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/best-sightseeing-pair/discuss/2429468/python-oror-DP-oror-constant-space-oror-constant-time-oror-fast-oror | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
i = 0
score = 0
for j in range(1, len(values)):
score = max(score, values[i] + values[j] + (i - j))
if values[j] >= values[i]:
i = j
elif values[i] - values[j] < j... | best-sightseeing-pair | python || DP || constant space || constant time || fast || | Yared_betsega | 0 | 48 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,600 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/2204185/Python3-or-Clear-Explanation-with-Illustration-or-Faster-than-93.63-or-Less-than-87.31 | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
Max = res = 0
for i in range(1, len(values)):
Max = max(Max-1, values[i-1]-1)
res = max(res, values[i]+Max)
return res | best-sightseeing-pair | ✅[Python3] | Clear Explanation with Illustration | Faster than 93.63% | Less than 87.31% | chanchishen | 0 | 35 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,601 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/2185730/python-3-or-very-simple-solution-or-O(n)O(1) | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
score = i = 0
for j in range(1, len(values)):
score = max(score, values[i] + values[j] + i - j)
if values[j] + j - i > values[i]:
i = j
return score | best-sightseeing-pair | python 3 | very simple solution | O(n)/O(1) | dereky4 | 0 | 80 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,602 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/2060075/Python-or-DP | class Solution:
def maxScoreSightseeingPair(self, va: List[int]) -> int:
ma = float("-inf")
ans = float("-inf")
for i in range(len(va)):
if i!=0:
ans = max(ans,va[i]-i+ma)
ma = max(ma,va[i]+i)
return ans | best-sightseeing-pair | Python | DP | Shivamk09 | 0 | 50 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,603 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/2007189/ororPYTHON-SOLoror-FASTER-THAN-99-oror-EASY-oror-LINEAR-TIME-oror-EXPLAINED-oror-INTUTIVE-oror | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
ans = 0
n = len(values)
highest = values[0]
for i in range(1,n):
sub = values[i] - i
if sub + highest > ans : ans = sub + highest
if values[i] + i > highest: highest = val... | best-sightseeing-pair | ||PYTHON SOL|| FASTER THAN 99% || EASY || LINEAR TIME || EXPLAINED || INTUTIVE || | reaper_27 | 0 | 35 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,604 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1898262/Python3%3A-DP-Solution-%3A-Easy-to-Understand | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
"""
If you look closely on the requirement:
it asks for chosing a previous index(i)
from current index(j) by which the summation is maximum.
So, for a current index(j), our task is to figure... | best-sightseeing-pair | Python3: DP Solution : Easy to Understand | showing_up_each_day | 0 | 71 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,605 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1849875/Python-One-Pass-or-Clean-Code | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
currScore = maxScore = 0
for i in range(len(values)-2,-1,-1):
value1 = values[i] + values[i+1] -1
value2 = currScore - values[i+1] + values[i] -1
currSc... | best-sightseeing-pair | Python One Pass | Clean Code | deepanksinghal | 0 | 46 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,606 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1793193/Super-Clear-And-Easy-Understanding-Python3-Solution | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
ans=0
curMax=values[0]-1
for i in range(1,len(values)):
ans=max(ans,values[i]+curMax)
if values[i]>=curMax:
curMax=values[i]-1
else:
curMax-=1
... | best-sightseeing-pair | ♠️ Super Clear And Easy Understanding Python3 Solution | edwardchor | 0 | 45 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,607 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1784258/Python-easy-to-read-and-understand-or-DP | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
n = len(values)
t1, t2 = [values[0]], [values[0]]
for i in range(1, n):
t1.append(max(values[i]+i, t1[i-1]))
t2.append(values[i]-i)
t1, t2 = t1[:-1], t2[1:]
#print(t1, t2)
... | best-sightseeing-pair | Python easy to read and understand | DP | sanial2001 | 0 | 56 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,608 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1692717/Keep-calculating-best-fit-sightseeing-and-maximum-score | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
valuable_point_pair = (values[0], 0)
maxiumum_score = 0
def get_score(p, value):
return valuable_point_pair[0] + value - abs(p-valuable_point_pair[1])
for p, value in enumerate(... | best-sightseeing-pair | Keep calculating best fit sightseeing and maximum score | puremonkey2001 | 0 | 43 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,609 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1689682/One-Pass-Python-O(n) | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
best = 0
high, spot = 0, -1
for i, x in enumerate(values):
high -= 1 # old spot loses value due to increase in travel distance
score = high + x
best = max(score, best... | best-sightseeing-pair | One Pass Python [O(n)] | briancoj | 0 | 49 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,610 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1493819/Simple-Python-Solution-w-explanation-o(1)-space-o(n)-runtime | class Solution(object):
def maxScoreSightseeingPair(self, nums):
starter = nums[0]+0
ender = 0
bestPair = 0
for i in range(1,len(nums)):
ender = starter +nums[i]-i
starter = max(starter,nums[i]+i)
bestPair = max(bestPair,ender)
return bestP... | best-sightseeing-pair | Simple Python Solution w/ explanation, o(1) space, o(n) runtime | CesarDN | 0 | 79 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,611 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1480982/Python3-solution-with-comment | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
