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/count-number-of-special-subsequences/discuss/2589744/Clean-Simple-Python3-or-Iterative-DP-or-O(n)-Time-O(1)-Space | class Solution:
def countSpecialSubsequences(self, nums: List[int]) -> int:
dp0 = dp1 = dp2 = 0
for num in nums:
if num == 2:
dp2 += dp1 + dp2
elif num == 1:
dp1 += dp0 + dp1
else:
dp0 += dp0 + 1
... | count-number-of-special-subsequences | Clean, Simple Python3 | Iterative DP | O(n) Time, O(1) Space | ryangrayson | 0 | 15 | count number of special subsequences | 1,955 | 0.513 | Hard | 27,500 |
https://leetcode.com/problems/count-number-of-special-subsequences/discuss/1505276/Python-DP-O(N)-easy-solution | class Solution:
def countSpecialSubsequences(self, nums: List[int]) -> int:
MOD = 10**9 + 7
count = collections.Counter()
for n in nums:
count[n] += (count[n] + count[n - 1]) % MOD + (not n)
return count[2] % MOD | count-number-of-special-subsequences | [Python] DP O(N) easy solution | licpotis | 0 | 87 | count number of special subsequences | 1,955 | 0.513 | Hard | 27,501 |
https://leetcode.com/problems/count-number-of-special-subsequences/discuss/1375488/Python3-dp-ish | class Solution:
def countSpecialSubsequences(self, nums: List[int]) -> int:
MOD = 1_000_000_007
s0 = s1 = s2 = 0
for x in nums:
if x == 0: s0 = (1 + 2*s0) % MOD
elif x == 1: s1 = (s0 + 2*s1) % MOD
else: s2 = (s1 + 2*s2) % MOD
return s2 | count-number-of-special-subsequences | [Python3] dp-ish | ye15 | 0 | 29 | count number of special subsequences | 1,955 | 0.513 | Hard | 27,502 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2714159/Python-or-Easy-Solution | class Solution:
def makeFancyString(self, s: str) -> str:
stack = []
for letter in s:
if len(stack) > 1 and letter == stack[-1] == stack[-2]:
stack.pop()
stack.append(letter)
return ''.join(stack) | delete-characters-to-make-fancy-string | Python | Easy Solution✔ | manayathgeorgejames | 6 | 113 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,503 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1389361/Python3-stack | class Solution:
def makeFancyString(self, s: str) -> str:
stack = []
for ch in s:
if len(stack) > 1 and stack[-2] == stack[-1] == ch: continue
stack.append(ch)
return "".join(stack) | delete-characters-to-make-fancy-string | [Python3] stack | ye15 | 6 | 328 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,504 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1389309/This-my-simple-solution | class Solution:
def makeFancyString(self, s: str) -> str:
if len(s) < 3:
return s
ans = ''
ans += s[0]
ans += s[1]
for i in range(2,len(s)):
if s[i] != ans[-1] or s[i] != ans[-2]:
ans += s[i]
return ans | delete-characters-to-make-fancy-string | This my simple solution | youbou | 4 | 454 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,505 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1455438/Python3-Simple-O(n)-solution-faster-than-94 | class Solution:
def makeFancyString(self, s: str) -> str:
t = ''
ct = 1
ans = ''
for c in s:
if c == t:
ct += 1
else:
ct = 1
if ct < 3:
ans += c
t = c
return ans | delete-characters-to-make-fancy-string | [Python3] Simple O(n) solution, faster than 94% | terrencetang | 2 | 255 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,506 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2848150/Python-stack-easy-understandable-solution | class Solution:
def makeFancyString(self, s: str) -> str:
stack = []
count = 0
while count < len(s):
if stack == []:
stack.append(s[count])
count += 1
continue
elif len(stack) >= 2:
if s[count] == stack[... | delete-characters-to-make-fancy-string | Python stack easy understandable solution | youssefyzahran | 0 | 1 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,507 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2800707/Python-Easy-Solution | class Solution:
def makeFancyString(self, s: str) -> str:
s = list(s)
f, sec = '', ''
for index, i in enumerate(s):
if f == i and sec == i:
s[index] = ''
else:
sec = f
f = i
return ''.join(s)
``` | delete-characters-to-make-fancy-string | Python Easy Solution | AliAlievMos | 0 | 2 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,508 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2372059/Python3-Fast-and-Easy-solution | class Solution:
def makeFancyString(self, s: str) -> str:
final=[]
c=1
prev=""
l=len(s)
i=0
while i<l:
if s[i]==prev:
c+=1
if c<=2:
final.append(s[i])
else:
c=1
... | delete-characters-to-make-fancy-string | [Python3] Fast & Easy solution | sunakshi132 | 0 | 70 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,509 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2347688/Easy-solution-beating-95-submissions-by-memory-usage | class Solution:
