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| python_solutions
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| post_title
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| user
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int64 -20
1.2k
| views
int64 0
60.9k
| problem_title
stringlengths 3
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| number
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| acceptance
float64 0.14
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int64 0
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|---|---|---|---|---|---|---|---|---|---|---|---|
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
return dp2 % (10**9 + 7)
|
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[-1] and s[count] == stack[-2]:
stack.pop(-1)
stack.append(s[count])
count += 1
return ''.join(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
final.append(s[i])
prev=s[i]
i+=1
return "".join(final)
|
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:
stack.append(s[i])
curr_count += 1
elif stack[len(stack)-1] != s[i]:
stack.append(s[i])
curr_count = 1
s = "".join(stack)
return s
|
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:
consecutive = 1
return ''.join(s)
|
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:
ans += c
l +=1
else:
current_c = c
ans += c
l = 1
return ans
|
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:
if board[i][j] == color and step: return True
if board[i][j] == "." or board[i][j] == color and not step: break
i, j = i+di, j+dj
step += 1
return False
|
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 range(8):
r, c = rMove + moves[d][0], cMove + moves[d][1]
if 0 <= r < 8 and 0 <= c < 8 and board[r][c] == opposite_color:
directions[d] = True
for step in range(2, 8):
if not any(d for d in directions):
return False
for d in range(8):
if directions[d]:
r, c = rMove + step * moves[d][0], cMove + step * moves[d][1]
if 0 <= r < 8 and 0 <= c < 8:
if board[r][c] == color:
return True
elif board[r][c] == ".":
directions[d] = False
else:
directions[d] = False
return False
|
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 range(8):
r, c = rMove + moves[d][0], cMove + moves[d][1]
if 0 <= r < 8 and 0 <= c < 8 and board[r][c] == opposite_color:
directions[d] = True
for step in range(2, 8):
if not any(d for d in directions):
return False
for d in range(8):
if directions[d]:
r, c = rMove + step * moves[d][0], cMove + step * moves[d][1]
if 0 <= r < 8 and 0 <= c < 8:
if board[r][c] == color:
return True
elif board[r][c] == ".":
directions[d] = False
else:
directions[d] = False
return False
|
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, c_offset in offsets:
valid = False
r, c = row, col
while True:
r, c = r + r_offset, c + c_offset
if r < 0 or r >= rows or c < 0 or c >= cols:
break
elif board[r][c] == color:
if valid:
return True
else:
break
elif board[r][c] == '.':
break
else:
valid = True
return False
|
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
for j in range(i, len(nums)):
rmx = max(rmx, nums[j])
rsm += nums[j]
ans = min(ans, rmx*(j-i+1) - rsm + fn(j+1, k-1))
return ans
return fn(0, k)
|
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 in range(k+1)]
wasted = [[0 for i in range(n)] for j in range(n)]
for i in range(0, n):
prev_max = nums[i]
for j in range(i, n):
if prev_max >= nums[j]:
wasted[i][j] = wasted[i][j-1] + prev_max - nums[j]
else:
diff = nums[j] - prev_max
wasted[i][j] = diff * (j - i) + wasted[i][j-1]
prev_max = nums[j]
for i in range(n):
dp[0][i] = wasted[0][i]
for j in range(1, k+1):
for i in range(j, n):
for l in range(j, i+1):
dp[j][i] = min(dp[j][i], dp[j-1][l-1] + wasted[l][i])
return dp[k][n-1]
|
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:
return MAX
if (i, k) in memoize:
return memoize[(i, k)]
_max = A[i]
r = MAX
for j in range(i-1, -2, -1):
r = min(r, dp(j, k-1) + waste(j+1, i, _max))
_max = max(_max, A[j])
memoize[(i, k)] = r
return r
sums = list(accumulate(A))
n = len(A)
MAX = 10**6*200
memoize = {}
return dp(n-1, K)
|
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 greater or equal to index, then we can just resize for all remaining indices, meaning the loss is zer0.