# Keeps ith sightseeing spot max value so far
current_max_i = values[0]
# The maximum score of a pair of sightseeing splots to return
answer = 0
for i in range(1, len(values)):
# Image v... | best-sightseeing-pair | Python3 solution with comment | yukikitayama | 0 | 52 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,612 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1006939/Python3-linear-scan | class Solution:
def maxScoreSightseeingPair(self, A: List[int]) -> int:
ans = val = 0
for i, x in enumerate(A):
ans = max(ans, x - i + val)
val = max(val, x + i)
return ans | best-sightseeing-pair | [Python3] linear scan | ye15 | 0 | 89 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,613 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/398015/Python-3 | class Solution:
def maxScoreSightseeingPair(self, A):
a, b = A[0], 0
for i in range(1, len(A)):
b, a = max(a + A[i] - i, b), max(A[i] + i, a)
return b | best-sightseeing-pair | Python 3 | slight_edge | 0 | 274 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,614 |
https://leetcode.com/problems/best-sightseeing-pair/discuss/1598504/Python3-One-pass-O(1)-space | class Solution:
def maxScoreSightseeingPair(self, values: List[int]) -> int:
prev, res = values[0], 0
for i in range(1, len(values)):
res = max(res, prev + values[i] - i)
prev = max(prev, values[i] + i)
return res | best-sightseeing-pair | [Python3] One pass, O(1) space | maosipov11 | -1 | 117 | best sightseeing pair | 1,014 | 0.595 | Medium | 16,615 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/1655649/Python3-Less-Math-More-Intuition-or-2-Accepted-Solutions-or-Intuitive | class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if not k % 2 or not k % 5: return -1
n = length = 1
while True:
if not n % k: return length
length += 1
n = 10*n + 1 | smallest-integer-divisible-by-k | [Python3] ✔️ Less Math, More Intuition ✔️ | 2 Accepted Solutions | Intuitive | PatrickOweijane | 26 | 1,900 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,616 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/1655649/Python3-Less-Math-More-Intuition-or-2-Accepted-Solutions-or-Intuitive | class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if not k % 2 or not k % 5: return -1
r = length = 1
while True:
r = r % k
if not r: return length
length += 1
r = 10*r + 1 | smallest-integer-divisible-by-k | [Python3] ✔️ Less Math, More Intuition ✔️ | 2 Accepted Solutions | Intuitive | PatrickOweijane | 26 | 1,900 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,617 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/1655706/Using-while-loop-and-hashmap-in-Python | class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
#edge case
if k % 2 == 0 or k % 5 == 0: return -1
#keep track of the remainder
remain, length = 0, 0
found_so_far = set()
while remain not in found_so_far:
found_so_far.add(remain)
remain =... | smallest-integer-divisible-by-k | Using while loop and hashmap in Python | kryuki | 2 | 118 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,618 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/948652/Python-Simple-Solutiom | class Solution:
def smallestRepunitDivByK(self, K):
if K % 2 == 0 or K % 5 == 0:
return -1
r = 0
for N in range(1, K + 1):
r = (r * 10 + 1) % K
if r==0:
return N | smallest-integer-divisible-by-k | Python Simple Solutiom | lokeshsenthilkumar | 1 | 175 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,619 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/2007261/oror-PYTHON-SOL-oror-REMAINDER-FIND-oror-HASHMAP-oror-EASY-oror-EXPLAINED | class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if k % 2 == 0: return -1
n = 1
leng = 1
mapp = {}
while True:
rem = n % k
if rem == 0: return leng
if rem in mapp : return -1
mapp[rem] = True
n = n*10 ... | smallest-integer-divisible-by-k | || PYTHON SOL || REMAINDER FIND || HASHMAP || EASY || EXPLAINED | reaper_27 | 0 | 73 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,620 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/1657500/Python3-Performant-and-streamlined-code(payload-4-lines!)-if-you-have-minimum-lines-obsession-%3A) | class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if math.gcd(k,10) != 1: return -1 # intuitative after some observation: any number divisable by 2 or 5 will never be a divisor of preunit.
reminder = 0
for length in range(1, k+1):
if (reminder := (1+reminder*10) % k... | smallest-integer-divisible-by-k | ✅ [Python3] Performant and streamlined code(payload 4 lines!) if you have minimum lines obsession :) | win-9527 | 0 | 21 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,621 |
https://leetcode.com/problems/smallest-integer-divisible-by-k/discuss/352334/Solution-in-Python-3-(beats-~98)-(With-Explanation) | class Solution:
def smallestRepunitDivByK(self, K: int) -> int:
if K % 2 == 0 or K % 5 == 0: return -1
i = n = 1
while n % K != 0: n, i = (10*n + 1) % K, i + 1
return i
- Junaid Mansuri
(LeetCode ID)@hotmail.com | smallest-integer-divisible-by-k | Solution in Python 3 (beats ~98%) (With Explanation) | junaidmansuri | 0 | 375 | smallest integer divisible by k | 1,015 | 0.47 | Medium | 16,622 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/1106296/Python3-2-approaches | class Solution:
def queryString(self, S: str, N: int) -> bool:
for x in range(N, 0, -1):
if bin(x)[2:] not in S: return False
return True | binary-string-with-substrings-representing-1-to-n | [Python3] 2 approaches | ye15 | 6 | 387 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,623 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/1106296/Python3-2-approaches | class Solution:
def queryString(self, S: str, N: int) -> bool:
ans = set()
for i in range(len(S)):
for ii in range(i, i + N.bit_length()):
x = int(S[i:ii+1], 2)
if 1 <= x <= N: ans.add(x)
return len(ans) == N | binary-string-with-substrings-representing-1-to-n | [Python3] 2 approaches | ye15 | 6 | 387 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,624 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/2841590/2-LINES-ororor-PYTHON-EASY-SOLUTIONoror-USING-STRING | class Solution:
def queryString(self, s: str, n: int) -> bool:
for i in range(1,n+1):
if bin(i)[2:] not in s:return 0
return 1 | binary-string-with-substrings-representing-1-to-n | 2 LINES ||| PYTHON EASY SOLUTION|| USING STRING | thezealott | 1 | 3 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,625 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/1798033/Python3-solution-or-Using-python-bin()-function-or-88-lesser-memory | class Solution:
def queryString(self, s: str, n: int) -> bool:
for i in range(1,n+1):
if (bin(i)[2:]) not in s:
return False
return True | binary-string-with-substrings-representing-1-to-n | ✔Python3 solution | Using python bin() function | 88% lesser memory | Coding_Tan3 | 1 | 112 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,626 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/2007280/ororPYTHON-SOL-oror-SIMPLE-oror-EXPLAINED-oror-STRINGS-oror | class Solution:
def queryString(self, s: str, n: int) -> bool:
leng_s = len(s)
for i in range(1,n+1):
binary = str(bin(i)[2:])
leng_b = len(binary)
flag = False
for j in range(leng_s - leng_b + 1):
if s[j:j + leng_b] == binary:
... | binary-string-with-substrings-representing-1-to-n | ||PYTHON SOL || SIMPLE || EXPLAINED || STRINGS || | reaper_27 | 0 | 95 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,627 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/1408126/24ms-or-96-fasteror-Simple-python3-solution. | class Solution:
def queryString(self, s: str, n: int) -> bool:
while(n):
a=bin(n)
if(a.replace('0b','') not in s):
return 0
n-=1
return 1
Improved version
class Solution:
def queryString(self, s: str, n: int) -> bool:
while(n):
if(bin(n)[2:] not in s):
return 0
n-=1
re... | binary-string-with-substrings-representing-1-to-n | 24ms | 96% faster| Simple python3 solution. | kavikidadumbe | 0 | 111 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,628 |
https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/discuss/597247/Super-easy-python-solution | class Solution:
def queryString(self, S: str, N: int) -> bool:
for i in range(1,N+1):
x=str(bin(i).replace("0b", ""))
if S.find(x)==-1:
return False
return True | binary-string-with-substrings-representing-1-to-n | Super easy python solution | Ayu-99 | 0 | 86 | binary string with substrings representing 1 to n | 1,016 | 0.575 | Medium | 16,629 |
https://leetcode.com/problems/convert-to-base-2/discuss/2007392/PYTHON-SOL-oror-EASY-oror-BINARY-CONVERSION-oror-WELL-EXPLAINED-oror | class Solution:
def baseNeg2(self, n: int) -> str:
ans = ""
while n != 0:
if n%-2 != 0 :
ans = '1' + ans
n = (n-1)//-2
else:
ans = '0' + ans
n = n//-2
return ans if ans !="" else '0' | convert-to-base-2 | PYTHON SOL || EASY || BINARY CONVERSION || WELL EXPLAINED || | reaper_27 | 1 | 172 | convert to base 2 | 1,017 | 0.61 | Medium | 16,630 |
https://leetcode.com/problems/convert-to-base-2/discuss/2482057/Python-siolution | class Solution:
def baseNeg2(self, n: int) -> str:
result = ""
while n != 0:
if n%2 != 0 :
result = '1' + result
n = (n-1)//-2
else:
result = '0' + result
n = n//-2
return result if result != "" else '0' | convert-to-base-2 | Python siolution | Yauhenish | 0 | 57 | convert to base 2 | 1,017 | 0.61 | Medium | 16,631 |
https://leetcode.com/problems/convert-to-base-2/discuss/1015354/Python3-similar-to-base-2 | class Solution:
def baseNeg2(self, N: int) -> str:
ans = []
while N:
ans.append(N & 1)
N = (1-N) >> 1
return "".join(map(str, ans[::-1] or [0])) | convert-to-base-2 | [Python3] similar to base-2 | ye15 | 0 | 159 | convert to base 2 | 1,017 | 0.61 | Medium | 16,632 |
https://leetcode.com/problems/convert-to-base-2/discuss/1015354/Python3-similar-to-base-2 | class Solution:
def baseNeg2(self, N: int) -> str:
ans = []
while N:
ans.append(N & 1)
N = -(N >> 1)
return "".join(map(str, ans[::-1] or [0])) | convert-to-base-2 | [Python3] similar to base-2 | ye15 | 0 | 159 | convert to base 2 | 1,017 | 0.61 | Medium | 16,633 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/356289/Solution-in-Python-3-(beats-~98)-(three-lines)-(-O(1)-space-) | class Solution:
def prefixesDivBy5(self, A: List[int]) -> List[bool]:
n = 0
for i in range(len(A)): A[i], n = (2*n + A[i]) % 5 == 0, (2*n + A[i]) % 5
return A
- Junaid Mansuri
(LeetCode ID)@hotmail.com | binary-prefix-divisible-by-5 | Solution in Python 3 (beats ~98%) (three lines) ( O(1) space ) | junaidmansuri | 4 | 349 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,634 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/791074/Python-Simple-solution | class Solution:
def prefixesDivBy5(self, A: List[int]) -> List[bool]:
s='';l=[]
for i in A:
s+=str(i)
l.append(int(s,2)%5==0)
return l | binary-prefix-divisible-by-5 | Python Simple solution | lokeshsenthilkumar | 3 | 275 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,635 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1231086/Python3-simple-solution | class Solution:
def prefixesDivBy5(self, nums: List[int]) -> List[bool]:
res = []
n = 0
for i in nums:
n *= 2
if i == 1:
n += 1
res.append(n % 5 == 0)
return res | binary-prefix-divisible-by-5 | Python3 simple solution | EklavyaJoshi | 2 | 59 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,636 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1060101/Time-O(n)-Space-O(1)-Python3-solution | class Solution:
def prefixesDivBy5(self, A: List[int]) -> List[bool]:
# time O(n)
# space O(1)
output = []
last_bit = 0
for i in range(len(A)):
new_bit = last_bit*2 + A[i]
output.append(new_bit % 5 == 0)
last_bit = new_bit
return ou... | binary-prefix-divisible-by-5 | Time O(n) Space O(1) Python3 solution | mhviraf | 1 | 110 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,637 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/2524509/python | class Solution:
def prefixesDivBy5(self, nums: List[int]) -> List[bool]:
s = ''
res = []
for i in nums:
s += str(i)
decimal = int(s , 2)
if decimal % 5 == 0:
res.append(True)
else:
res.append(False)
