigzag[i] += ch
i += 1
if i == numRows:
goingUp = True
... | zigzag-conversion | Easy to understand Python 3 Solution | thenewkiller | 0 | 5 | zigzag conversion | 6 | 0.432 | Medium | 254 |
https://leetcode.com/problems/zigzag-conversion/discuss/2719268/Easy-python-3-solutionBeats-89 | class Solution:
def convert(self, s: str, numRows: int) -> str:
b=[''] * numRows
i=0
for j in s:
b[i]+=j
if i==0 and i==(numRows-1):
return s
elif i==0:
swap=0
elif i==(numRows-1):
swap=1
... | zigzag-conversion | Easy python 3 solution,Beats 89% | joeljoby111 | 0 | 2 | zigzag conversion | 6 | 0.432 | Medium | 255 |
https://leetcode.com/problems/zigzag-conversion/discuss/2676023/Basic-Python-solution-easy-explanation | class Solution:
def convert(self, s: str, numRows: int) -> str:
newstr = ""
if numRows == 1: #one row -> thats just the input string
return s
#first row
j = 0
while j < len(s):
newstr += s[j]
j += 2 * (numRows - 1)
#middle rows
... | zigzag-conversion | Basic Python solution - easy explanation | mat1124 | 0 | 7 | zigzag conversion | 6 | 0.432 | Medium | 256 |
https://leetcode.com/problems/zigzag-conversion/discuss/2671340/Simple-Python-Solution-for-6.-Zigzag-Conversion | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows==1:
return s
ans = ['' for i in range(numRows)]
curr = 0
for i in s:
ans[curr] += i
if curr>=0:
curr+=1
if curr==numRows:
... | zigzag-conversion | Simple Python Solution for 6. Zigzag Conversion | Akkishanu | 0 | 4 | zigzag conversion | 6 | 0.432 | Medium | 257 |
https://leetcode.com/problems/zigzag-conversion/discuss/2667076/Simple-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1 or len(s) == 1:
return s
res = ""
for r in range(numRows):
inc = 2 * (numRows - 1)
for i in range(r, len(s), inc):
res += s[i]
if r > 0 and r < ... | zigzag-conversion | Simple solution | CCsobu | 0 | 1 | zigzag conversion | 6 | 0.432 | Medium | 258 |
https://leetcode.com/problems/zigzag-conversion/discuss/2640621/python-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if len(s) <= numRows:
return s
arr = []
s = list(s)
for i in range(numRows):
arr.append([s.pop(0)])
i -= 1
while i > 0 and s:
arr[i].append(s.pop(0))
... | zigzag-conversion | python solution | sarthakchawande14 | 0 | 5 | zigzag conversion | 6 | 0.432 | Medium | 259 |
https://leetcode.com/problems/zigzag-conversion/discuss/2628689/Python-double-99 | class Solution:
def convert(self, s: str, numRows: int) -> str:
arr = [""] * numRows
if numRows == 1:
return s
direction = 1
row = 0
for i in range(len(s)):
arr[row] += s[i]
row += direction
if row == numRows - 1:
... | zigzag-conversion | Python double 99% | babyplutokurt | 0 | 44 | zigzag conversion | 6 | 0.432 | Medium | 260 |
https://leetcode.com/problems/zigzag-conversion/discuss/2598914/Python-9-line-Simple-Solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
track = ["" for _ in range(numRows)]
period = numRows + max(0, numRows-2)
for index, char in enumerate(s):
rest = index%period
row = rest if rest<numRows else numRows-(rest-numRows)-2
track[ro... | zigzag-conversion | Python 9-line Simple Solution | ParkerMW | 0 | 146 | zigzag conversion | 6 | 0.432 | Medium | 261 |
https://leetcode.com/problems/zigzag-conversion/discuss/2564910/Easy-python-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
elif numRows ==2:
rows=['']*numRows # Stores the letters of each rows
