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/reverse-string-ii/discuss/1477226/Simple-or-Python-3-or-24-ms-faster-than-97.61 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
return ''.join(s[i:i+k][::-1] + s[i+k:i+k*2] for i in range(0, len(s), k*2)) | reverse-string-ii | Simple | Python 3 | 24 ms, faster than 97.61% | deep765 | 2 | 205 | reverse string ii | 541 | 0.505 | Easy | 9,500 |
https://leetcode.com/problems/reverse-string-ii/discuss/1400209/One-multi-line-97-speed | class Solution:
def reverseStr(self, s: str, k: int) -> str:
return "".join(w if i % 2 else w[::-1] for i, w in
enumerate([s[i * k: (i + 1) * k]
for i in range(len(s) // k + 1)])) | reverse-string-ii | One multi-line, 97% speed | EvgenySH | 2 | 200 | reverse string ii | 541 | 0.505 | Easy | 9,501 |
https://leetcode.com/problems/reverse-string-ii/discuss/1371012/Python-20ms-faster-than-99.64 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
result = ""
for i in range(0, len(s), k*2):
result += s[i:i+k][::-1] + s[i+k:i+k*2]
return result | reverse-string-ii | Python, 20ms, faster than 99.64% | SG5 | 2 | 172 | reverse string ii | 541 | 0.505 | Easy | 9,502 |
https://leetcode.com/problems/reverse-string-ii/discuss/1272171/Easy-Python-Solution(97.20) | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s=list(s)
for i in range(0,len(s),2*k):
# print(s[i:i+k:-1])
s[i:i+k]=reversed(s[i:i+k])
return "".join(s) | reverse-string-ii | Easy Python Solution(97.20%) | Sneh17029 | 2 | 361 | reverse string ii | 541 | 0.505 | Easy | 9,503 |
https://leetcode.com/problems/reverse-string-ii/discuss/1145073/Easy-python-solution-O(N)-Time-O(N)-Space | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if k >= len(s):
return s[::-1]
i = 0
s = list(s)
while i < len(s):
l = i
h = (i + k - 1) if (i + k - 1) < len(s) else len(s) - 1
while l < len(s) and l < h:
... | reverse-string-ii | Easy python solution O(N) Time, O(N) Space | vanigupta20024 | 2 | 194 | reverse string ii | 541 | 0.505 | Easy | 9,504 |
https://leetcode.com/problems/reverse-string-ii/discuss/2559705/Simple-Python-recursion-solution-(Beginner-Friendly) | class Solution:
def reverseStr(self, s: str, k: int) -> str:
def reverse1(s, k):
if len(s)<k:
return s[::-1]
if len(s)>=k and len(s)<=2*k:
s1 = s[:k]; s2 = s[k:]
s1 = list(s1)
i = 0; j = len(s1)-1
while(i... | reverse-string-ii | Simple Python recursion solution (Beginner Friendly) | Arana | 1 | 125 | reverse string ii | 541 | 0.505 | Easy | 9,505 |
https://leetcode.com/problems/reverse-string-ii/discuss/1969603/easy-python-solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s_rev, i = '', 0
while i<len(s):
s_rev += (s[i:i+k])[::-1] + s[i+k:i+2*k]
i += 2*k
return s_rev | reverse-string-ii | easy python solution | wezel | 1 | 159 | reverse string ii | 541 | 0.505 | Easy | 9,506 |
https://leetcode.com/problems/reverse-string-ii/discuss/1902410/Python3-or-2-solution-or-Recursively-or-Iteratively | class Solution:
def reverseStr(self, s: str, k: int) -> str:
return s[:k][::-1] + s[k:2*k] + self.reverseStr(s[2*k:], k) if s else "" | reverse-string-ii | ✔Python3 | 2 solution | Recursively | Iteratively | Anilchouhan181 | 1 | 128 | reverse string ii | 541 | 0.505 | Easy | 9,507 |
https://leetcode.com/problems/reverse-string-ii/discuss/1902410/Python3-or-2-solution-or-Recursively-or-Iteratively | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if k>len(s):
return s[::-1]
for i in range(0,len(s),2*k):
s=s[:i]+s[i:i+k][::-1]+s[k+i:]
return s | reverse-string-ii | ✔Python3 | 2 solution | Recursively | Iteratively | Anilchouhan181 | 1 | 128 | reverse string ii | 541 | 0.505 | Easy | 9,508 |
https://leetcode.com/problems/reverse-string-ii/discuss/1821576/Python3-28ms-faster-than-96.94 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
out = ''
reverse = True
for i in range(0, len(s)+1, k):
out += s[i:i+k][::-1 if reverse else 1]
reverse = not reverse
return out | reverse-string-ii | Python3 28ms, faster than 96.94% | Minh4893IT | 1 | 160 | reverse string ii | 541 | 0.505 | Easy | 9,509 |
https://leetcode.com/problems/reverse-string-ii/discuss/1546351/Python3-Simple-Solution-but-not-fast | class Solution:
def reverseStr(self, s: str, k: int) -> str:
for idx in range(0, len(s), k * 2):
s = s[:idx] + s[idx:idx+k][::-1] + s[idx+k:]
return s | reverse-string-ii | [Python3] Simple Solution but not fast | JingMing_Chen | 1 | 122 | reverse string ii | 541 | 0.505 | Easy | 9,510 |
https://leetcode.com/problems/reverse-string-ii/discuss/1222858/Python3-simple-solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
flag = True
res = ''
x = ''
c = 0
for i in range(len(s)):
if flag:
if c < k:
x = s[i] + x
elif c == k:
flag = False
... | reverse-string-ii | Python3 simple solution | EklavyaJoshi | 1 | 76 | reverse string ii | 541 | 0.505 | Easy | 9,511 |
https://leetcode.com/problems/reverse-string-ii/discuss/1222858/Python3-simple-solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
flag = True
res = ''
for i in range(0,len(s),k):
if flag:
res += (s[i:i+k])[::-1]
flag = False
else:
res += s[i:i+k]
flag = True
return... | reverse-string-ii | Python3 simple solution | EklavyaJoshi | 1 | 76 | reverse string ii | 541 | 0.505 | Easy | 9,512 |
https://leetcode.com/problems/reverse-string-ii/discuss/1222858/Python3-simple-solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res = ''
for i in range(0,len(s),2*k):
x = s[i:i+2*k]
res += x[:k][::-1] + x[k:]