retur... | binary-prefix-divisible-by-5 | python | akashp2001 | 0 | 25 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,638 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/2021039/Python-Clean-and-Simple!-Bitwise | class Solution:
def prefixesDivBy5(self, nums):
total, result = 0, []
for num in nums:
total <<= 1
total += num
result.append(total % 5 == 0)
return result | binary-prefix-divisible-by-5 | Python - Clean and Simple! Bitwise | domthedeveloper | 0 | 64 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,639 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1568866/Python3-dollarolution | class Solution:
def prefixesDivBy5(self, nums: List[int]) -> List[bool]:
x, l = 0, []
for i in nums:
n = x * 2 + i
l.append(n%5 == 0)
x = n
return l | binary-prefix-divisible-by-5 | Python3 $olution | AakRay | 0 | 64 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,640 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1248954/Python-or-O(n) | class Solution:
def prefixesDivBy5(self, nums: List[int]) -> List[bool]:
len_nums=len(nums)
result=[False]*len_nums
prev = nums[0]
result[0] = True if prev%5==0 else False
for i in range(1, len_nums):
prev = 2*prev+nums[i] # previous = 2*(2^0num[0] + 2^1*num[1]) + nums[i] (0 or 1)
result[i] = True ... | binary-prefix-divisible-by-5 | Python | O(n) | rksharma19896 | 0 | 53 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,641 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1106199/python-direct-approach | class Solution:
def prefixesDivBy5(self, A: List[int]) -> List[bool]:
lis = []
st = ""
for i in range(len(A)):
st = st+str(A[i])
lis.append(int(st,2)%5 == 0)
return lis | binary-prefix-divisible-by-5 | python direct approach | abhisek_ | 0 | 69 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,642 |
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/1085391/Deterministic-Finite-Automaton-Python3-(-96-time-80-memory-) | class Solution:
def prefixesDivBy5(self, A: List[int]) -> List[bool]:
state = 0
answer = []
for a in A:
if a == 0:
state = ( 2*state ) % 5
else:
state = ( 2*state+1 ) % 5
answer.append(state==0)
return answe... | binary-prefix-divisible-by-5 | Deterministic Finite Automaton Python3 ( 96% time 80% memory ) | rafic | 0 | 47 | binary prefix divisible by 5 | 1,018 | 0.473 | Easy | 16,643 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/283607/Clean-Python-Code | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
result = []
stack = []
for i, current in enumerate(self.value_iterator(head)):
result.append(0)
while stack and stack[-1][0] < current:
_, index = stack.pop()
resul... | next-greater-node-in-linked-list | Clean Python Code | aquafie | 3 | 817 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,644 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/303575/Python-Solution | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
res, stack, idx = [], [], 0
while head:
while stack and stack[-1][0] < head.val:
_, i = stack.pop()
res[i] = head.val
res.append(0)
stack.append((h... | next-greater-node-in-linked-list | Python Solution | tahir3 | 2 | 761 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,645 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1554340/Python3-Two-version-of-solutions-with-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
res = []
stack = []
idx = 0
while head:
res.append(0)
while stack and stack[-1][0] < head.val:
_, index = stack.pop()
res[index] = hea... | next-greater-node-in-linked-list | [Python3] Two version of solutions with using stack | maosipov11 | 1 | 81 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,646 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1554340/Python3-Two-version-of-solutions-with-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
lst = []
stack = []
res = []
while head:
lst.append(head.val)
head = head.next
for i in range(len(lst) - 1, -1, -1):
max_prev = 0
... | next-greater-node-in-linked-list | [Python3] Two version of solutions with using stack | maosipov11 | 1 | 81 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,647 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1320814/Python3-solution-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
values = []
temp = head
while temp:
values.append(temp.val)
temp = temp.next
ans = [0]*len(values)
stack = []
for i,j in enumerate(values):
if not stack or stac... | next-greater-node-in-linked-list | Python3 solution using stack | EklavyaJoshi | 1 | 104 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,648 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/2782042/Easy-python-solution-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
arr = []
while head:
arr.append(head.val)
head = head.next
l=[]
stk=[]
n=len(arr)
for i in range(n-1,-1,-1):
if(len(stk)<=0):
l.append(0)
... | next-greater-node-in-linked-list | Easy python solution using stack | liontech_123 | 0 | 2 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,649 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/2781493/Next-Greater-Node-In-Linked-List-(Python) | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
nodeValue=[]
current = head
while current :
nodeValue.append(current.val)
current = current.next
output = [0] * len(nodeValue)
stack = []
for index , value in enu... | next-greater-node-in-linked-list | Next Greater Node In Linked List (Python) | abdullah956 | 0 | 1 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,650 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/2577113/Python-92-or-using-deque-as-a-stack | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
#reverse it
r = None
curr = head
while curr:
tmp = curr.next
curr.next = r
r = curr
curr = tmp
curr = r
'''
#use double ende... | next-greater-node-in-linked-list | Python 92% | using deque as a stack | pandish | 0 | 16 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,651 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/2333834/Python-monotonic-stack | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
result = []
stack = []
idx = 0
node = head
while node:
while stack and node.val > stack[-1][0]:
_, i = stack.pop()
result[i] = node.val
... | next-greater-node-in-linked-list | Python, monotonic stack | blue_sky5 | 0 | 14 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,652 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1975543/python-simple-easy-small-(Time-On-space-On) | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
l = [head]
w = head.next
while w != None :
while len(l) != 0 and l[-1].val < w.val :
l[-1].val = w.val
l.pop()
l.append(w)
w = w.next
while len(l) != 0 :
l[-1].val = 0
... | next-greater-node-in-linked-list | python - simple, easy, small (Time On, space On) | ZX007java | 0 | 53 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,653 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1764531/Python-oror-Recursion-oror-Stack | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
def recur(head):
if head.next is None:
return [0], [head.val]
ans, stack = recur(head.next)
while stack and stack[-1] <= head.val:
stack.pop(-1)
i... | next-greater-node-in-linked-list | Python || Recursion || Stack | kalyan_yadav | 0 | 74 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,654 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1624165/WEEB-DOES-PYTHON-(BEATS-98.48) | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
arr = []
pointer = head
while pointer:
arr.append(pointer.val)
pointer = pointer.next
result = [0] * len(arr)
stack = [] # stores index
for i in range(len(arr)):
# implement decreasing stack
while stack and arr... | next-greater-node-in-linked-list | WEEB DOES PYTHON (BEATS 98.48%) | Skywalker5423 | 0 | 127 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,655 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1452133/In-O(N)-easy-to-undestand-python3 | class Solution:
def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
if not head:
return head
arr = []
while head:
arr.append(head.val)
head = head.next
n = len(arr)
stack = []
... | next-greater-node-in-linked-list | In O(N) - easy to undestand - python3 | Shubham_Muramkar | 0 | 112 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,656 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1387313/Python3-Stacks-89-Faster-with-less-Memory | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
stack = []
while head:
stack.append(head.val)
head = head.next
tmp = []
i = len(stack) - 1
ans = [0] * len(stack)
while stack:
curr = stack.p... | next-greater-node-in-linked-list | [Python3] Stacks 89%, Faster with less Memory | whitehatbuds | 0 | 163 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,657 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1387313/Python3-Stacks-89-Faster-with-less-Memory | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
stack = []
while head:
stack.append(head.val)
head = head.next
tmp = []
i = len(stack) - 1
ans = [0] * len(stack)
for i in range(i, -1, -1):
while tm... | next-greater-node-in-linked-list | [Python3] Stacks 89%, Faster with less Memory | whitehatbuds | 0 | 163 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,658 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1387299/Python3-Recursive-Dynamic-Programming-Slow-but-Accepted-One-Pass. | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
if head is None:
return []
def recurse(node, mem, ans):
if node.next is None:
ans.append(0)
mem[node.val] = 0
return
... | next-greater-node-in-linked-list | [Python3] Recursive Dynamic Programming, Slow but Accepted, One Pass. | whitehatbuds | 0 | 59 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,659 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1291620/python3-O(N)-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
if head==None:
return []
if head.next==None:
return [0]
L=[]
while head:
L.append(head.val)
head=head.next
stack=[L[-1]]
t=[None for _ in range(len(... | next-greater-node-in-linked-list | python3 O(N) using stack | ketan_raut | 0 | 104 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,660 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1074486/python-3-simple-solution-using-stack | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
if not head:
return []
stack=[0]
li=[]
prev=head
head=head.next
prev.next=None
while(head):
node=head
head=head.next
node.next=prev
... | next-greater-node-in-linked-list | python 3 simple solution using stack | AchalGupta | 0 | 419 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,661 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1007891/Python3-forward-and-backward-approaches | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
ans, stack = [], []
while head:
while stack and stack[-1][1] < head.val: ans[stack.pop()[0]] = head.val
stack.append((len(ans), head.val))
ans.append(0)
head = head.next
... | next-greater-node-in-linked-list | [Python3] forward & backward approaches | ye15 | 0 | 46 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,662 |
https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/1007891/Python3-forward-and-backward-approaches | class Solution:
def nextLargerNodes(self, head: ListNode) -> List[int]:
prev, node = None, head
while node: node.next, node, prev = prev, node.next, node
node = prev
ans, stack = [], []
while node:
while stack and stack[-1] <= node.val: stack.pop()
... | next-greater-node-in-linked-list | [Python3] forward & backward approaches | ye15 | 0 | 46 | next greater node in linked list | 1,019 | 0.599 | Medium | 16,663 |
https://leetcode.com/problems/number-of-enclaves/discuss/1040282/Python-BFS-and-DFS-by-yours-truly | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
row, col = len(A), len(A[0])
if not A or not A[0]:
return 0
boundary1 = deque([(i,0) for i in range(row) if A[i][0]==1]) + deque([(i,col-1) for i in range(row) if A[i][col-1]==1])
boundary2 = deque([(0,i) for i in range(... | number-of-enclaves | Python BFS and DFS by yours truly | Skywalker5423 | 14 | 1,100 | number of enclaves | 1,020 | 0.65 | Medium | 16,664 |
https://leetcode.com/problems/number-of-enclaves/discuss/1040282/Python-BFS-and-DFS-by-yours-truly | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