def makeFancyString(self, s: str) -> str:
fancy_string = ''
for letter in s:
if fancy_string[-2:] != letter*2:
fancy_string += letter
else:
pass
return fancy_string | delete-characters-to-make-fancy-string | Easy solution beating 95% submissions by memory usage | IvanKorsakov | 0 | 27 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,510 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2288324/Python3-or-Stack-or-O(NorN) | class Solution:
def makeFancyString(self, s: str) -> str:
stack = []
curr_count = 1
deletion = 0
stack.append(s[0])
for i in range(1,len(s)):
if stack[len(stack)-1] == s[i]:
if curr_count == 2:
continue
else:
... | delete-characters-to-make-fancy-string | Python3 | Stack | O(N|N) | user7457RV | 0 | 24 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,511 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2065330/Concise-simple-straight-forward-python-solution | class Solution:
def makeFancyString(self, s: str) -> str:
res = s[:2]
for i in range(2,len(s)):
if s[i]*2 != s[i-2:i]:
res+=s[i]
return res | delete-characters-to-make-fancy-string | Concise simple straight-forward python solution | Lexcapital | 0 | 31 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,512 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2003958/Python-Clean-and-Simple! | class Solution:
def makeFancyString(self, s):
s = list(s)
a, b = '*', '*'
for i in range(len(s)-1,-1,-1):
c = s[i]
if a == b == c: del s[i]
a, b = b, c
return "".join(s) | delete-characters-to-make-fancy-string | Python - Clean and Simple! | domthedeveloper | 0 | 88 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,513 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/2003958/Python-Clean-and-Simple! | class Solution:
def makeFancyString(self, s):
x = ""
a, b = '*', '*'
for c in s:
if not a == b == c:
x += c
a, b = b, c
return x | delete-characters-to-make-fancy-string | Python - Clean and Simple! | domthedeveloper | 0 | 88 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,514 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1989025/Python-easy-solution-for-beginners | class Solution:
def makeFancyString(self, s: str) -> str:
if len(s) < 3:
return s
res = s[0] + s[1]
for i in range(2, len(s)):
if s[i] != res[-1] or s[i] != res[-2]:
res += s[i]
return res | delete-characters-to-make-fancy-string | Python easy solution for beginners | alishak1999 | 0 | 61 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,515 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1967465/python-3-oror-simple-O(n)-solution | class Solution:
def makeFancyString(self, s: str) -> str:
consecutive = 1
s = list(s)
for i in range(1, len(s)):
if s[i] == s[i - 1]:
consecutive += 1
if consecutive >= 3:
s[i - 1] = ''
else:
consecut... | delete-characters-to-make-fancy-string | python 3 || simple O(n) solution | dereky4 | 0 | 47 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,516 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1875390/Python-dollarolution | class Solution:
def makeFancyString(self, s: str) -> str:
l = s[0]
count = 1
for i in s[1:]:
if l[-1] == i:
count += 1
else:
count = 1
if count > 2:
continue
l += i
return l | delete-characters-to-make-fancy-string | Python $olution | AakRay | 0 | 31 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,517 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1869196/Python3-or-Simple | class Solution:
def makeFancyString(self, s: str) -> str:
ans = ""
l = 0
current_c = ""
for c in s:
# if current character == previous character
if c == current_c:
# only append the character if consecutive character length < 2
if l < 2:
... | delete-characters-to-make-fancy-string | Python3 | Simple | user0270as | 0 | 35 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,518 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1857937/Python-Solution-or-88-Lesser-Memory-or-O(n)-Solution-or-5-lines-of-code | class Solution:
def makeFancyString(self, s: str) -> str:
s = list(s)
for i in range(1,len(s)-1):
if (s[i-1] == s[i]) and (s[i+1] == s[i]):
s[i-1] = ""
return "".join(s) | delete-characters-to-make-fancy-string | ✔Python Solution | 88% Lesser Memory | O(n) Solution | 5 lines of code | Coding_Tan3 | 0 | 41 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,519 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1645081/Python-solution-not-the-fastest-but-kind-of-interesting | class Solution:
def makeFancyString(self, s: str) -> str:
ranges, r = [], []
for v in s:
if v not in r:
r = []
ranges.append(r)
r[1:] = (v,)