return res | reverse-string-ii | Python3 simple solution | EklavyaJoshi | 1 | 76 | reverse string ii | 541 | 0.505 | Easy | 9,513 |
https://leetcode.com/problems/reverse-string-ii/discuss/2833237/Python3 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res = ''
i = 0
while i < len(s):
res += s[i : i + k][::-1]
res += s[i + k: 2 * k + i]
i += 2 * k
return res | reverse-string-ii | Python3 | adilet_1864 | 0 | 3 | reverse string ii | 541 | 0.505 | Easy | 9,514 |
https://leetcode.com/problems/reverse-string-ii/discuss/2806881/String-Function-Based-Recursive-Python-Solution-45-faster-than-rest-O(n2k)-Complexity | class Solution:
def reverseStr(self, s: str, k: int) -> str:
n=len(s)
if n<k:
return s[::-1]
elif n<(2*k) and n>=k:
s1=s[:k]
s1=s1[::-1]+s[k:]
return s1
else:
s1=s[:2*k]
s2=s1[:k]
s2=s2[::-1]+s1[k:]
... | reverse-string-ii | String Function Based Recursive Python Solution 45% faster than rest O(n/2k) Complexity | SnehaGanesh | 0 | 2 | reverse string ii | 541 | 0.505 | Easy | 9,515 |
https://leetcode.com/problems/reverse-string-ii/discuss/2786225/Reverse-String-2 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
return ''.join([s[i*k:(i+1)*k][::((-1)**(i+1))] for i in range(len(s)//k+1)]) | reverse-string-ii | Reverse String 2 | Luna-martinez | 0 | 2 | reverse string ii | 541 | 0.505 | Easy | 9,516 |
https://leetcode.com/problems/reverse-string-ii/discuss/2749767/O(N)-time-or-O(1)-space-complexity | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if len(s) < k:
return s[::-1]
i=0
ans = ""
while i<len(s):
l = 2*k+i
# print(l)
if l <= len(s):
r = s[i:i+k]
ans += r[::-1]
... | reverse-string-ii | O(N) time | O(1) space complexity | wakadoodle | 0 | 9 | reverse string ii | 541 | 0.505 | Easy | 9,517 |
https://leetcode.com/problems/reverse-string-ii/discuss/2742606/Python-3-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
for i in range(0,len(s),2*k):
if(i+k<len(s)):
s=s[0:i]+s[i:i+k][::-1]+s[i+k:]
else:
s=s[0:i]+s[i:i+k][::-1]
return s | reverse-string-ii | Python 3 Solution | dnvavinash | 0 | 5 | reverse string ii | 541 | 0.505 | Easy | 9,518 |
https://leetcode.com/problems/reverse-string-ii/discuss/2739968/Python3-Recursive-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if len(s)<k:
return s[::-1]
if len(s)<2*k:
return s[:k][::-1]+s[k:]
return s[:k][::-1]+s[k:2*k]+self.reverseStr(s[2*k:],k) | reverse-string-ii | Python3 Recursive Solution | jsheng | 0 | 1 | reverse string ii | 541 | 0.505 | Easy | 9,519 |
https://leetcode.com/problems/reverse-string-ii/discuss/2710486/Python-Tri-liner | class Solution:
def reverseStr(self, s: str, k: int) -> str:
new_s = ""
for i in range(0, len(s), 2*k): new_s += s[i : i+k][::-1] + s[i+k : i+2*k]
return new_s | reverse-string-ii | Python Tri liner | code_snow | 0 | 10 | reverse string ii | 541 | 0.505 | Easy | 9,520 |
https://leetcode.com/problems/reverse-string-ii/discuss/2697524/Python-simple-solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s = [s[i:i+k] for i in range(0, len(s), k)]
for i in range(0, len(s), 2):
s[i] = s[i][::-1]
return ''.join(s) | reverse-string-ii | Python simple solution | kruzhilkin | 0 | 8 | reverse string ii | 541 | 0.505 | Easy | 9,521 |
https://leetcode.com/problems/reverse-string-ii/discuss/2675256/Easy-Python-Solution-oror-(Slicing) | class Solution:
def reverseStr(self, s: str, k: int) -> str:
for i in range(0,len(s),2*k):
if(i+k<len(s)):
s=s[0:i]+s[i:i+k][::-1]+s[i+k:]
else:
s=s[0:i]+s[i:i+k][::-1]
return s | reverse-string-ii | Easy Python Solution || (Slicing) | Kiran_Rokkam | 0 | 4 | reverse string ii | 541 | 0.505 | Easy | 9,522 |
https://leetcode.com/problems/reverse-string-ii/discuss/2650125/Python | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s = list(s)
fin = []
counter=0
def rev(res):
l,r =0,len(res)-1
while(l<r):
res[l],res[r] = res[r], res[l]
l+=1
r-=1
return res
for ... | reverse-string-ii | Python | naveenraiit | 0 | 6 | reverse string ii | 541 | 0.505 | Easy | 9,523 |
https://leetcode.com/problems/reverse-string-ii/discuss/2611829/Simple-Implementation-or-Python | class Solution:
def reverseStr(self, s: str, k: int) -> str:
words = [s[i:i+2*k] for i in range(0, len(s),2*k)]
for key, val in enumerate(words):
if len(val)>= k:
words[key] = val[:k][::-1] + val[k:]
else:
words[key] = val[::-1]
... | reverse-string-ii | Simple Implementation | Python | Abhi_-_- | 0 | 30 | reverse string ii | 541 | 0.505 | Easy | 9,524 |
https://leetcode.com/problems/reverse-string-ii/discuss/2608278/Python-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
string = list(s)
n = len(string)
reverse = True
for i in range(0, n, k):
if reverse:
left = i
right = min(i + k - 1, n - 1)
while left < right:
... | reverse-string-ii | Python Solution | mansoorafzal | 0 | 82 | reverse string ii | 541 | 0.505 | Easy | 9,525 |
https://leetcode.com/problems/reverse-string-ii/discuss/2598984/Python-oror-Easy-and-well-explain-solution | class Solution(object):
def reverseStr(self, s, k):
"""
:type s: str
:type k: int
:rtype: str
"""
for i in range(0,len(s), 2*k):
if i==0:
temp = s[i:k]
temp = temp[::-1]+s[k:]
s = temp
else:
... | reverse-string-ii | Python || Easy and well explain solution | ride-coder | 0 | 52 | reverse string ii | 541 | 0.505 | Easy | 9,526 |
https://leetcode.com/problems/reverse-string-ii/discuss/2519474/Python-O(n)-solution-simple-readable-code | class Solution:
def reverseStr(self, s: str, k: int) -> str:
counter = 0
ans = [c for c in s]