row, col = len(A), len(A[0])
if not A or not A[0]:
return 0
def dfs(x,y,A):
if 0<=x<row and 0<=y<col and A[x][y] ==1:
A[x][y] = "T"
dfs(x+1,y,A)
dfs(x-1,y,A)
dfs(x,y+1,... | number-of-enclaves | Python BFS and DFS by yours truly | Skywalker5423 | 14 | 1,100 | number of enclaves | 1,020 | 0.65 | Medium | 16,665 |
https://leetcode.com/problems/number-of-enclaves/discuss/2519350/Python-Elegant-and-Short-or-In-place-or-DFS | class Solution:
"""
Time: O(n^2)
Memory: O(n^2)
"""
WATER = 0
LAND = 1
def numEnclaves(self, grid: List[List[int]]) -> int:
n, m = len(grid), len(grid[0])
for i in range(n):
self.sink_island(i, 0, grid)
self.sink_island(i, m - 1, grid)
for j in range(m):
self.sink_island(0, j, grid)
self.... | number-of-enclaves | Python Elegant & Short | In-place | DFS | Kyrylo-Ktl | 2 | 85 | number of enclaves | 1,020 | 0.65 | Medium | 16,666 |
https://leetcode.com/problems/number-of-enclaves/discuss/2274986/Python3-clean-DFS-solution | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
m,n = len(grid), len(grid[0])
visited = set()
result = [0]
def dfs(i,j, isBoundary):
if i < 0 or i >= m or j < 0 or j>= n or grid[i][j]!=1 or (i,j) in visited:
return
... | number-of-enclaves | 📌 Python3 clean DFS solution | Dark_wolf_jss | 1 | 19 | number of enclaves | 1,020 | 0.65 | Medium | 16,667 |
https://leetcode.com/problems/number-of-enclaves/discuss/1853713/Python-DFS-Solution | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
if not grid:
return 0
rows, cols = len(grid), len(grid[0])
def dfs(r, c, value):
if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] == 0:
return
grid[r][c]... | number-of-enclaves | [Python] DFS Solution | tejeshreddy111 | 1 | 91 | number of enclaves | 1,020 | 0.65 | Medium | 16,668 |
https://leetcode.com/problems/number-of-enclaves/discuss/2822959/DFS-approach-with-similar-question | class Solution:
def dfs_util(self,grid: List[List[int]],i:int,j:int,r:int,c:int) -> int:
# if any land cell touches boundary return -1
if i<0 or j<0 or i>=r or j>=c:
return -1
if grid[i][j]==0:
return 0
grid[i][j]=0
left = self.dfs_util(grid,i,j-1,r,c)... | number-of-enclaves | DFS approach with similar question | Sakshamji | 0 | 4 | number of enclaves | 1,020 | 0.65 | Medium | 16,669 |
https://leetcode.com/problems/number-of-enclaves/discuss/2772377/Python-or-Easy-Solution-or-Matrix | class Solution(object):
def numEnclaves(self, mat):
"""
:type grid: List[List[int]]
:rtype: int
"""
def solve(r, c):
if r not in range(len(mat)):
return float("inf")
if c not in range(len(mat[0])):
return float("inf")
... | number-of-enclaves | Python | Easy Solution | Matrix | atharva77 | 0 | 3 | number of enclaves | 1,020 | 0.65 | Medium | 16,670 |
https://leetcode.com/problems/number-of-enclaves/discuss/2690795/DFS-solution-in-python | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
#this bfs function will make all the boundary 1's and the 1's connected to them as 0
def dfs(i,j):
if i<0 or i>=m or j<0 or j>=n or grid[i][j]==0:
return 0
grid[i][j]=0
dfs(i+1,j)... | number-of-enclaves | DFS solution in python | shashank_2000 | 0 | 5 | number of enclaves | 1,020 | 0.65 | Medium | 16,671 |
https://leetcode.com/problems/number-of-enclaves/discuss/2642340/Python-BFS | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
R, C = len(grid), len(grid[0])
ans = 0
q = collections.deque([])
for r in range(R):
if r == 0 or r == R-1:
for c in range(C):
if grid[r][c] == 1:
... | number-of-enclaves | Python BFS | stanleyyuen_pang | 0 | 2 | number of enclaves | 1,020 | 0.65 | Medium | 16,672 |
https://leetcode.com/problems/number-of-enclaves/discuss/2589109/Python-really-easy-to-understand-approach....... | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
visited=set()
for i in range(0,len(grid)):
for j in range(0,len(grid[0])):
if i==0:
if grid[i][j]==1:
visited.add((i,j))
elif j==0:
... | number-of-enclaves | Python really easy to understand approach....... | guneet100 | 0 | 18 | number of enclaves | 1,020 | 0.65 | Medium | 16,673 |
https://leetcode.com/problems/number-of-enclaves/discuss/2407906/Optimal-python3-solution | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
n = len(grid)
m = len(grid[0])
def getNeighbours(root):
x, y = root
neighbours = []
if x > 0 and grid[x-1][y] == 1:
neighbours.append((x-1, y))
... | number-of-enclaves | Optimal python3 solution | destifo | 0 | 5 | number of enclaves | 1,020 | 0.65 | Medium | 16,674 |
https://leetcode.com/problems/number-of-enclaves/discuss/2356427/Python-3-or-O(rows*cols)-runtime-solution(Straightforward-BFS-%2B-Queue) | class Solution:
#Time-Complexity: O(rows*cols + rows*cols), since for loop must run through each and every cell grid! Our bfs
#helper in worst case has to run while loop for each entry if our grid is all land cells (rows*cols)!
#-> O(rows*cols)
#Space: O(rows*cols + rows*cols),worst case each and every ... | number-of-enclaves | Python 3 | O(rows*cols) runtime solution(Straightforward BFS + Queue) | JOON1234 | 0 | 10 | number of enclaves | 1,020 | 0.65 | Medium | 16,675 |
https://leetcode.com/problems/number-of-enclaves/discuss/2285869/Python3-DFS.-Remove-border-islands-and-count-%221%22s | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
C = len(grid[0])
R = len(grid)
if C <= 1 or R <= 1:
return 0
def dfs(i,j):
if (i>=0 and j>=0 and i<R and j<C and grid[i][j] == 1):
#if 0<i<R-1 and 0<j<C-1:
... | number-of-enclaves | Python3 DFS. Remove border islands and count "1"s | devmich | 0 | 12 | number of enclaves | 1,020 | 0.65 | Medium | 16,676 |
https://leetcode.com/problems/number-of-enclaves/discuss/2186157/python-3-or-simple-dfs-or-O(mn)O(1) | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
def dfs(i, j):
if grid[i][j] == 0:
return
grid[i][j] = 0
if i: dfs(i - 1, j)
if i != m - 1: dfs(i + 1, j)
if j: df... | number-of-enclaves | python 3 | simple dfs | O(mn)/O(1) | dereky4 | 0 | 35 | number of enclaves | 1,020 | 0.65 | Medium | 16,677 |
https://leetcode.com/problems/number-of-enclaves/discuss/2166254/Simple-DFS-Solution | class Solution:
def isSafe(self,i,j,grid):
n = len(grid)
m = len(grid[0])
if 0 <= i < n and 0 <= j < m:
return True
else:
return False
def dfs(self,i,j,grid):
if not self.isSafe(i,j,grid) or grid[i][j] != 1:
return
... | number-of-enclaves | Simple DFS Solution | Vaibhav7860 | 0 | 56 | number of enclaves | 1,020 | 0.65 | Medium | 16,678 |
https://leetcode.com/problems/number-of-enclaves/discuss/2107023/Python3-or-DFS | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
def dfs(grid, i, j):
if i<0 or j<0 or i >= len(grid) or j >= len(grid[0]):
return
if grid[i][j] == 0:
return
grid[i][j] = 0... | number-of-enclaves | Python3 | DFS | iamirulofficial | 0 | 18 | number of enclaves | 1,020 | 0.65 | Medium | 16,679 |
https://leetcode.com/problems/number-of-enclaves/discuss/2070141/Python3-fill-from-the-edges-and-then-count-ones | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
# 0 = sea, 1 = land
# move = adjacent cells 4-directionally
# can move walk off boundary
# we want number of land cells in grid
# which you cannot move off grid
# aka just fill in from edges and coun... | number-of-enclaves | Python3 fill from the edges and then count ones | normalpersontryingtopayrent | 0 | 20 | number of enclaves | 1,020 | 0.65 | Medium | 16,680 |
https://leetcode.com/problems/number-of-enclaves/discuss/2007519/PYTHON-SOL-oror-EASY-TO-READ-oror-BFS-SOL-oror-SIMPLE-oror-EXPLAINED-oror | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
rows = len(grid)
cols = len(grid[0])
ones = 0
queue = []
vis =[[False for i in range(cols)] for j in range(rows)]
for i in range(rows):
for j in range(cols):
if grid[i][j]... | number-of-enclaves | PYTHON SOL || EASY TO READ || BFS SOL || SIMPLE || EXPLAINED || | reaper_27 | 0 | 25 | number of enclaves | 1,020 | 0.65 | Medium | 16,681 |
https://leetcode.com/problems/number-of-enclaves/discuss/2006908/Python-easy-to-read-and-understand-or-DFS | class Solution:
def dfs(self, grid, row, col):
if row < 0 or col < 0 or row == len(grid) or col == len(grid[0]) or grid[row][col] != 1:
return 0
grid[row][col] = 2
t = self.dfs(grid, row-1, col)
l = self.dfs(grid, row, col-1)
d = self.dfs(grid, row+1, col)
... | number-of-enclaves | Python easy to read and understand | DFS | sanial2001 | 0 | 27 | number of enclaves | 1,020 | 0.65 | Medium | 16,682 |
https://leetcode.com/problems/number-of-enclaves/discuss/1960865/faster-than-98.63-of-Python3-online-submissions-for-Number-of-Enclaves. | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
m,n=len(grid),len(grid[0])
def Util(r,c):
grid[r][c]=0
for i,j in [(r-1,c),(r+1,c),(r,c-1),(r,c+1)]:
if i<0 or i>=m or j<0 or j>=n:
continue
if grid[i][... | number-of-enclaves | faster than 98.63% of Python3 online submissions for Number of Enclaves. | Neerajbirajdar | 0 | 30 | number of enclaves | 1,020 | 0.65 | Medium | 16,683 |
https://leetcode.com/problems/number-of-enclaves/discuss/1891956/python-easy-to-understand-bfs-solution | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
enclaves = 0
rows = len(grid)
cols = len(grid[0])
visited = [[False for _ in range(cols)] for _ in range(rows)]
for i in range(rows):
for j in range(cols):
if not visited[i][j] an... | number-of-enclaves | python easy to understand bfs solution | karthik2265 | 0 | 14 | number of enclaves | 1,020 | 0.65 | Medium | 16,684 |
https://leetcode.com/problems/number-of-enclaves/discuss/1853496/Dye-mainland-and-remain-enclaves-or-DFS-or-clear-and-with-explanation | class Solution:
def numEnclaves(self, grid):
"""
consider cells out side of grid as 'Mainland', like surrounding by 1s
so any 1 in the border is connect to Mainland, so as that island
since it's not a part of enclave, let's dye it dfs as water
and then use another dfs we can ... | number-of-enclaves | Dye mainland and remain enclaves | DFS | clear and with explanation | steve-jokes | 0 | 25 | number of enclaves | 1,020 | 0.65 | Medium | 16,685 |
https://leetcode.com/problems/number-of-enclaves/discuss/1808963/Python-Recursive-DFS | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
rows, cols = len(grid), len(grid[0])
Position = namedtuple('Position', ['row', 'col'])
def withinBounds(cell):
return 0 <= cell.row < rows and 0 <= cell.col < cols
def dfs(cell... | number-of-enclaves | Python Recursive DFS | Rush_P | 0 | 52 | number of enclaves | 1,020 | 0.65 | Medium | 16,686 |
https://leetcode.com/problems/number-of-enclaves/discuss/1665558/Python-DFS-Readable-with-Comments | class Solution:
def numEnclaves(self, grid: List[List[int]]) -> int:
#we will start dfs from the 1's on the boundary and will keep looking for 1's if we can visit them
#the final result is total number of 1's - visited 1's
M = len(grid)
N = len(grid[0])
visit = set()
count = 0
def dfs(i,j):
i... | number-of-enclaves | Python DFS Readable with Comments | Jazzyb1999 | 0 | 56 | number of enclaves | 1,020 | 0.65 | Medium | 16,687 |
https://leetcode.com/problems/number-of-enclaves/discuss/1659172/Python-simple-bfs-solution-(O(mn)-time-O(mn)-space) | class Solution:
from collections import deque
def numEnclaves(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
def valid(x, y):
return x>=0 and x<=m-1 and y>=0 and y<=n-1
def bfs(p, q):
queue = deque([(p, q)])
while... | number-of-enclaves | Python simple bfs solution (O(mn) time, O(mn) space) | byuns9334 | 0 | 30 | number of enclaves | 1,020 | 0.65 | Medium | 16,688 |
https://leetcode.com/problems/number-of-enclaves/discuss/1062883/Easy-Python-Solution-or-DFS-or-Python | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
# [[0,0,0,0],
# [1,0,1,0],
# [0,1,1,0],
# [0,0,0,0]]
def dfs(i,j,A):
if i < 0 or i > len(A) - 1 or j < 0 or j > len(A[0]) - 1 or A[i][j] != 1:
return
if A[i][j] == 1... | number-of-enclaves | Easy Python Solution | DFS | Python | Ayush87 | 0 | 129 | number of enclaves | 1,020 | 0.65 | Medium | 16,689 |
https://leetcode.com/problems/number-of-enclaves/discuss/1015528/Python-faster-than-93-DFS | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
'''
1) We need number of land squares, so a result variable count.