return ''.join(v for r in ranges for v in r) | delete-characters-to-make-fancy-string | Python solution, not the fastest, but kind of interesting | emwalker | 0 | 54 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,520 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1403427/PYTHON-or-simple-solution | class Solution:
def makeFancyString(self, s: str) -> str:
x = s[:2]
for i in range(2,len(s)) :
if x[-1] != s[i] or x[-2] != s[i] :
x += s[i]
return x | delete-characters-to-make-fancy-string | PYTHON | simple solution | rohitkhairnar | 0 | 117 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,521 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1396210/Java-%2B-Python3 | class Solution:
def makeFancyString(self, s: str) -> str:
if len(s) < 3: return s
idx = 2
res = s[0] + s[1]
while idx < len(s):
if s[idx] != res[-1] or s[idx] != res[-2]:
res += s[idx]
idx += 1
return res | delete-characters-to-make-fancy-string | Java + Python3 | jlee9077 | 0 | 49 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,522 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1392515/Python-oror-Very-Easy-Solution | class Solution:
def makeFancyString(self, s: str) -> str:
from itertools import groupby
p = ""
for i, j in groupby(s):
q = len(list(j))
if q >= 3:
p += i * 2
else:
p += i * q
return (p) | delete-characters-to-make-fancy-string | Python || Very Easy Solution | naveenrathore | 0 | 103 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,523 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1392271/Python-one-liner | class Solution:
def makeFancyString(self, s):
return s[:2] + ''.join(
c2 for c0, c1, c2 in zip(s[:-2], s[1:-1], s[2:]) if not c0 == c1 == c2
) | delete-characters-to-make-fancy-string | Python one-liner | greengum | 0 | 43 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,524 |
https://leetcode.com/problems/delete-characters-to-make-fancy-string/discuss/1426380/Python-create-a-new-array.-Time%3A-O(N)-Space%3A-O(N) | class Solution:
def makeFancyString(self, s: str) -> str:
fancy = []
for idx in range(len(s)):
if idx < 2 or not s[idx] == s[idx-1] == s[idx-2]:
fancy.append(s[idx])
return "".join(fancy) | delete-characters-to-make-fancy-string | Python, create a new array. Time: O(N), Space: O(N) | blue_sky5 | -1 | 67 | delete characters to make fancy string | 1,957 | 0.567 | Easy | 27,525 |
https://leetcode.com/problems/check-if-move-is-legal/discuss/1389250/Python3-check-8-directions | class Solution:
def checkMove(self, board: List[List[str]], rMove: int, cMove: int, color: str) -> bool:
for di, dj in (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0), (-1, 1):
i, j = rMove+di, cMove+dj
step = 0
while 0 <= i < 8 and 0 <= j < 8:
... | check-if-move-is-legal | [Python3] check 8 directions | ye15 | 5 | 312 | check if move is legal | 1,958 | 0.445 | Medium | 27,526 |
https://leetcode.com/problems/check-if-move-is-legal/discuss/1450810/Step-in-all-legal-directions-96-speed | class Solution:
def checkMove(self, board: List[List[str]], rMove: int, cMove: int, color: str) -> bool:
directions = [False] * 8
moves = [(1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0),
(-1, -1), (0, -1), (1, -1)]
opposite_color = "W" if color == "B" else "B"
for d in ra... | check-if-move-is-legal | Step in all legal directions, 96% speed | EvgenySH | 1 | 101 | check if move is legal | 1,958 | 0.445 | Medium | 27,527 |
https://leetcode.com/problems/check-if-move-is-legal/discuss/2502150/Best-Python3-implementation-(Top-93.9) | class Solution:
def checkMove(self, board: List[List[str]], rMove: int, cMove: int, color: str) -> bool:
directions = [False] * 8
moves = [(1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0),
(-1, -1), (0, -1), (1, -1)]
opposite_color = "W" if color == "B" else "B"
for d in ra... | check-if-move-is-legal | ✔️ Best Python3 implementation (Top 93.9%) | Kagoot | 0 | 10 | check if move is legal | 1,958 | 0.445 | Medium | 27,528 |
https://leetcode.com/problems/check-if-move-is-legal/discuss/1389381/Clean-Code-oror-Well-Explained-oror-100-Faster-oror-Simple | class Solution:
def checkMove(self, board: List[List[str]], row: int, col: int, color: str) -> bool:
rows, cols = 8, 8
offsets = ((-1, -1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, 0), (-1, 1))
if board[row][col] != '.' or row < 0 or row >= rows or col < 0 or col >= cols:
return False
for r_offset, ... | check-if-move-is-legal | 🐍 Clean-Code || Well-Explained || 100% Faster || Simple 📌 | abhi9Rai | 0 | 36 | check if move is legal | 1,958 | 0.445 | Medium | 27,529 |
https://leetcode.com/problems/minimum-total-space-wasted-with-k-resizing-operations/discuss/1389260/Python3-dp | class Solution:
def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int:
@cache
def fn(i, k):
"""Return min waste from i with k ops."""