# number of iterations that we will make in s
iterations = len(s) // (2 * k)
# if s is not perfectly divisible by 2k, add 1 extra iteration
if len(s) % (2 * k):
... | reverse-string-ii | Python O(n) solution - simple readable code | ishaan06 | 0 | 42 | reverse string ii | 541 | 0.505 | Easy | 9,527 |
https://leetcode.com/problems/reverse-string-ii/discuss/2283319/Python-Fast-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
i = 0
res = ''
while(i<len(s)):
if i+k<len(s):
res += (s[i:i+k])[::-1]
i+=k
if i+k<len(s):
res+= s[i:i+k]
i+=k
... | reverse-string-ii | Python Fast Solution | harsh30199 | 0 | 77 | reverse string ii | 541 | 0.505 | Easy | 9,528 |
https://leetcode.com/problems/reverse-string-ii/discuss/2281269/Python-fast-(beats-98.4-)-and-short-(almost-1-line)-solution-with-Python-3.8-features-(PEP572) | class Solution:
def reverseStr(self, s: str, k: int) -> str:
rev = False # ↓ PEP572 (assignment expressions + list comprehension, see Xavier Guihot's answer at https://stackoverflow.com/questions/16632124/how-to-emulate-sum-using-a-list-comprehension)
return "".join([s[i:i... | reverse-string-ii | Python fast (beats 98.4 %) and short (almost 1 line) solution with Python 3.8 features (PEP572) | amaargiru | 0 | 100 | reverse string ii | 541 | 0.505 | Easy | 9,529 |
https://leetcode.com/problems/reverse-string-ii/discuss/2155136/Python-or-Simple-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
ans = ""
for i in range(0,len(s),2*k): ans+=s[i:i+k][::-1]+s[i+k:i+2*k]
return ans; | reverse-string-ii | Python | Simple Solution | hemanthkakumanu1 | 0 | 117 | reverse string ii | 541 | 0.505 | Easy | 9,530 |
https://leetcode.com/problems/reverse-string-ii/discuss/1918234/Python-Faster-Than-97.32-Easy-To-Understand-Explained-Solution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res, flip, i = "", True, 0
while i <= len(s):
if not flip:
res += s[i : i + k]
i += k
flip ^= True
else:
res += s[i : i + k][::-1]
... | reverse-string-ii | Python Faster Than 97.32% Easy To Understand Explained Solution | Hejita | 0 | 120 | reverse string ii | 541 | 0.505 | Easy | 9,531 |
https://leetcode.com/problems/reverse-string-ii/discuss/1762990/Simple-Python-Solution-oror-Faster-than-96 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if len(s)<k:
return s[::-1]
elif len(s)>k and len(s)< 2*k:
first_part = s[:k]
second_part = s[k:]
return first_part[::-1] + second_part
else:
ou... | reverse-string-ii | Simple Python Solution || Faster than 96% | Akhilesh_Pothuri | 0 | 140 | reverse string ii | 541 | 0.505 | Easy | 9,532 |
https://leetcode.com/problems/reverse-string-ii/discuss/1723905/Python-Simple-or-Beats-100-in-Memory | class Solution:
def reverseStr(self, s: str, k: int) -> str:
op = []
for i in range(0,len(s),2*k):
op.append(s[i:k+i][::-1])
op.append(s[k+i:i + (2*k)])
return "".join(op) | reverse-string-ii | Python Simple | Beats 100% in Memory | veerbhansari | 0 | 114 | reverse string ii | 541 | 0.505 | Easy | 9,533 |
https://leetcode.com/problems/reverse-string-ii/discuss/1568905/Python-faster-than-98 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res = ''
for i in range(0, len(s), 2 * k):
a = i + k
b = i + 2*k
res += s[i:a][::-1] + s[a:b]
return res | reverse-string-ii | Python faster than 98% | dereky4 | 0 | 183 | reverse string ii | 541 | 0.505 | Easy | 9,534 |
https://leetcode.com/problems/reverse-string-ii/discuss/1482829/Literal-two-pointer-approach-in-Python | class Solution:
def reverseStr(self, s: str, k: int) -> str:
n, s = len(s), list(s)
def tow_pointer_gen():
i, j = 0, k - 1
while i < n:
yield i, j if j < n else n - 1
i += 2 * k
j += 2 * k
for l... | reverse-string-ii | Literal two-pointer approach in Python | mousun224 | 0 | 117 | reverse string ii | 541 | 0.505 | Easy | 9,535 |
https://leetcode.com/problems/reverse-string-ii/discuss/1298992/Python3-dollarolution | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s = list(s)
if len(s) < k:
return (''.join(reversed(s)))
for i in range(0,len(s),2*k):
v = s[i:i+k]
c = 1
for j in range(i,i+len(v)):
s[j] = v[-c]
c +=... | reverse-string-ii | Python3 $olution | AakRay | 0 | 118 | reverse string ii | 541 | 0.505 | Easy | 9,536 |
https://leetcode.com/problems/reverse-string-ii/discuss/1265018/Python3-Solution-Straight-forward | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s = list(s)
for i in range(0, len(s), 2*k):
s[i:i+k] = reversed(s[i:i+k])
return "".join(s) | reverse-string-ii | Python3 Solution Straight forward | Sanyamx1x | 0 | 80 | reverse string ii | 541 | 0.505 | Easy | 9,537 |
https://leetcode.com/problems/reverse-string-ii/discuss/1211865/Python-or-3-liner | class Solution:
def reverseStr(self, s: str, k: int) -> str:
n = len(s)
ans = [
s[i:i+k][::-1] + s[i+k:i+2*k]
for i in range(0,n,2*k)
]
return ''.join(ans) | reverse-string-ii | Python | 3 liner | Sanjaychandak95 | 0 | 54 | reverse string ii | 541 | 0.505 | Easy | 9,538 |
https://leetcode.com/problems/reverse-string-ii/discuss/1204568/Python-3-solution-Runtime%3A-20-ms-faster-than-99.26 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res = ''
for i in range(0, len(s), 2*k):
# print(i, i+k, i+2*k, s[i: i+k], s[i+k: i+2*k])
res += s[i: i+k][::-1] + s[i+k: i+2*k]
return res | reverse-string-ii | Python 3 solution Runtime: 20 ms, faster than 99.26% | hyiche | 0 | 43 | reverse string ii | 541 | 0.505 | Easy | 9,539 |