for i in range(0,len(s),2):
rows[0]+=s[i]
for i in range(1,len(... | zigzag-conversion | Easy python solution | guojunfeng1998 | 0 | 53 | zigzag conversion | 6 | 0.432 | Medium | 262 |
https://leetcode.com/problems/zigzag-conversion/discuss/2564490/easy-python | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
res = [''] * numRows
# with function call
'''
cur = 0
def increase(num):
return num + 1
def decrease(num):
return num - 1
patt... | zigzag-conversion | easy python | MajimaAyano | 0 | 35 | zigzag conversion | 6 | 0.432 | Medium | 263 |
https://leetcode.com/problems/zigzag-conversion/discuss/2552704/Python3-O(n)-with-explanation | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
res = ""
for i in range(numRows):
col = i
while col < len(s):
res += s[col]
col_offset = 2 * (numRows - 1)
col_next =... | zigzag-conversion | [Python3] O(n) with explanation | gdm | 0 | 90 | zigzag conversion | 6 | 0.432 | Medium | 264 |
https://leetcode.com/problems/zigzag-conversion/discuss/2483938/PYTHON-3-Simple-Clean-approach-beat-99 | class Solution:
def convert(self, s: str, numRows: int) -> str:
all_list = ['']*numRows
counter = 0
forward, backward = True, False
for i in s:
all_list[counter] = all_list[counter] + i
if forward:
if (counter < numRows-1):
... | zigzag-conversion | ✅ [PYTHON 3] Simple, Clean approach beat 99% 🏆 | girraj_14581 | 0 | 170 | zigzag conversion | 6 | 0.432 | Medium | 265 |
https://leetcode.com/problems/zigzag-conversion/discuss/2416228/Python-Accurate-Fast-Solution-Matrix-Beats-96-oror-Documented | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1: return s # return s, given number of rows is 1
m = [[] for _ in range(numRows)] # create matrix with given rows
row = 0 # current row in which char to be add
... | zigzag-conversion | [Python] Accurate Fast Solution - Matrix - Beats 96% || Documented | Buntynara | 0 | 75 | zigzag conversion | 6 | 0.432 | Medium | 266 |
https://leetcode.com/problems/zigzag-conversion/discuss/2373621/Easy-Python-3-Solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
i = 0
ls = ['']*numRows
sign = 1
for ch in s:
ls[i]+= ch
i+=1*sign
if i == numRows:
... | zigzag-conversion | Easy Python 3 Solution | harsh30199 | 0 | 171 | zigzag conversion | 6 | 0.432 | Medium | 267 |
https://leetcode.com/problems/zigzag-conversion/discuss/2324247/O(n)-time-O(n)-space-decent-array-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
arr = [[]for _ in range(numRows)]
i, row = 0, -1 # row offset by one for proper zigzgging
while i < len(s):
while row < numRows-1 and i <... | zigzag-conversion | O(n) time, O(n) space - decent array solution | mfarrill | 0 | 52 | zigzag conversion | 6 | 0.432 | Medium | 268 |
https://leetcode.com/problems/zigzag-conversion/discuss/2204988/Python3-solution-write-by-AI(Github-Copilot) | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
# create a list of lists
rows = [[] for _ in range(numRows)]
# print(rows)
# iterate through the string
# if the index is even, append to the first row
# if t... | zigzag-conversion | Python3 solution write by AI(Github Copilot) | aaron17 | 0 | 43 | zigzag conversion | 6 | 0.432 | Medium | 269 |
https://leetcode.com/problems/zigzag-conversion/discuss/2114039/Zigzag-Conversion | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
if len(s) == numRows:
return s
arr = [[] for i in range(numRows)]
row = 0
direction = 1
i = 0
while i < len(s):
... | zigzag-conversion | Zigzag Conversion | somendrashekhar2199 | 0 | 137 | zigzag conversion | 6 | 0.432 | Medium | 270 |
https://leetcode.com/problems/zigzag-conversion/discuss/2060194/Simple-FSM-or-python | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
temp = [[] for i in range(numRows)]
cnt = 0
turnBack = False
for i in s:
temp[cnt].append(i)
if not turnBack:
if cnt == numRows - 1:... | zigzag-conversion | Simple FSM | python | skrrtttt | 0 | 95 | zigzag conversion | 6 | 0.432 | Medium | 271 |
https://leetcode.com/problems/zigzag-conversion/discuss/2005125/8-Lines-Python-Solution-oror-90-Faster-oror-Memory-less-than-80 | class Solution:
def convert(self, S: str, N: int) -> str:
if N==1 or N>=len(S): return S
ans=['']*N ; idx=0 ; step=1
for s in S:
ans[idx]+=s
if idx==0: step=1
elif idx==N-1: step=-1
idx+=step
return ''.join(ans) | zigzag-conversion | 8-Lines Python Solution || 90% Faster || Memory less than 80% | Taha-C | 0 | 82 | zigzag conversion | 6 | 0.432 | Medium | 272 |
https://leetcode.com/problems/zigzag-conversion/discuss/1987779/Python3-Simple-linear-solution-with-explanations | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1: return s
res = ""
for r in range(numRows):
increment = 2*(numRows - 1) # offset added to the next element on the same row
for i in range(r, len(s), increment): # for each element... | zigzag-conversion | [Python3] Simple linear solution with explanations | leqinancy | 0 | 59 | zigzag conversion | 6 | 0.432 | Medium | 273 |
https://leetcode.com/problems/zigzag-conversion/discuss/1924411/Zigzag-easily-understandable-Python3-solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
#Creating lines list with length numRows
lines = []
for i in range(numRows):
lines.append([])
#Seperating chars to their corresponding lines
idx =... | zigzag-conversion | Zigzag easily understandable Python3 solution | 2255141 | 0 | 43 | zigzag conversion | 6 | 0.432 | Medium | 274 |
https://leetcode.com/problems/zigzag-conversion/discuss/1874454/Python3-Solution-using-Dictionary | class Solution:
def convert(self, s: str, numRows: int) -> str:
dicto = {}
row = 1
rowPrev = row
for i in range(len(s)):
if row in dicto:
dicto[row] = dicto[row] + s[i]
else:
dicto[row] = s[i]
if row == numRows or ... | zigzag-conversion | Python3 Solution using Dictionary | mrpuffie | 0 | 65 | zigzag conversion | 6 | 0.432 | Medium | 275 |
https://leetcode.com/problems/zigzag-conversion/discuss/1873827/Simple-Python3-Solution | class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
a = ['' for i in range(numRows)]
c = numRows
inc = False
for i in s:
a[numRows-c] += i # If numRows = 3, the range of numRows-c is [0, 2] as c is in the range of ... | zigzag-conversion | Simple Python3 Solution | rjnkokre | 0 | 54 | zigzag conversion | 6 | 0.432 | Medium | 276 |
https://leetcode.com/problems/reverse-integer/discuss/1061403/Clean-pythonic-solution | class Solution:
def reverse(self, x: int) -> int:
retval = int(str(abs(x))[::-1])
if(retval.bit_length()>31):
return 0
if x<0:
return -1*retval
else:
return retval | reverse-integer | Clean pythonic solution | njain07 | 20 | 3,300 | reverse integer | 7 | 0.273 | Medium | 277 |
https://leetcode.com/problems/reverse-integer/discuss/2803440/Python-or-Easy-Solution | class Solution:
def reverse(self, x: int) -> int:
if x not in range(-9,9):
x = int(str(x)[::-1].lstrip('0')) if x >= 0 else int(f"-{str(x)[:0:-1]}".lstrip('0'))
return x if (x < 2**31-1 and x > -2**31) else 0 | reverse-integer | Python | Easy Solution✔ | manayathgeorgejames | 5 | 1,500 | reverse integer | 7 | 0.273 | Medium | 278 |
https://leetcode.com/problems/reverse-integer/discuss/1071721/My-python3-solution | class Solution:
def reverse(self, x: int) -> int:
number = "".join(reversed(list(str(abs(x)))))
result = int("-" + number) if x < 0 else int(number)
if -2**31 <= result <= (2**31)-1:
return result
else:
return 0 | reverse-integer | My python3 solution | DmitriyL02 | 5 | 810 | reverse integer | 7 | 0.273 | Medium | 279 |
https://leetcode.com/problems/reverse-integer/discuss/670499/Python-line-by-line-explanation | class Solution:
def reverse(self, x: int) -> int:
# first convert it to string
x = str(x)
# if less than zero
if (int(x)<0) :
# first character is "-", so let's retain it
# reverse the rest of the characters, then add it up using "+"
# convert it back to integer
x = int(x[0]+x[... | reverse-integer | Python line by line explanation | derekchia | 5 | 738 | reverse integer | 7 | 0.273 | Medium | 280 |
https://leetcode.com/problems/reverse-integer/discuss/1645810/Python3-Checks-Overflow-or-Time-O(logx)-or-Space-O(1)-or-Abiding-by-all-Rules | class Solution:
def reverse(self, x: int) -> int:
sign = -1 if x < 0 else 1
x = abs(x)
maxInt = (1 << 31) - 1
res = 0
while x:
if res > (maxInt - x % 10) // 10: return 0
res = res * 10 + x % 10
x //= 10
return sign * res | reverse-integer | [Python3] Checks Overflow | Time O(logx) | Space O(1) | Abiding by all Rules | PatrickOweijane | 4 | 367 | reverse integer | 7 | 0.273 | Medium | 281 |
https://leetcode.com/problems/reverse-integer/discuss/1270724/Python-Simple-Solution-Easy-to-Read | class Solution:
def reverse(self, x: int) -> int:
negative = x < 0
ans = 0
if negative:
x = x*-1
while x > 0:
rem = x % 10
ans = (ans*10)+rem
x = x // 10
if negative:
ans = ans... | reverse-integer | Python Simple Solution, Easy to Read | Khonshu | 3 | 436 | reverse integer | 7 | 0.273 | Medium | 282 |
https://leetcode.com/problems/reverse-integer/discuss/1048717/Clean-and-Elegant-Python-Solution-with-Overflow-Handled | class Solution:
def reverse(self, x: int) -> int:
if x == 0:
return 0
reversed_integer = 0
sign = 1 if x > 0 else -1
x = abs(x)
while x != 0:
current_number = x % 10
if reversed_integer * 10 + current_number > (2 **31) or reversed_... | reverse-integer | Clean and Elegant Python Solution with Overflow Handled | jdshah | 3 | 268 | reverse integer | 7 | 0.273 | Medium | 283 |
https://leetcode.com/problems/reverse-integer/discuss/972410/Python3 | class Solution:
def reverse(self, x: int) -> int:
x = ','.join(str(x)).split(',')
sign = x.pop(0) if not x[0].isalnum() else None
x = ''.join(x[::-1])
res = sign+x if sign else x
res = int(res)
return res if -1*pow(2, 31) < res < pow(2, 31)-1 else 0 | reverse-integer | Python3 | mailolumide | 3 | 493 | reverse integer | 7 | 0.273 | Medium | 284 |
https://leetcode.com/problems/reverse-integer/discuss/2387574/Solution-that-respects-the-signed-32-bit-integer-constraint | class Solution:
def reverse(self, x: int) -> int:
if x == -2147483648: # -2**31
return 0