if i == len(nums): return 0
if k < 0: return inf
ans = inf
rmx = rsm = 0
... | minimum-total-space-wasted-with-k-resizing-operations | [Python3] dp | ye15 | 14 | 987 | minimum total space wasted with k resizing operations | 1,959 | 0.42 | Medium | 27,530 |
https://leetcode.com/problems/minimum-total-space-wasted-with-k-resizing-operations/discuss/1703531/Python-DP-Faster-than-86 | class Solution:
def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int:
n = len(nums)
if k == 0:
mx = max(nums)
return sum(list(map(lambda x: mx-x, nums)))
if k >= n-1:
return 0
dp = [[math.inf for i in range(n)] for j ... | minimum-total-space-wasted-with-k-resizing-operations | Python DP Faster than 86% | tantianx777 | 1 | 108 | minimum total space wasted with k resizing operations | 1,959 | 0.42 | Medium | 27,531 |
https://leetcode.com/problems/minimum-total-space-wasted-with-k-resizing-operations/discuss/1391539/python3-DP-with-memoization-beat-100-python-submissions | class Solution:
def minSpaceWastedKResizing(self, A: List[int], K: int) -> int:
def waste(i, j, h):
sumI = sums[i-1] if i > 0 else 0
return (j-i+1)*h - sums[j] + sumI
def dp(i, k):
if i <= k:
return 0
if k < 0:
... | minimum-total-space-wasted-with-k-resizing-operations | [python3] DP with memoization, beat 100% python submissions | hieuvpm | 1 | 136 | minimum total space wasted with k resizing operations | 1,959 | 0.42 | Medium | 27,532 |
https://leetcode.com/problems/minimum-total-space-wasted-with-k-resizing-operations/discuss/1780594/python3-DP-solution-for-newbies | class Solution:
def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int:
MAX = 200e6
# if nums = [n1, n2, n3, ...] then sums = [n1, n1 + n2, n1 + n2+ n3, ...]
sums = list(accumulate(nums))
d = {}
def dynamicprogramming(i, k):
# Break condition one: if the no. of resizes left is g... | minimum-total-space-wasted-with-k-resizing-operations | [python3] DP - solution for newbies | yuwei-1 | 0 | 133 | minimum total space wasted with k resizing operations | 1,959 | 0.42 | Medium | 27,533 |
https://leetcode.com/problems/maximum-product-of-the-length-of-two-palindromic-substrings/discuss/1393832/Python3-Manacher | class Solution:
def maxProduct(self, s: str) -> int:
n = len(s)
# Manacher's algo
hlen = [0]*n # half-length
center = right = 0
for i in range(n):
if i < right: hlen[i] = min(right - i, hlen[2*center - i])
while 0 <= i-1-hlen[i] and i+1+hlen... | maximum-product-of-the-length-of-two-palindromic-substrings | [Python3] Manacher | ye15 | 3 | 200 | maximum product of the length of two palindromic substrings | 1,960 | 0.293 | Hard | 27,534 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1390199/Python3-move-along-s | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
i = 0
for word in words:
if s[i:i+len(word)] != word: return False
i += len(word)
if i == len(s): return True
return False | check-if-string-is-a-prefix-of-array | [Python3] move along s | ye15 | 22 | 1,200 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,535 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1390509/Python-Easy | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
a = ''
for i in words:
a += i
if a == s:
return True
if not s.startswith(a):
break
ret... | check-if-string-is-a-prefix-of-array | Python Easy | lokeshsenthilkumar | 8 | 466 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,536 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2003861/Python-Most-Simple-Solution-One-Liner! | class Solution:
def isPrefixString(self, s, words):
return s in accumulate(words) | check-if-string-is-a-prefix-of-array | Python - Most Simple Solution - One Liner! | domthedeveloper | 3 | 163 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,537 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2221696/Python-solution-for-beginners-by-beginner. | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
ans = ''
for i in words:
ans += i
if ans == s :
return True
return False | check-if-string-is-a-prefix-of-array | Python solution for beginners by beginner. | EbrahimMG | 2 | 93 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,538 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1403347/PYTHON-or-simple-solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
x = ''
for i in words :
x += i
if x == s : return True
if len(x) > len(s) : return False | check-if-string-is-a-prefix-of-array | PYTHON | simple solution | rohitkhairnar | 1 | 136 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,539 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1402845/Straightforward-O(n)-approach | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
prefix = ""
n = len(s)
for w in words:
prefix+=w
if(prefix==s):
return True
elif(len(prefix)>n):
return False
return False | check-if-string-is-a-prefix-of-array | Straightforward O(n) approach | krishna_sharma_s | 1 | 53 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,540 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1397000/Easy-Python-Solution-or-Faster-than-98 | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
for word in words:
if s:
if s.startswith(word):
s = s[len(word):]
else:
return False
if not s:
# for the case when len(s) > len(len(w... | check-if-string-is-a-prefix-of-array | Easy Python Solution | Faster than 98% | the_sky_high | 1 | 181 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,541 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2804883/Simple-Python-Approach | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
news = ""
for i in range(0,len(words)):
news += words[i]
if news == s:
return True
return False | check-if-string-is-a-prefix-of-array | Simple Python Approach | Shagun_Mittal | 0 | 1 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,542 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2690352/Simple-Python-Solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
j=0
ind=0
n=len(s)
m=len(words)
for i in range(n):
if s[j:i+1]==words[ind]:
ind+=1
j=i+1
if ind==m:
break
if j==n:
... | check-if-string-is-a-prefix-of-array | Simple Python Solution | Siddharth_singh | 0 | 6 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,543 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2601297/Python-Simple-Solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
# get a start index
start = 0
# save the length of the string s as we will need it repeatedly
s_length = len(s)
# go over all words
for word in words:
... | check-if-string-is-a-prefix-of-array | [Python] - Simple Solution | Lucew | 0 | 11 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,544 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2289502/Python-or-Easy-and-Understood-or-Fast-and-Simple | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
st=0
end=len(words)
while(end):
k=''.join(words[:end])
if k==s :
return True
end-=1
return False | check-if-string-is-a-prefix-of-array | Python | Easy and Understood | Fast and Simple | ajay_keelu | 0 | 26 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,545 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2104759/Python-%3A-Beginner-Friendly | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
string=''
for word in words:
string+=word
if string==s:
return True
return False | check-if-string-is-a-prefix-of-array | Python : Beginner Friendly | kushal2201 | 0 | 27 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,546 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/2072711/Extremely-simple-and-easy | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
sentence = ""
for word in words:
sentence += word
if s == sentence:
return True
return False
``` | check-if-string-is-a-prefix-of-array | Extremely simple and easy | Okosa | 0 | 33 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,547 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1793722/2-Lines-Python-Solution-oror-98-Faster-(28ms)-oror-Memory-less-than-90 | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
for i in range(len(words)):
if ''.join(words[:i+1]) == s: return True
return False | check-if-string-is-a-prefix-of-array | 2-Lines Python Solution || 98% Faster (28ms) || Memory less than 90% | Taha-C | 0 | 63 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,548 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1552557/One-pass-96-speed | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
pos = i = 0
len_s, len_w = len(s), len(words)
while pos < len_s and i < len_w:
if s[pos: pos + len(words[i])] == words[i]:
pos += len(words[i])
i += 1
else:
... | check-if-string-is-a-prefix-of-array | One pass, 96% speed | EvgenySH | 0 | 131 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,549 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1513193/Python-O(n)-time-solution.-(Compare-only-once) | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
n = len(s)
prefix = ''
length = 0
for word in words:
if length < n:
prefix += word
length += len(word)
return length == n and prefix == s | check-if-string-is-a-prefix-of-array | Python O(n) time solution. (Compare only once) | byuns9334 | 0 | 113 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,550 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1466274/Simple-and-Easy-Python-Solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
tmp=""
for i in words:
tmp+=i
if tmp==s:
return True
return False | check-if-string-is-a-prefix-of-array | Simple and Easy Python Solution | sangam92 | 0 | 75 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,551 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1422416/Python3-36ms-Create-S-from-Words | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
ss, i = "", 0
while(True):
if len(ss) == len(s) or i == len(words):
break
ss += words[i]
if len(ss) > len(s):
return False
... | check-if-string-is-a-prefix-of-array | Python3 36ms Create S from Words | Hejita | 0 | 72 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,552 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1392572/Python-oror-Basic-Solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
p = ""
count = 0
for i in words:
p += i
count += len(i)
if len(s) == count:
if p == s:
return (True)
else:
return (False)
elif len(s) < count:
... | check-if-string-is-a-prefix-of-array | Python || Basic Solution | naveenrathore | 0 | 66 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,553 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1392109/Java-%2B-Python | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
ans = ""
for w in words:
ans += w
if ans == s: return True
if ans > s: return False | check-if-string-is-a-prefix-of-array | Java + Python | jlee9077 | 0 | 16 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,554 |
https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/discuss/1390225/Python-Simple-solution | class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
n = len(s)
m = 0
c = ""
for i in words:
m+=len(i)
c+=i
if m > n:
return False
elif m == n and s == c:
return True
return... | check-if-string-is-a-prefix-of-array | [Python] Simple solution | ritika99 | 0 | 41 | check if string is a prefix of array | 1,961 | 0.541 | Easy | 27,555 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/1390207/Python3-priority-queue | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
pq = [-x for x in piles]
heapify(pq)
for _ in range(k): heapreplace(pq, pq[0]//2)
return -sum(pq) | remove-stones-to-minimize-the-total | [Python3] priority queue | ye15 | 7 | 413 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,556 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/1390561/Python-3-or-Heap-or-Explanation | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
heap = [-p for p in piles]
heapq.heapify(heap)
for _ in range(k):
cur = -heapq.heappop(heap)
heapq.heappush(heap, -(cur-cur//2))
return -sum(heap) | remove-stones-to-minimize-the-total | Python 3 | Heap | Explanation | idontknoooo | 6 | 287 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,557 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/2366593/Python-with-explanation-maxheap | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
for i, val in enumerate(piles): #we need maxheap, so negate all values
piles[i]=-val
heapify(piles)
while k:
x = -heappop(piles) #pop largest absolute no.
