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/number-of-provinces/discuss/2543862/Python-Elegant-and-Short-or-DSU | class Solution:
"""
Time: O(n^2)
Memory: O(n)
"""
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
dsu = DSU(n)
for i in range(n):
for j in range(n):
if isConnected[i][j]:
dsu.union(i, j)
return len({dsu.find(i) for i in range(n)}) | number-of-provinces | Python Elegant & Short | DSU | Kyrylo-Ktl | 2 | 392 | number of provinces | 547 | 0.634 | Medium | 9,600 |
https://leetcode.com/problems/number-of-provinces/discuss/2159710/Number-of-Provinces-or-Python-or-Union-Find-or-Count-of-connected-components | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
# set all parents value to 1 that is node is a parent of itself
par=[i for i in range(len(isConnected))]
# rank is the length of the particular component at that index, initially all are disjoint components so default to 1
... | number-of-provinces | Number of Provinces | Python | Union Find | Count of connected components | glimloop | 2 | 160 | number of provinces | 547 | 0.634 | Medium | 9,601 |
https://leetcode.com/problems/number-of-provinces/discuss/1781252/Python-Union-by-rank-and-path-compression-O(N) | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
class UnionFind:
# Constructor of Union-find. The size is the length of the root array.
def __init__(self, size):
self.root = [i for i in range(size)]
self.rank = [1]*size
... | number-of-provinces | [Python] Union by rank and path compression O(N) | haydarevren | 2 | 121 | number of provinces | 547 | 0.634 | Medium | 9,602 |
https://leetcode.com/problems/number-of-provinces/discuss/1726574/Python-3-DFS-with-meaningful-variable-names | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
nb_provinces = 0
def dfs(city_a: int) -> None:
for city_b in range(n):
if isConnected[city_a][city_b] == 1:
isConnected[city_a][city_b] ... | number-of-provinces | Python 3 DFS with meaningful variable names | thomasthiebaud | 2 | 76 | number of provinces | 547 | 0.634 | Medium | 9,603 |
https://leetcode.com/problems/number-of-provinces/discuss/2435895/Number-of-provinces-oror-Python3-oror-Union-Find | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
parent = [i for i in range(0, n)]
rank = [0] * n
# Initially there are n components
components = n
# Only traversing half of 2d array, since edges are bidirect... | number-of-provinces | Number of provinces || Python3 || Union-Find | vanshika_2507 | 1 | 27 | number of provinces | 547 | 0.634 | Medium | 9,604 |
https://leetcode.com/problems/number-of-provinces/discuss/1615407/Python-Solution-Faster-than-99 | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def dfsHelper(v,visited,isConnected):
# dfs search for all the connected components
visited[v]=True
for neigh in range(len(isConnected[v])):
if visited[neigh]==False and... | number-of-provinces | Python Solution Faster than 99% | Zach0787 | 1 | 245 | number of provinces | 547 | 0.634 | Medium | 9,605 |
https://leetcode.com/problems/number-of-provinces/discuss/1586277/Python-or-DFS-or-Simple-Solution-or-Easy-To-Understand | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def find_connected_cities(row):
for column in range(len(isConnected[0])):
if isConnected[row][column] == 1 and self.visited[column] != True:
self.visited[column] = True
find_connected_cities(column)
self.vi... | number-of-provinces | Python | DFS | Simple Solution | Easy To Understand | Call-Me-AJ | 1 | 207 | number of provinces | 547 | 0.634 | Medium | 9,606 |
https://leetcode.com/problems/number-of-provinces/discuss/874019/Python3-dfs-(99.75) | class Solution:
def findCircleNum(self, M: List[List[int]]) -> int:
# graph as adjacency matrix
n = len(M) # number of people in total
def fn(i):
"""Group (direct & indirect) friends."""