https://leetcode.com/problems/reverse-string-ii/discuss/474545/Python3-93.30-(24-ms)100.00-(13.1-MB)-O(n)-time-O(n)-space-(output)-using-deque | class Solution:
def reverseStr(self, s: str, k: int) -> str:
ret = collections.deque()
substring = collections.deque()
counter = 0
to_reverse = True
for letter in s:
counter += 1
if (to_reverse):
... | reverse-string-ii | Python3 93.30% (24 ms)/100.00% (13.1 MB) -- O(n) time / O(n) space (output) -- using deque | numiek_p | 0 | 114 | reverse string ii | 541 | 0.505 | Easy | 9,540 |
https://leetcode.com/problems/reverse-string-ii/discuss/438681/Python-3-100-time-100-memory | class Solution:
def reverseStr(self, s: str, k: int) -> str:
res = ''
for i in range(len(s)//(2*k)):
res += s[2*k*i:2*k*i+k][::-1] + s[2*k*i+k:2*k*(i+1)]
remainder = len(s) % (2*k)
if remainder <= k and remainder > 0:
res += s[-remainder:][::... | reverse-string-ii | Python 3 100% time, 100% memory | ilee102 | 0 | 104 | reverse string ii | 541 | 0.505 | Easy | 9,541 |
https://leetcode.com/problems/01-matrix/discuss/1556018/WEEB-DOES-PYTHON-BFS | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
row, col = len(mat), len(mat[0])
queue = deque([])
for x in range(row):
for y in range(col):
if mat[x][y] == 0:
queue.append((x, y, 1))
return self.bfs(row, col, queue, mat)
def bfs(self, row, col, queue, grid):
... | 01-matrix | WEEB DOES PYTHON BFS | Skywalker5423 | 7 | 414 | 01 matrix | 542 | 0.442 | Medium | 9,542 |
https://leetcode.com/problems/01-matrix/discuss/2604794/Python-Solution-Faster-then-99.63 | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
r=len(mat)
c=len(mat[0])
for i in range(r):
for j in range(c):
if mat[i][j]!=0:
top=mat[i-1][j] if i>0 else float('inf')
left=mat[i][j-1] if j>... | 01-matrix | Python Solution Faster then 99.63% | pranjalmishra334 | 3 | 408 | 01 matrix | 542 | 0.442 | Medium | 9,543 |
https://leetcode.com/problems/01-matrix/discuss/741536/Easy-Python-BFS-Solution-with-Comments! | class Solution:
def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
if not matrix: return
directions = ((1, 0), (-1, 0), (0, 1), (0, -1))
rows = len(matrix)
cols = len(matrix[0])
# Iterate through our mtrx til we find vals != 0
for... | 01-matrix | Easy Python BFS Solution with Comments! | Pythagoras_the_3rd | 3 | 204 | 01 matrix | 542 | 0.442 | Medium | 9,544 |
https://leetcode.com/problems/01-matrix/discuss/1919229/Python-BFS-Solution | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
queue = []
for i in range(len(mat)):
for j in range(len(mat[0])):
if mat[i][j] == 0:
queue.append((i,j))
else:
mat[i]... | 01-matrix | Python BFS Solution | Brillianttyagi | 2 | 186 | 01 matrix | 542 | 0.442 | Medium | 9,545 |
https://leetcode.com/problems/01-matrix/discuss/1786045/Python-3-(700ms)-or-2-Traversal-Approach | class Solution:
def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
m, n = len(matrix), len(matrix and matrix[0])
for i in range(m):
for j in range(n):
if matrix[i][j] != 0:
matrix[i][j] = float("inf")
if i > 0 and m... | 01-matrix | Python 3 (700ms) | 2 Traversal Approach | MrShobhit | 2 | 281 | 01 matrix | 542 | 0.442 | Medium | 9,546 |
https://leetcode.com/problems/01-matrix/discuss/874077/Python3-BFS-by-frontier | class Solution:
def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
m, n = len(matrix), len(matrix[0])
front = set((i, j) for i in range(m) for j in range(n) if not matrix[i][j]) # frontier
seen = front.copy() # visited cell
k = 0
while front: # bfs... | 01-matrix | [Python3] BFS by frontier | ye15 | 2 | 144 | 01 matrix | 542 | 0.442 | Medium | 9,547 |
https://leetcode.com/problems/01-matrix/discuss/874077/Python3-BFS-by-frontier | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
ans = [[inf]*n for _ in range(m)]
for i in range(m):
for j in range(n):
if not mat[i][j]: ans[i][j] = 0
else:
... | 01-matrix | [Python3] BFS by frontier | ye15 | 2 | 144 | 01 matrix | 542 | 0.442 | Medium | 9,548 |
https://leetcode.com/problems/01-matrix/discuss/2387807/Python-Solution-Dynamic-Programmming-95-faster | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
rows = len(mat)
cols = len(mat[0])
for i in range(rows):
for j in range(cols):
if mat[i][j] != 0:
top = mat[i-1][j] if i>0 else float('inf')
... | 01-matrix | Python Solution Dynamic Programmming, 95% faster | prachetshah26 | 1 | 167 | 01 matrix | 542 | 0.442 | Medium | 9,549 |
https://leetcode.com/problems/01-matrix/discuss/2075926/Iterative-Tabulation-oror-EXPLAINED-oror-SIMPLE-FAST | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m=len(mat)
n=len(mat[0])
dist = [[float('inf')]* n for _ in range(m)]
for i in range(m):
for j in range(n):
if mat[i][j]==0:
dist[i][j]=0
... | 01-matrix | Iterative Tabulation || EXPLAINED || SIMPLE FAST | karan_8082 | 1 | 79 | 01 matrix | 542 | 0.442 | Medium | 9,550 |
https://leetcode.com/problems/01-matrix/discuss/1370034/DP-Solution-Explanation-with-Diagrams-%2B-Code-in-Python3 | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
mat = mat
n = len(mat)
m = len(mat[0])
# init with +inf
ans = [[float('inf')]*m for _ in range(n)]
# from top left to bottom right, as we usually do
for i in range(n):
... | 01-matrix | DP Solution Explanation with Diagrams + Code in Python3 | chaudhary1337 | 1 | 143 | 01 matrix | 542 | 0.442 | Medium | 9,551 |
https://leetcode.com/problems/01-matrix/discuss/1356594/Elegant-Python-Iterative-BFS | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
ROWS, COLUMNS = len(mat), len(mat[0])
grid = [[0 for _ in range(COLUMNS)] for _ in range(ROWS)]