x-=(x//2)
heapp... | remove-stones-to-minimize-the-total | Python with explanation - maxheap | sunakshi132 | 0 | 16 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,558 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/2347622/Python3-or-O(klogn)-or-Priority-Queue | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
newPiles=[-1*i for i in piles]
heapify(newPiles)
while k>0:
currEle=-1*heappop(newPiles)
currEle-=(currEle//2)
heappush(newPiles,-1*currEle)
k-=1
return -1*sum(newP... | remove-stones-to-minimize-the-total | [Python3] | O(klogn) | Priority Queue | swapnilsingh421 | 0 | 11 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,559 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/2007136/Simple-solution-using-priority-queue-(python) | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
# since python provides only min heap, we need to make all the elements negative
# so that it can be utilized as a max heap
for i in range(len(piles)):
piles[i]=-piles[i]
q=heapq.hea... | remove-stones-to-minimize-the-total | Simple solution using priority queue (python) | pbhuvaneshwar | 0 | 46 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,560 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/1635504/python-simple-heap-solution | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
import heapq
piles = [-x for x in piles]
heapq.heapify(piles)
def convert(x):
y = -x
y = (y+1)//2
return -y
for _ in range(k):
v = heapq.h... | remove-stones-to-minimize-the-total | python simple heap solution | byuns9334 | 0 | 78 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,561 |
https://leetcode.com/problems/remove-stones-to-minimize-the-total/discuss/1390611/Clean-and-Simple-oror-94-faster-oror-Easy-understandable | class Solution:
def minStoneSum(self, piles: List[int], k: int) -> int:
local = [-p for p in piles]
heapq.heapify(local)
for i in range(k):
tmp = -1*heapq.heappop(local)
tmp = (tmp+1)//2
heapq.heappush(local,-tmp)
return -sum(local) | remove-stones-to-minimize-the-total | 📌 Clean & Simple || 94% faster || Easy-understandable 🐍 | abhi9Rai | 0 | 65 | remove stones to minimize the total | 1,962 | 0.593 | Medium | 27,562 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1390576/Two-Pointers | class Solution:
def minSwaps(self, s: str) -> int:
res, bal = 0, 0
for ch in s:
bal += 1 if ch == '[' else -1
if bal == -1:
res += 1
bal = 1
return res | minimum-number-of-swaps-to-make-the-string-balanced | Two Pointers | votrubac | 19 | 2,000 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,563 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1948127/Easy-and-Intuitive-Python-Solution-With-Images-and-Walkthrough | class Solution:
def minSwaps(self, s: str) -> int:
count = 0
for i in s:
if i == "[":
count += 1 # increment only if we encounter an open bracket.
else:
if count > 0: #decrement only if count is positive. Else do nothing and move on. This is... | minimum-number-of-swaps-to-make-the-string-balanced | Easy and Intuitive Python Solution With Images and Walkthrough | danielkua | 9 | 406 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,564 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1390209/Python3-greedy | class Solution:
def minSwaps(self, s: str) -> int:
ans = prefix = 0
for ch in s:
if ch == "[": prefix += 1
else: prefix -= 1
if prefix == -1:
ans += 1
prefix = 1
return ans | minimum-number-of-swaps-to-make-the-string-balanced | [Python3] greedy | ye15 | 3 | 110 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,565 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/2807126/Python3-Solution-or-Clean-and-Concise | class Solution:
def minSwaps(self, s):
cur, ans = 0, 0
for i in s:
if i == ']' and cur == 0: ans += 1
if i == '[' or cur == 0: cur += 1
else: cur -= 1
return ans | minimum-number-of-swaps-to-make-the-string-balanced | ✔ Python3 Solution | Clean & Concise | satyam2001 | 1 | 97 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,566 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1444539/Python3-solution-or-O(n)-or-explained-pattern | class Solution:
def minSwaps(self, s: str) -> int:
l = 0
r = 0
for i in range(len(s)):
if s[i] == ']':
if l == 0:
r += 1
else:
l -= 1
else:
l += 1
if l % 2 == 0:
res = l // ... | minimum-number-of-swaps-to-make-the-string-balanced | Python3 solution | O(n) | explained pattern | FlorinnC1 | 1 | 368 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,567 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1397368/Python-Using-a-counter.-or-89.58-runtime-99.07-memory | class Solution:
def minSwaps(self, s: str) -> int:
count=0
for char in s:
if count and char != '[':
count -= 1
else:
count += 1
return (count + 1) //2 | minimum-number-of-swaps-to-make-the-string-balanced | [Python] Using a counter. | 89.58% runtime, 99.07% memory | Archangel21809 | 1 | 170 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,568 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/2794004/Python-Solution-Easy-to-understand | class Solution:
def minSwaps(self, s: str) -> int:
extraClose, maxClose = 0, 0
for i in s:
if i == "[":
extraClose -=1
else:
extraClose +=1
maxClose = max(maxClose, extraClose)