# -8463847412 is not a valid input
# hence result will never be -2147483648
# so we can work with positiv integers and multiply by the sign at the end
s = (x > 0) - (x < 0) # ... | reverse-integer | Solution that respects the signed 32-bit integer constraint | SpacePower | 2 | 243 | reverse integer | 7 | 0.273 | Medium | 285 |
https://leetcode.com/problems/reverse-integer/discuss/2323570/Python-oror-Two-Solutions-(Math-and-String)-with-explanations | class Solution:
def reverse(self, x: int) -> int:
pn = 1
if x < 0 :
pn = -1
x *= -1
x = int(str(x)[::-1]) * pn
#^Convert integer into string and reverse using slicing and convert it back to integer
return 0 if x < 2**31 * -1 or x > 2**31 -1 else x | reverse-integer | Python || Two Solutions (Math and String) with explanations | zhibin-wang09 | 2 | 264 | reverse integer | 7 | 0.273 | Medium | 286 |
https://leetcode.com/problems/reverse-integer/discuss/2323570/Python-oror-Two-Solutions-(Math-and-String)-with-explanations | class Solution:
def reverse(self, x: int) -> int:
ans,negative = 0,False
if x < 0:
negative = True
x *= -1
while x > 0:
ans *= 10 #Increse the length of answer
mod = int(x % 10) #Obtain last digit
ans +=... | reverse-integer | Python || Two Solutions (Math and String) with explanations | zhibin-wang09 | 2 | 264 | reverse integer | 7 | 0.273 | Medium | 287 |
https://leetcode.com/problems/reverse-integer/discuss/2318304/Python-Solution-Faster-than-96.24 | class Solution:
def reverse(self, x: int) -> int:
sign = 1
if x < 0:
sign = -1
x *= -1
x = int(str(x)[::-1])
if x > 2**31-1 or x < 2**31 * -1:
return 0
return sign * x | reverse-integer | Python Solution Faster than 96.24% | zip_demons | 2 | 313 | reverse integer | 7 | 0.273 | Medium | 288 |
https://leetcode.com/problems/reverse-integer/discuss/1486100/Python3-oror-Simplest-and-Valid-oror-With-Explanation | class Solution:
def reverse(self, x: int) -> int:
# initiate answer as zero
# the way we are gonna solve this is by understanding the simple math behind reversing an integer
# the objective is to be able to code the mathematical logic; which is why the string approach is totally rubbish and ... | reverse-integer | Python3 || Simplest and Valid || With Explanation | ajinkya2021 | 2 | 471 | reverse integer | 7 | 0.273 | Medium | 289 |
https://leetcode.com/problems/reverse-integer/discuss/1364007/99.55-Faster-Python-O(n)-Easily-understandable | class Solution:
def reverse(self, x: int) -> int:
if(x>0):
a = str(x)
a = a[::-1]
return int(a) if int(a)<=2**31-1 else 0
else:
x=-1*x
a = str(x)
a = a[::-1]
return int(a)*-1 if int(a)<=2**31 else 0 | reverse-integer | 99.55% Faster, Python, O(n), Easily understandable | unKNOWN-G | 2 | 1,200 | reverse integer | 7 | 0.273 | Medium | 290 |
https://leetcode.com/problems/reverse-integer/discuss/2796506/Python-Simple-Python-Solution-36-ms-faster-than-91.38 | class Solution:
def reverse(self, x: int) -> int:
def reverse_signed(num):
sum=0
sign=1
if num<0:
sign=-1
num=num*-1
while num>0:
rem=num%10
sum=sum*10+rem
num=num//10
... | reverse-integer | [ Python ] 🐍🐍 Simple Python Solution ✅✅36 ms, faster than 91.38% | sourav638 | 1 | 103 | reverse integer | 7 | 0.273 | Medium | 291 |
https://leetcode.com/problems/reverse-integer/discuss/2659747/Python-O(n)-Time-Solution-with-full-working-explanation | class Solution: # Time: O(n) and Space: O(1)