queue = deque()
visited = [[False for _ in range(COLUMNS)] for _ in range(ROWS)]
for row ... | 01-matrix | Elegant Python Iterative BFS | soma28 | 1 | 163 | 01 matrix | 542 | 0.442 | Medium | 9,552 |
https://leetcode.com/problems/01-matrix/discuss/2831267/Python-Like-62.-Unique-Paths-but-also-from-Bottom-%2B-Right | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
'''
The idea is similar to Unique Paths, https://leetcode.com/problems/unique-paths/
in that, we get the minimum of the accumulated path totals from TOP + LEFT
but here, we also then compare against accumulate... | 01-matrix | [Python] Like 62. Unique Paths, but also from Bottom + Right | graceiscoding | 0 | 3 | 01 matrix | 542 | 0.442 | Medium | 9,553 |
https://leetcode.com/problems/01-matrix/discuss/2810966/Python-(Simple-Dynamic-Programming) | class Solution:
def updateMatrix(self, mat):
m, n, ans, visited = len(mat), len(mat[0]), [], set()
for i in range(m):
for j in range(n):
if mat[i][j] == 0:
ans.append((i,j,0))
visited.add((i,j))
dp = [[0]*n for _ ... | 01-matrix | Python (Simple Dynamic Programming) | rnotappl | 0 | 4 | 01 matrix | 542 | 0.442 | Medium | 9,554 |
https://leetcode.com/problems/01-matrix/discuss/2801282/Python-3-BFS-simple-solution | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
def cell_n_neighbours (r, c):
visited.add((r, c))
for dr_r, dr_c in drt:
stack.append([r + dr_r, c + dr_c])
stack = deque()
visited = set()
drt = ... | 01-matrix | Python 3 - BFS - simple solution | noob_in_prog | 0 | 5 | 01 matrix | 542 | 0.442 | Medium | 9,555 |
https://leetcode.com/problems/01-matrix/discuss/2726325/Python3-Intuitive-BFS | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
ROWS, COLS = len(mat), len(mat[0])
res = [[0] * COLS for _ in range(ROWS)]
# get starting points
q = deque()
for r in range(ROWS):
for c in range(COLS):
if mat[r][c] ... | 01-matrix | Python3 Intuitive BFS | jonathanbrophy47 | 0 | 15 | 01 matrix | 542 | 0.442 | Medium | 9,556 |
https://leetcode.com/problems/01-matrix/discuss/2695576/Multi-source-BFS-Python-Easy-Solve | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
self.X = [1, 0, -1, 0]
self.Y = [0, 1, 0, -1]
vis = [[99999 for i in range(len(mat[0]))] for i in range(len(mat))]
q = deque()
for i in range(len(mat)):
for j in range(len(mat[0])):
... | 01-matrix | Multi source BFS Python Easy Solve | anu1rag | 0 | 5 | 01 matrix | 542 | 0.442 | Medium | 9,557 |
https://leetcode.com/problems/01-matrix/discuss/2677674/BFS-solution-in-python | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
visited=set()
queue=deque()
for i in range(len(mat)):
for j in range(len(mat[0])):
if mat[i][j]==0:
visited.add((i,j))
queue.append((i,j))
... | 01-matrix | BFS solution in python | shashank_2000 | 0 | 74 | 01 matrix | 542 | 0.442 | Medium | 9,558 |
https://leetcode.com/problems/01-matrix/discuss/2627439/python3-oror-easy-oror-bfs-solution | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
if not mat:
return
rowSize=len(mat)
colSize=len(mat[0])
visited=[[0]*colSize for i in range(rowSize)]
distanceArray=[[None]*colSize for i in range(len(mat)... | 01-matrix | python3 || easy || bfs solution | _soninirav | 0 | 30 | 01 matrix | 542 | 0.442 | Medium | 9,559 |
https://leetcode.com/problems/01-matrix/discuss/2607637/Python-Simple-solution-easy-to-understand-with-explanation-faster-than-89 | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
M = len(mat)
N = len(mat[0])
def update(i,j,d):
if not visited[i][j]:
dist[i][j] = d
q.append((i,j))
visited[i][j] = 1
def e... | 01-matrix | [Python] Simple solution, easy to understand, with explanation [faster than 89%] | alexion1 | 0 | 34 | 01 matrix | 542 | 0.442 | Medium | 9,560 |
https://leetcode.com/problems/01-matrix/discuss/2475891/Most-Optimal-Solution-with-Comments! | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
rows, cols = len(mat), len(mat[0])
# Note: it doesn't matter which direction you start,
# you just have to understand that in the first go,
# the 1's ARE NOT the distance, so don't take the mi... | 01-matrix | Most Optimal Solution with Comments! | EdwinJagger | 0 | 105 | 01 matrix | 542 | 0.442 | Medium | 9,561 |
https://leetcode.com/problems/01-matrix/discuss/2354503/Python-runtime-77.32-memory-90.18 | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
min_steps = {}
for row in range(len(mat)):
for col in range(len(mat[0])):
if mat[row][col] != 0:
mat[row][col] = self.updateCell(mat, min_steps, {}, row, col)
... | 01-matrix | Python, runtime 77.32%, memory 90.18% | tsai00150 | 0 | 182 | 01 matrix | 542 | 0.442 | Medium | 9,562 |
https://leetcode.com/problems/01-matrix/discuss/2354503/Python-runtime-77.32-memory-90.18 | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
# start from top-left
for row in range(len(mat)):
for col in range(len(mat[0])):
if mat[row][col] != 0:
left = top = float('inf')
if row-1 >= 0:
... | 01-matrix | Python, runtime 77.32%, memory 90.18% | tsai00150 | 0 | 182 | 01 matrix | 542 | 0.442 | Medium | 9,563 |
https://leetcode.com/problems/01-matrix/discuss/2248155/Python-DP.-O(mn)O(1) | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
for r in range(m):
for c in range(n):
if mat[r][c] == 1:
mat[r][c] = math.inf
if r > 0:
mat[r][c] = min(ma... | 01-matrix | Python, DP. O(mn)/O(1) | blue_sky5 | 0 | 28 | 01 matrix | 542 | 0.442 | Medium | 9,564 |