return (maxClose +1)//2 | minimum-number-of-swaps-to-make-the-string-balanced | Python Solution Easy to understand | agnishwar29 | 0 | 3 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,569 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1916065/Python-easy-to-read-and-understand | class Solution:
def minSwaps(self, s: str) -> int:
mx_cnt, cnt = 0, 0
for i in range(len(s)):
if s[i] == "]":
cnt += 1
mx_cnt = max(mx_cnt, cnt)
else:
cnt -= 1
return (mx_cnt + 1) // 2 | minimum-number-of-swaps-to-make-the-string-balanced | Python easy to read and understand | sanial2001 | 0 | 137 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,570 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1907952/Python-3-Solution | class Solution:
def minSwaps(self, s: str) -> int:
cnt =0
for i in s:
if i == '[':
cnt += 1
elif i == ']' and cnt > 0:
cnt -= 1
return ceil(cnt/2) | minimum-number-of-swaps-to-make-the-string-balanced | Python 3 Solution | DietCoke777 | 0 | 52 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,571 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1475094/Python3-Solution-with-using-stack | class Solution:
def minSwaps(self, s: str) -> int:
stack = []
for idx in range(len(s)):
if s[idx] == ']' and stack and stack[-1] == '[':
stack.pop()
else:
stack.append(s[idx])
bracket_count_left_part = len(stack) // 2
... | minimum-number-of-swaps-to-make-the-string-balanced | [Python3] Solution with using stack | maosipov11 | 0 | 134 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,572 |
https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/discuss/1437465/Python3-1-Line | class Solution:
def minSwaps(self, s: str) -> int:
return (reduce(lambda mismatch, ch: mismatch + (-1 if (mismatch > 0 and ch == ']') else 1), s, 0) + 1) // 2 | minimum-number-of-swaps-to-make-the-string-balanced | Python3 - 1 Line ✅ | Bruception | -2 | 118 | minimum number of swaps to make the string balanced | 1,963 | 0.684 | Medium | 27,573 |
https://leetcode.com/problems/find-the-longest-valid-obstacle-course-at-each-position/discuss/1390573/Clean-and-Simple-oror-98-faster-oror-Easy-Code | class Solution:
def longestObstacleCourseAtEachPosition(self, obs: List[int]) -> List[int]:
local = []
res=[0 for _ in range(len(obs))]
for i in range(len(obs)):
n=obs[i]
if len(local)==0 or local[-1]<=n:
local.append(n)
res[i]=len(local)
else:
ind... | find-the-longest-valid-obstacle-course-at-each-position | 📌 Clean & Simple || 98% faster || Easy-Code 🐍 | abhi9Rai | 1 | 48 | find the longest valid obstacle course at each position | 1,964 | 0.469 | Hard | 27,574 |
https://leetcode.com/problems/find-the-longest-valid-obstacle-course-at-each-position/discuss/1390219/Python3-LIS | class Solution:
def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
ans, vals = [], []
for i, x in enumerate(obstacles):
k = bisect_right(vals, x)
ans.append(k+1)
if k == len(vals): vals.append(x)
else: vals[k] = x
... | find-the-longest-valid-obstacle-course-at-each-position | [Python3] LIS | ye15 | 1 | 59 | find the longest valid obstacle course at each position | 1,964 | 0.469 | Hard | 27,575 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1404073/Python3-1-line | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(x in word for x in patterns) | number-of-strings-that-appear-as-substrings-in-word | [Python3] 1-line | ye15 | 19 | 1,400 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,576 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2642630/Python-O(N) | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count=0
for i in patterns:
if i in word:
count+=1
return count | number-of-strings-that-appear-as-substrings-in-word | Python O(N) | Sneh713 | 1 | 118 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,577 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2364475/Python-one-liner | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return len([pattern for pattern in patterns if pattern in word]) | number-of-strings-that-appear-as-substrings-in-word | Python one-liner | Arrstad | 1 | 44 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,578 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1611033/python-simple-solution-using-map | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
li = list(map(lambda x: x in word, patterns))
return li.count(True) | number-of-strings-that-appear-as-substrings-in-word | python simple solution using map | apgokul | 1 | 110 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,579 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1597820/Python-one-liner-making-use-of-bool-to-int | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(i in word for i in patterns)
``` | number-of-strings-that-appear-as-substrings-in-word | [Python] one liner making use of bool to int | nelsarrag | 1 | 76 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,580 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1473916/O(n-%2B-m)-T-or-O(n-%2B-m)-S-or-suffix-tree-or-Python-3 | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
def ukkonen(string: str) -> 'Node':
"""Build suffix tree by Ukkonen algorithm.
:param string: string with last character is terminator.