def reverse(self, x: int) -> int:
MIN = -2147483648 # -2^31
MAX = 2147483647 # 2^31 - 1
res = 0
while x: # Loop will run till x have some value either than 0 and None
digit = int(math.fmod(x, 10))
x = int(... | reverse-integer | Python O(n) Time Solution with full working explanation | DanishKhanbx | 1 | 233 | reverse integer | 7 | 0.273 | Medium | 292 |
https://leetcode.com/problems/reverse-integer/discuss/2275852/Python3-Math-solution | class Solution:
def reverse(self, x: int) -> int:
if x == 0:
return 0
elif x < 0:
sign = -1
x = abs(x)
flag = True
else:
sign = 1
flag = False
ans = 0
digits = int(math.log10(x)) # nubmer of ... | reverse-integer | Python3 Math solution | frolovdmn | 1 | 56 | reverse integer | 7 | 0.273 | Medium | 293 |
https://leetcode.com/problems/reverse-integer/discuss/2082744/Python3-Runtime%3A-O(n)-oror-O(n)-Runtime%3A-34ms-86.01 | class Solution:
def reverse(self, x: int) -> int:
return self.reverseIntegerByList(x)
# O(n) || O(1) 34ms 86.01%
def reverseIntegerByList(self, value):
if value == 0:return value
isNeg = value < 0
value = abs(value)
numList = list()
newReverseNumber = 0
... | reverse-integer | Python3 Runtime: O(n) || O(n) Runtime: 34ms 86.01% | arshergon | 1 | 101 | reverse integer | 7 | 0.273 | Medium | 294 |
https://leetcode.com/problems/reverse-integer/discuss/2025235/Python-compact-solution | class Solution:
def reverse(self, x: int) -> int:
if x > -1:
return int(str(x)[::-1]) if int(str(x)[::-1]) < 2147483647 else 0
else:
return int("-"+str(abs(x))[::-1]) if int("-"+str(abs(x))[::-1]) > -2147483647 else 0 | reverse-integer | Python compact solution | Yodawgz0 | 1 | 185 | reverse integer | 7 | 0.273 | Medium | 295 |
https://leetcode.com/problems/reverse-integer/discuss/1863788/python3-O(N) | class Solution:
def reverse(self, x: int) -> int:
sign = 1 if x >= 0 else -1
s = str(x*sign)
res = int(s[::-1])*sign
return 0 if(-2**31 > res or res > (2**31)-1) else res | reverse-integer | [python3] O(N) | Rahul-Krishna | 1 | 70 | reverse integer | 7 | 0.273 | Medium | 296 |
https://leetcode.com/problems/reverse-integer/discuss/1837832/Python-Simple-Slution | class Solution:
def reverse(self, x: int) -> int:
a = 1
if x < 0:
a = -1
x = abs(x)
s = 0
while x:
s *= 10
s += x%10
x //= 10
s = a*s
if s < -2**31 or s>2**31-1:
return 0
return s | reverse-integer | [Python] Simple Slution | zouhair11elhadi | 1 | 82 | reverse integer | 7 | 0.273 | Medium | 297 |
https://leetcode.com/problems/reverse-integer/discuss/1791782/Python-Solution-(only-4-Lines-of-Code) | class Solution:
def reverse(self, x: int) -> int:
flage=False
if(x<0):
flage=True
return((-int(str(abs(x))[::-1]) if flage else int(str(abs(x))[::-1])) if(-231<int(str(abs(x))[::-1])<231) else 0) | reverse-integer | Python Solution (only 4 Lines of Code) | aswin_thumati | 1 | 178 | reverse integer | 7 | 0.273 | Medium | 298 |
https://leetcode.com/problems/reverse-integer/discuss/1645357/Python-runtime-99.84-memory-98.89 | class Solution(object):
def reverse(self, x):
if x < 0:
y = 0-int(str(abs(x))[::-1])
if y <= -(pow(2,31)):
return 0
return y
else:
y = int(str(x)[::-1])
if y >= pow(2,31)-1:
return 0
... | reverse-integer | Python runtime 99.84%, memory 98.89% | cwrli | 1 | 205 | reverse integer | 7 | 0.273 | Medium | 299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.