https://leetcode.com/problems/01-matrix/discuss/2246707/Python-BFS-level-order-traversal.-Time%3A-O(m*n)-Space%3A-O(m*n) | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
q = deque()
for r in range(m):
for c in range(n):
if mat[r][c] == 0:
q.append((r, c))
else:
mat[r]... | 01-matrix | Python, BFS level-order traversal. Time: O(m*n), Space: O(m*n) | blue_sky5 | 0 | 49 | 01 matrix | 542 | 0.442 | Medium | 9,565 |
https://leetcode.com/problems/01-matrix/discuss/2012854/Python-easy-to-read-and-understand-or-BFS | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
visit = set()
q = []
for i in range(m):
for j in range(n):
if mat[i][j] == 0:
visit.add((i, j))
q.append((... | 01-matrix | Python easy to read and understand | BFS | sanial2001 | 0 | 196 | 01 matrix | 542 | 0.442 | Medium | 9,566 |
https://leetcode.com/problems/01-matrix/discuss/1863061/Python3-or-Using-DP-or-Two-Pass | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m,n = len(mat), len(mat[0])
dp = [[float("inf") for i in range(n)] for j in range(m)]
for i in range(m):
for j in range(n):
if mat[i][j] == 0:
dp[i][j] = 0
... | 01-matrix | Python3 | Using DP | Two Pass | goyaljatin9856 | 0 | 71 | 01 matrix | 542 | 0.442 | Medium | 9,567 |
https://leetcode.com/problems/01-matrix/discuss/1373278/DP-with-thinking-process | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
max_v = max(m, n)
ans = [[0 if c == 0 else max_v for c in mat[i]] for i in range(m)]
for i in range(1, n):
ans[0][i] = min(ans[0][i], ans[0][i-1]+1 if ans[0][i-1]... | 01-matrix | DP with thinking process | ytb_algorithm | 0 | 65 | 01 matrix | 542 | 0.442 | Medium | 9,568 |
https://leetcode.com/problems/01-matrix/discuss/1370668/Python3-DP-solution-O(1)-space | class Solution:
def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
mx = len(mat) * len(mat[0])
for i in range(len(mat)):
for j in range(len(mat[i])):
if mat[i][j] == 0:
continue
top = mx
... | 01-matrix | [Python3] DP solution O(1) space | maosipov11 | 0 | 80 | 01 matrix | 542 | 0.442 | Medium | 9,569 |
https://leetcode.com/problems/01-matrix/discuss/1159451/Python-3-Getting-wrong-answer | class Solution:
def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
h, w = len(matrix), len(matrix[0])
output = [ [ float('inf') for _ in range(w) ] for _ in range (h) ]
for x in range(h):
for y in range(w):
if ma... | 01-matrix | [Python 3] Getting wrong answer | jaipoo | 0 | 78 | 01 matrix | 542 | 0.442 | Medium | 9,570 |
https://leetcode.com/problems/01-matrix/discuss/391652/Solution-in-Python-3-(Alternative-Approach) | class Solution:
def updateMatrix(self, A: List[List[int]]) -> List[List[int]]:
M, N, Q = len(A), len(A[0]), []
B = [[0]*N for _ in range(M)]
for i,j in itertools.product(range(M),range(N)):
if not A[i][j]: continue
b = 1
for k,l in [[i-1,j],[i,j+1],[i+1,j],[i,j-1]]:
if not (k... | 01-matrix | Solution in Python 3 (Alternative Approach) | junaidmansuri | -2 | 258 | 01 matrix | 542 | 0.442 | Medium | 9,571 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1515564/Python-Easy-to-understand-solution-w-Explanation | class Solution:
def __init__(self):
self.diameter = 0 # stores the maximum diameter calculated
def depth(self, node: Optional[TreeNode]) -> int:
"""
This function needs to do the following:
1. Calculate the maximum depth of the left and right sides of the given node
... | diameter-of-binary-tree | Python Easy-to-understand solution w Explanation | zayne-siew | 56 | 3,100 | diameter of binary tree | 543 | 0.561 | Easy | 9,572 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/480877/543.-Diameter-of-Binary-Tree-and-124.-Binary-Tree-Maximum-Path-Sum | class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
def height(root):
nonlocal diameter
if not root:
return 0
left = height(root.left)
right = height(root.right)
diameter = max(diameter, left + right)... | diameter-of-binary-tree | 543. Diameter of Binary Tree and 124. Binary Tree Maximum Path Sum | shchshhappy | 51 | 8,300 | diameter of binary tree | 543 | 0.561 | Easy | 9,573 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/480877/543.-Diameter-of-Binary-Tree-and-124.-Binary-Tree-Maximum-Path-Sum | class Solution:
def maxPathSum(self, root: TreeNode) -> int:
def maxPath(root):
nonlocal maxSum
if not root:
return 0
left = maxPath(root.left)
right = maxPath(root.right)
maxSum = max(maxSum, left + right + root.val)
... | diameter-of-binary-tree | 543. Diameter of Binary Tree and 124. Binary Tree Maximum Path Sum | shchshhappy | 51 | 8,300 | diameter of binary tree | 543 | 0.561 | Easy | 9,574 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/574083/Python3-post-order-dfs | class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
def fn(node):
"""Return length+1 and diameter rooted at node"""
if not node: return (0, 0)
l1, d1 = fn(node.left)
l2, d2 = fn(node.right)
return 1 + max(l1, l2), max(d1... | diameter-of-binary-tree | [Python3] post-order dfs | ye15 | 14 | 1,500 | diameter of binary tree | 543 | 0.561 | Easy | 9,575 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2678310/python-or-recursion | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
self.max_diameter = 0
self.getDiameter(root)
return self.max_diameter
def getDiameter(self, root):
if not root:
return 0
left_depth = self.getDiameter(root.left)
... | diameter-of-binary-tree | python | recursion | MichelleZou | 11 | 1,400 | diameter of binary tree | 543 | 0.561 | Easy | 9,576 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2253278/Python-DFS-%2B-2-Solutions-Explained-Time-O(N)-or-Space-O(N) | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