:return:
"""
class Node:
... | number-of-strings-that-appear-as-substrings-in-word | O(n + m) T | O(n + m) S | suffix tree | Python 3 | CiFFiRO | 1 | 300 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,581 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1405851/Python-or-Simple-1-liner | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(pattern in word for pattern in patterns) | number-of-strings-that-appear-as-substrings-in-word | Python | Simple 1 liner | leeteatsleep | 1 | 103 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,582 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2808148/Python3-one-line-solution | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum([p in word for p in patterns]) | number-of-strings-that-appear-as-substrings-in-word | Python3 one-line solution | sipi09 | 0 | 2 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,583 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2806359/simple-python-solution | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
ans=0
for i in patterns:
ans+=1 if i in word else 0
return ans | number-of-strings-that-appear-as-substrings-in-word | simple python solution | Nikhil2532 | 0 | 2 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,584 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2767294/Python-or-LeetCode-or-1967.-Number-of-Strings-That-Appear-as-Substrings-in-Word | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
x = 0
for i in patterns:
if i in word:
x += 1
return x | number-of-strings-that-appear-as-substrings-in-word | Python | LeetCode | 1967. Number of Strings That Appear as Substrings in Word | UzbekDasturchisiman | 0 | 4 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,585 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2767294/Python-or-LeetCode-or-1967.-Number-of-Strings-That-Appear-as-Substrings-in-Word | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(x in word for x in patterns) | number-of-strings-that-appear-as-substrings-in-word | Python | LeetCode | 1967. Number of Strings That Appear as Substrings in Word | UzbekDasturchisiman | 0 | 4 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,586 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2764901/Python-solution-by-checking-if-the-pattern-exist-in-word | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count = 0
for pattern in patterns:
if pattern in word:
count += 1
return count | number-of-strings-that-appear-as-substrings-in-word | Python solution by checking if the pattern exist in word | samanehghafouri | 0 | 2 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,587 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2631668/Solution-using-For-loop | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count = 0
for match in patterns:
if match in word:
count+=1
return count | number-of-strings-that-appear-as-substrings-in-word | Solution using For loop | akankshanagar | 0 | 4 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,588 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2336532/Python-easy-beginner-solution | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count = 0
for i in patterns:
if i in word:
count+=1
return count | number-of-strings-that-appear-as-substrings-in-word | Python easy beginner solution | EbrahimMG | 0 | 11 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,589 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2099744/Python-simple-solution | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
ans = 0
for i in patterns:
if i in word:
ans += 1
return ans | number-of-strings-that-appear-as-substrings-in-word | Python simple solution | StikS32 | 0 | 62 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,590 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2094955/PYTHON-or-Simple-python-solution | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
res = 0
for i in patterns:
if i in word:
res += 1
return res | number-of-strings-that-appear-as-substrings-in-word | PYTHON | Simple python solution | shreeruparel | 0 | 19 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,591 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/2003825/Python-One-Liner! | class Solution:
def numOfStrings(self, patterns, word):
return sum(p in word for p in patterns) | number-of-strings-that-appear-as-substrings-in-word | Python - One-Liner! | domthedeveloper | 0 | 61 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,592 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1860657/Python-solution-memory-less-than-99 | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count = 0
for i in patterns:
if i in word:
count += 1
return count | number-of-strings-that-appear-as-substrings-in-word | Python solution, memory less than 99% | alishak1999 | 0 | 49 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,593 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1664385/Python-using-str.__contains__ | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(map(word.__contains__, patterns)) | number-of-strings-that-appear-as-substrings-in-word | Python, using str.__contains__ | emwalker | 0 | 29 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,594 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1662485/Python3-one-liner | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(1 for x in patterns if x in word) | number-of-strings-that-appear-as-substrings-in-word | Python3 one-liner | denizen-ru | 0 | 49 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,595 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1580868/python-sol-or-faster-than-99 | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count = 0
for i in range(len(patterns)):
if patterns[i] in word: count += 1
return count | number-of-strings-that-appear-as-substrings-in-word | python sol | faster than 99% | anandanshul001 | 0 | 81 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,596 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1468038/Python3-Faster-Than-91.40-Memory-Less-Than-88.68 | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
c = 0
for i in patterns:
if i in word:
c += 1
return c | number-of-strings-that-appear-as-substrings-in-word | Python3 Faster Than 91.40%, Memory Less Than 88.68% | Hejita | 0 | 72 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,597 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1405465/PYTHON-3-%3A-EASY-SOLUTION | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
return sum(1 for i in patterns if i in word) | number-of-strings-that-appear-as-substrings-in-word | PYTHON 3 : EASY SOLUTION | rohitkhairnar | 0 | 94 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,598 |
https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/discuss/1404444/Easy-Python-Solution(28ms) | class Solution:
def numOfStrings(self, patterns: List[str], word: str) -> int:
count=0
for i in range(len(patterns)):
if patterns[i] in word:
count+=1
return count | number-of-strings-that-appear-as-substrings-in-word | Easy Python Solution(28ms) | Sneh17029 | 0 | 127 | number of strings that appear as substrings in word | 1,967 | 0.799 | Easy | 27,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.