if i <= k:
return 0
# Break condition 2: if there are no resizes left
elif k < 0:
return MAX
elif (i, k) in d.keys():
return d[(i, k)]
loss = MAX
curr_max = nums[i]
for j in range(i - 1, -2, -1):
loss = min(loss, dynamicprogramming(j, k-1) + waste(j + 1, i, curr_max))
d[(i, k)] = loss
curr_max = max(curr_max, nums[j])
return loss
def waste(i, j, curr_max):
if i - 1 < 0:
sumi = 0
else:
sumi = sums[i - 1]
return (j - i + 1)*curr_max - (sums[j] - sumi)
return dynamicprogramming(len(nums) - 1, k)
|
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[i] < len(s) and s[i-1-hlen[i]] == s[i+1+hlen[i]]:
hlen[i] += 1
if right < i+hlen[i]: center, right = i, i+hlen[i]
prefix = [0]*n
suffix = [0]*n
for i in range(n):
prefix[i+hlen[i]] = max(prefix[i+hlen[i]], 2*hlen[i]+1)
suffix[i-hlen[i]] = max(suffix[i-hlen[i]], 2*hlen[i]+1)
for i in range(1, n):
prefix[~i] = max(prefix[~i], prefix[~i+1]-2)
suffix[i] = max(suffix[i], suffix[i-1]-2)
for i in range(1, n):
prefix[i] = max(prefix[i-1], prefix[i])
suffix[~i] = max(suffix[~i], suffix[~i+1])
return max(prefix[i-1]*suffix[i] for i in range(1, n))
|
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
return False
|
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(words[0])......len(words[n-1])
return True
else:
return False
|
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:
return True
return False
|
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 whether we are out of scope for s
if start >= s_length:
return True
# check if current word is included in s
if s[start:start+len(word)] != word:
return False
# increase the start
start += len(word)
# check whether we reached the end of the string of the prefix
if start >= s_length:
return True
else:
return False
|
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:
return False
return pos == len_s
|
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
i += 1
return s == ss
|
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:
return (False)
|
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 False
|
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)
heappush(piles,-x) #negate value before pushing
k-=1
return sum(piles)*-1 #all values are negative so use abs or multiply with -1
|
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(newPiles)
|
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.heapify(piles)
while k>0:
p=heapq.heappop(piles)
# since the elements are negative, we need to use ceil instead of floor
p=p-(math.ceil(p/2))
heapq.heappush(piles,p)
k-=1
return -sum(piles)
|
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.heappop(piles)
heapq.heappush(piles, convert(v))
res = 0
while piles:
res += heapq.heappop(piles)
return -res
|
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 because for the case " ] [ [ ] " we do not need to in
count -= 1
return (count + 1) // 2
|
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 // 2
else:
res = (l+1)//2
return res
|
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
return bracket_count_left_part // 2 + (bracket_count_left_part & 1)
|
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 = bisect.bisect_right(local,n)
local[ind]=n
res[i]=ind+1
return res
|
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
return ans
|
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:
"""Class for suffix tree nodes."""
index = 0
def __init__(self, begin, end, parent):
"""Create node.
:param begin: begin index.
:param end: end index (not included).
:param parent:
"""
self.edges = dict()
self.begin = begin
self.end = end
self.parent = parent
self.id = self.index
self.index += 1
def __hash__(self) -> int:
"""Return hash value.
:return:
"""
return self.id
def __len__(self) -> int:
"""Return length of input edge.
:return:
"""
return self.end - self.begin
def __str__(self) -> str:
"""Return string interpretation suffix tree.
:return:
"""
result = []
def dfs(node):
nonlocal result
result.append(f'{string[node.begin:node.end]} -> ' + '{')
for character in node.edges:
result.append(f'{character}: -> ' + '{')
dfs(node.edges[character])
result.append('} ')
result.append('} ')
dfs(self)
return ''.join(result)
root = Node(0, 0, None)
suffix_links_table = dict()
def split(node: 'Node', position: int) -> 'Node':
"""Split edge by position (need for suffix links).
:param node:
:param position:
:return:
"""
if position == len(node):
return node
if position == 0:
return node.parent
new_node = Node(node.begin, node.begin + position, node.parent)
node.parent.edges[string[node.begin]] = new_node
new_node.edges[string[node.begin + position]] = node
node.parent = new_node
node.begin += position
return new_node
def suffix_link(node: 'Node') -> 'Node':
"""Return suffix link by node.
:param node:
:return:
"""
nonlocal suffix_links_table
if node in suffix_links_table:
return suffix_links_table[node]
if node is root:
result = root
else:
link = suffix_link(node.parent)
result = split(*next_state(
link, len(link),
node.begin + (1 if node.parent is root else 0),
node.end
))
suffix_links_table[node] = result
return result
def next_state(node: 'Node', position: int, begin: int, end: int) -> tuple['Node', int]:
"""Return next state.
:param node: node.
:param position: position on edge.
:param begin: begin index.
:param end: end index.
:return:
"""
while begin < end:
if position == len(node):
if string[begin] in node.edges:
node = node.edges[string[begin]]
position = 0
else:
result = (None, None)
break
else:
if string[node.begin + position] != string[begin]:
result = (None, None)
break
if end - begin < len(node) - position:
result = (node, position + end - begin)
break
begin += len(node) - position
position = len(node)
else:
result = (node, position)
return result
node = root
position = 0
for index, character in enumerate(string):
while True:
node_, position_ = next_state(node, position, index, index + 1)
if node_ is not None:
node = node_
position = position_
break
mid = split(node, position)
leaf = Node(index, len(string), mid)
mid.edges[string[index]] = leaf
node = suffix_link(mid)
position = len(node)
if mid is root:
break
return root
weight_pattern = collections.Counter(patterns)
patterns = list(set(patterns))
word += '$'
suffix_tree = ukkonen(word)
result = 0
for pattern in patterns:
position = 0
node = suffix_tree
for character in pattern:
if position < node.end:
if position >= len(word):
break
if character != word[position]:
break
position += 1
else:
if character not in node.edges:
break
node = node.edges[character]
position = node.begin + 1
else:
result += weight_pattern[pattern]
return result
|
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.