result = [0] # Global result variable
self.findDiameter(root, result)
return result[0]
def findDiameter(self, root, result):
if not root:
return 0
left = self.findDi... | diameter-of-binary-tree | [Python] DFS + 2 Solutions Explained - Time O(N) | Space O(N) | Symbolistic | 9 | 652 | diameter of binary tree | 543 | 0.561 | Easy | 9,577 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2253278/Python-DFS-%2B-2-Solutions-Explained-Time-O(N)-or-Space-O(N) | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
return self.findDiameter(root)[1]
def findDiameter(self, root):
if not root:
return (0, 0)
left = self.findDiameter(root.left)
right = self.findDiameter(root.right)
... | diameter-of-binary-tree | [Python] DFS + 2 Solutions Explained - Time O(N) | Space O(N) | Symbolistic | 9 | 652 | diameter of binary tree | 543 | 0.561 | Easy | 9,578 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1351854/Python-Iterative-solution-40-ms-faster-than-90.06-15.2-MB-less-than-99.95 | class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
ans = 0
hm = {}
stack = [(root, False)]
while stack:
node, visited = stack.pop()
if node:
if visited:
lh = 0 if node.left is None else hm.pop(no... | diameter-of-binary-tree | Python, Iterative solution, 40 ms, faster than 90.06%, 15.2 MB, less than 99.95% | MihailP | 8 | 1,100 | diameter of binary tree | 543 | 0.561 | Easy | 9,579 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1667695/Python-Simple-recursive-DFS-explained | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
self.diameter = 0
def dfs(root):
if not root: return 0
leftLongestPath = dfs(root.left)
rightLongestPath = dfs(root.right)
# keep track of the lo... | diameter-of-binary-tree | [Python] Simple recursive DFS explained | buccatini | 5 | 363 | diameter of binary tree | 543 | 0.561 | Easy | 9,580 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2222455/Recursive-dfs-solution-in-Python-with-detailed-comments-and-explanation | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
# Initialize a variable to keep track of the max diameter as we recursively traverse through each node in the tree
max_diameter = 0
# Define a depth-first search function to count the number of nodes along a giv... | diameter-of-binary-tree | Recursive dfs solution, in Python, with detailed comments & explanation | zachtheyek | 4 | 228 | diameter of binary tree | 543 | 0.561 | Easy | 9,581 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1542791/Python3-solution | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
def f(node: TreeNode) -> int: # return longest depth under node
L = 1 + f(node.left) if node.left else 0
R = 1 + f(node.right) if node.right else 0
self.ans = max(self.ans, L+R)
r... | diameter-of-binary-tree | Python3 solution | dalechoi | 1 | 157 | diameter of binary tree | 543 | 0.561 | Easy | 9,582 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2549163/Python-3-solution-pass-by-reference | class Solution:
def height(self, node, dia) -> int:
if node == None:
return 0
#print(node)
lh = self.height(node.left, self.dia)
rh = self.height(node.right, self.dia)
#print("hi", self.dia, lh, rh)
self.dia = max(self.dia, lh+rh)
#print("hi2", sel... | diameter-of-binary-tree | Python 3 solution - pass by reference | sumedha19129 | 0 | 45 | diameter of binary tree | 543 | 0.561 | Easy | 9,583 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2452208/Bottom-up-O(n)-solution-by-Python-without-global-variable | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
_, diameter = self.helper(root)
return diameter
def helper(self, root: Optional[TreeNode]) -> (int, int):
if root == None:
return 0, 0
leftHeight, leftDiameter = self.helper... | diameter-of-binary-tree | Bottom-up O(n) solution by Python without global variable | amikai | 0 | 123 | diameter of binary tree | 543 | 0.561 | Easy | 9,584 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2168815/Python-Solution-or-This-should-be-a-medium-tag-question-IMO | class Solution:
diameterLength = 0
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
self.findDiameter(root)
return self.diameterLength-1 # -1 because for N nodes there are N-1 edges
def findDiameter(self, root):
if root is None:
return 0
... | diameter-of-binary-tree | Python Solution | This should be a medium tag question IMO | sudonitin | 0 | 83 | diameter of binary tree | 543 | 0.561 | Easy | 9,585 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2144294/Simple-clean-recursive-solution-in-Python | class Solution:
maxDiameter = 0
def __init(self):
self.maxDiamter = maxDiameter
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
def maxDepth(root):
if (not root): return 0;
leftMax = maxDepth(root.left)
rightMax = maxDepth(roo... | diameter-of-binary-tree | Simple clean recursive solution in Python | leqinancy | 0 | 32 | diameter of binary tree | 543 | 0.561 | Easy | 9,586 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2073277/Python-Solution-MaxDepth-or-Height | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
diameter = [0]
def dfs(root):
if not root:
return 0
left = dfs(root.left)
right = dfs(root.right)
diameter[0] = max(diam... | diameter-of-binary-tree | Python Solution, MaxDepth or Height | Osvaldo11 | 0 | 110 | diameter of binary tree | 543 | 0.561 | Easy | 9,587 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/2073162/python3-solution | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