2) Return 0, if the size of A is 0.
3) If not, find a path to land from borders, extract borders and run dfs.
4) In dfs, mark each visited cell... | number-of-enclaves | Python, faster than 93%, DFS | Narasimhag | 0 | 62 | number of enclaves | 1,020 | 0.65 | Medium | 16,690 |
https://leetcode.com/problems/number-of-enclaves/discuss/1007898/Python3-flood-fill-via-dfs | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
m, n = len(A), len(A[0]) # dimensions
stack = []
for i in range(m):
if A[i][0]: stack.append((i, 0))
if A[i][n-1]: stack.append((i, n-1))
for j in range(n):
if A[0... | number-of-enclaves | [Python3] flood fill via dfs | ye15 | 0 | 39 | number of enclaves | 1,020 | 0.65 | Medium | 16,691 |
https://leetcode.com/problems/number-of-enclaves/discuss/650787/Python3-flood-fill-%2B-sum-Number-of-Enclaves | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
m = len(A)
n = len(A[0])
def floodFill(i: int, j:int) -> None:
nonlocal m, n
if not 0 <= i < m or not 0 <= j < n or not A[i][j]:
return
A[i][j] = 0
for x... | number-of-enclaves | Python3 flood fill + sum - Number of Enclaves | r0bertz | 0 | 100 | number of enclaves | 1,020 | 0.65 | Medium | 16,692 |
https://leetcode.com/problems/number-of-enclaves/discuss/479657/520ms-python3-using-stack | class Solution:
def numEnclaves(self, A: List[List[int]]) -> int:
m = len(A)
n = len(A[0])
stack = []
for i in range(m):
if A[i][0]==1:
A[i][0]=2
if 1<=i<=m-2 and A[i][1]==1 and n>=2:
A[i][1]=2
stack.... | number-of-enclaves | 520ms python3, using stack | felicia1994 | 0 | 53 | number of enclaves | 1,020 | 0.65 | Medium | 16,693 |
https://leetcode.com/problems/number-of-enclaves/discuss/471336/Python3-98.40-(496-ms)100.00-(13.9-MB)-O(n)-time-O(1)-space-recursion | class Solution:
def delete_valid_squares(self, A, row, column, max_row, max_column):
if (A[row][column]):
A[row][column] = 0
if (column < max_column):
self.delete_valid_squares(A, row, column + 1, max_row, max_column)
... | number-of-enclaves | Python3 98.40% (496 ms)/100.00% (13.9 MB) -- O(n) time / O(1) space -- recursion | numiek_p | 0 | 73 | number of enclaves | 1,020 | 0.65 | Medium | 16,694 |
https://leetcode.com/problems/number-of-enclaves/discuss/300833/Python%3A-Using-generators-to-make-the-code-easier-to-read-(beats-95) | class Solution:
# Generates all coordinates on boundaries
def boundary_coordinates(self, grid):
rows = len(grid)
cols = len(grid[0])
for row_index in range(rows):
yield (row_index, 0)
yield (row_index, cols - 1)
for col_index in range(1, cols - 1):
... | number-of-enclaves | Python: Using generators to make the code easier to read (beats 95%) | Hai_dee | 0 | 119 | number of enclaves | 1,020 | 0.65 | Medium | 16,695 |
https://leetcode.com/problems/remove-outermost-parentheses/discuss/1162269/Python-Simplest-Solution | class Solution:
def removeOuterParentheses(self, S: str) -> str:
stack=[]
counter=0
for i in S:
if i=='(':
counter=counter+1
if counter==1:
pass
else:
stack.append(i)
else... | remove-outermost-parentheses | Python Simplest Solution | aishwaryanathanii | 5 | 164 | remove outermost parentheses | 1,021 | 0.802 | Easy | 16,696 |
https://leetcode.com/problems/remove-outermost-parentheses/discuss/942888/Python-Simple-Solution | class Solution:
def removeOuterParentheses(self, S: str) -> str:
ans=[];o=0
for i in S:
if i=='(' and o>0:
ans.append(i)
if i==')' and o>1:
ans.append(')')
o+=1 if i=='(' else -1
return ''.join(ans) | remove-outermost-parentheses | Python Simple Solution | lokeshsenthilkumar | 2 | 534 | remove outermost parentheses | 1,021 | 0.802 | Easy | 16,697 |
https://leetcode.com/problems/remove-outermost-parentheses/discuss/2819789/Python-oror-96.77-Faster-oror-Without-Stack-oror-O(n)-Solution | class Solution:
def removeOuterParentheses(self, s: str) -> str:
c,j,n=0,0,len(s)
ans=[]
for i in range(n):
if s[i]=='(':
c+=1 #If there is opening paranthesis we increment the counter variable
else:
c-=1 #If there is closing paranthesi... | remove-outermost-parentheses | Python || 96.77% Faster || Without Stack || O(n) Solution | DareDevil_007 | 1 | 95 | remove outermost parentheses | 1,021 | 0.802 | Easy | 16,698 |
https://leetcode.com/problems/remove-outermost-parentheses/discuss/2819732/Python-Easy-Solution-Using-Stack-in-O(n)-Complexity | class Solution:
def removeOuterParentheses(self, s: str) -> str:
a,n=[],len(s)
i=j=0
t=''
while i<n:
print(s[i],"s[i]")
if a and a[-1]=='(' and s[i]==')':
a.pop()
elif len(a)==0 and s[i]=='(' and i>0:
t+=s[j+1:i-1]
... | remove-outermost-parentheses | Python Easy Solution Using Stack in O(n) Complexity | DareDevil_007 | 1 | 61 | remove outermost parentheses | 1,021 | 0.802 | Easy | 16,699 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.