def depth(root):
if root:
left, right = depth(root.left), depth(root.right)
return 1+ max(left, right)
return 0
if root:
return max(depth(root.left) + ... | diameter-of-binary-tree | python3 solution | chloee_icebear | 0 | 35 | diameter of binary tree | 543 | 0.561 | Easy | 9,588 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1997262/2-Solutions-or-O(N2)-or-O(N)-or-Simple | class Solution:
def height(self,node):
if node is None:
return 0
else:
return max(self.height(node.left), self.height(node.right)) + 1
def diameter(self,node):
if node is None:
return 0
dia1 = self.diameter(node.left)
dia2 = self.diame... | diameter-of-binary-tree | 2 Solutions | O(N^2) | O(N) | Simple | divyamohan123 | 0 | 118 | diameter of binary tree | 543 | 0.561 | Easy | 9,589 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1997262/2-Solutions-or-O(N2)-or-O(N)-or-Simple | class Solution:
def diameter(self,node):
if node is None:
return 0,0
dia1,height1 = self.diameter(node.left)
dia2,height2 = self.diameter(node.right)
height = max(height1 , height2) + 1
dia3 = height1 + height2 + 1
return max(dia1, dia2, dia3), height
... | diameter-of-binary-tree | 2 Solutions | O(N^2) | O(N) | Simple | divyamohan123 | 0 | 118 | diameter of binary tree | 543 | 0.561 | Easy | 9,590 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1972321/Python-Recursive-O(N)-Solution-Faster-Than-97.42-No-Memory | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
ans = [0]
def dfs(tree):
if not tree:
return 0
left, right = dfs(tree.left), dfs(tree.right)
ans[0] = max(ans[0], left + ri... | diameter-of-binary-tree | Python Recursive O(N) Solution, Faster Than 97.42%, No Memory | Hejita | 0 | 115 | diameter of binary tree | 543 | 0.561 | Easy | 9,591 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/1758225/Python3-oror-O(n)-solution-oror-40ms | class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
res = [0]
def height(root):
if not root:
return -1
lh = height(root.left)
rh = h... | diameter-of-binary-tree | Python3 || O(n) solution || 40ms | Dewang_Patil | 0 | 90 | diameter of binary tree | 543 | 0.561 | Easy | 9,592 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/986475/Beats-99.92-or-Dynamic-Programming-(General-Syntax)-or-Python-or-Very-easy-to-understand | class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if not root:
return 0
global res
res = -1
def solve(root):
global res
#BC
if not root:
return 0
#Hypothesis
... | diameter-of-binary-tree | Beats 99.92% | Dynamic Programming (General Syntax) | Python | Very easy to understand | vinit-dash | 0 | 134 | diameter of binary tree | 543 | 0.561 | Easy | 9,593 |
https://leetcode.com/problems/diameter-of-binary-tree/discuss/437084/Python3-Divide-Conquer | class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if not root:
return 0
return self.dfs(root)[1]
def dfs(self, root):
if not root:
return 0, -sys.maxsize
l_depth, l_diameter = self.dfs(root.left)
r_depth, r_diameter = self... | diameter-of-binary-tree | [Python3] Divide Conquer | Hornettao | 0 | 138 | diameter of binary tree | 543 | 0.561 | Easy | 9,594 |
https://leetcode.com/problems/remove-boxes/discuss/1379392/Python3-dp | class Solution:
def removeBoxes(self, boxes: List[int]) -> int:
@cache
def fn(lo, hi, k):
"""Return max score of removing boxes from lo to hi with k to the left."""
if lo == hi: return 0
while lo+1 < hi and boxes[lo] == boxes[lo+1]: lo, k = lo+1, k+1
... | remove-boxes | [Python3] dp | ye15 | 2 | 293 | remove boxes | 546 | 0.479 | Hard | 9,595 |
https://leetcode.com/problems/remove-boxes/discuss/2787368/Python-Recursive-DP%3A-92-time-64-space | class Solution:
def removeBoxes(self, boxes: List[int]) -> int:
@cache
def dp(l, r, count = 0):
if l > r: return 0
// Initial count for the letter at boxes[l]
count += 1
ptr = l + 1
while ptr <= r and boxes[l] == boxes[ptr]:
... | remove-boxes | Python Recursive DP: 92% time, 64% space | hqz3 | 0 | 17 | remove boxes | 546 | 0.479 | Hard | 9,596 |
https://leetcode.com/problems/number-of-provinces/discuss/727759/Python3-solution-with-detailed-explanation | class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""
n = len(M) #1
visited = [False]*n #2
count = 0 #3
if not M: #4
return 0 #5
def dfs(u): #6
for v in ... | number-of-provinces | Python3 solution with detailed explanation | peyman_np | 17 | 1,600 | number of provinces | 547 | 0.634 | Medium | 9,597 |
https://leetcode.com/problems/number-of-provinces/discuss/479621/Intuitive-Disjoint-Set-Union-Find-in-JavaPython3 | class Solution:
def findCircleNum(self, M: List[List[int]]) -> int:
dsu = DSU(len(M))
for i, r in enumerate(M):
for j, v in enumerate(r):
if j > i - 1: break
if v == 1:
dsu.union(i, j)
return dsu.numFrdCir
class DSU:
de... | number-of-provinces | Intuitive Disjoint Set Union Find in [Java/Python3] | BryanBoCao | 8 | 1,500 | number of provinces | 547 | 0.634 | Medium | 9,598 |
https://leetcode.com/problems/number-of-provinces/discuss/2670773/Python-3-Simple-DFS | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
visited = [False]*n
def trev(u):
if not visited[u]:
visited[u] = True
for i in range(n):
if isConnected[i][u]: #treveses all the ... | number-of-provinces | [Python 3] Simple DFS | user2667O | 2 | 376 | number of provinces | 547 | 0.634 | Medium | 9,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.