seen[i] = True # mark visited
for ii in range(n):
... | number-of-provinces | [Python3] dfs (99.75%) | ye15 | 1 | 255 | number of provinces | 547 | 0.634 | Medium | 9,607 |
https://leetcode.com/problems/number-of-provinces/discuss/874019/Python3-dfs-(99.75) | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
ans = 0
visited = [False]*n
for x in range(n):
if not visited[x]:
ans += 1
stack = [x]
visited[x] = True
whi... | number-of-provinces | [Python3] dfs (99.75%) | ye15 | 1 | 255 | number of provinces | 547 | 0.634 | Medium | 9,608 |
https://leetcode.com/problems/number-of-provinces/discuss/2833400/Python-BFS-approach | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
visited = set()
queue = []
num = 0
for i in range(len(isConnected)):
if isConnected[i][i] == 1 and i not in visited:
queue.append(i)
visited.add(i)
... | number-of-provinces | Python BFS approach | paul1202 | 0 | 2 | number of provinces | 547 | 0.634 | Medium | 9,609 |
https://leetcode.com/problems/number-of-provinces/discuss/2831177/BFSDFSUnion-Find-for-number-of-Components | class Solution:
"BFS"
def bfs(self, isConnected):
n = len(isConnected)
visited = set()
ans = 0
for i in range(n):
if i in visited:
continue
q = deque([i])
while q:
n = q.popleft()
visited.add(n)
... | number-of-provinces | BFS/DFS/Union-Find for number of Components | MACHMichael | 0 | 4 | number of provinces | 547 | 0.634 | Medium | 9,610 |
https://leetcode.com/problems/number-of-provinces/discuss/2824984/Python-BFS-Solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
graph = {i+1:[] for i in range(len(isConnected))}
for i in range(1, len(isConnected)+1):
for j in range(1, len(isConnected[i-1])+1):
if i == j:
continue
if is... | number-of-provinces | Python BFS Solution | vijay_2022 | 0 | 2 | number of provinces | 547 | 0.634 | Medium | 9,611 |
https://leetcode.com/problems/number-of-provinces/discuss/2818533/Python-DFS | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
ROW, COL = len(isConnected), len(isConnected[0])
count = 0
visited = set()
def dfs(i):
visited.add(i)
for j in range(ROW):
if isConnected[i][j] and j not in visited:
... | number-of-provinces | Python DFS | zananpech9 | 0 | 3 | number of provinces | 547 | 0.634 | Medium | 9,612 |
https://leetcode.com/problems/number-of-provinces/discuss/2812508/simple-Dfs-solution | class Solution:
def dfs(grid,visited,i):
for j in range(len(grid[i])):
if grid[i][j]==1 and visited[j]==0:
visited[j]=1
visited=Solution.dfs(grid,visited,j)
# print(visited)
return visited
def findCircleNum(self, isConnected: List[List... | number-of-provinces | simple Dfs solution | althrun | 0 | 2 | number of provinces | 547 | 0.634 | Medium | 9,613 |
https://leetcode.com/problems/number-of-provinces/discuss/2812499/python-solution-with-DFS | class Solution:
def dfs(grid,visited,i):
for j in range(len(grid[i])):
if grid[i][j]==1 and visited[j]==0:
visited[j]=1
Solution.dfs(grid,visited,j)
def findCircleNum(self, isConnected: List[List[int]]) -> int:
grid=isConnected
q=[]
m=... | number-of-provinces | python solution with DFS | althrun | 0 | 1 | number of provinces | 547 | 0.634 | Medium | 9,614 |
https://leetcode.com/problems/number-of-provinces/discuss/2811710/DFS-Python | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
adjMatrix = [ [] for i in range(n)]
for i in range(n):
for j in range(n):
if i != j and isConnected[i][j]:
adjMatrix[i].append(j)
#pr... | number-of-provinces | DFS - Python | ysreddy | 0 | 1 | number of provinces | 547 | 0.634 | Medium | 9,615 |
https://leetcode.com/problems/number-of-provinces/discuss/2739539/Simple-Python-BFS | class Solution:
def bfs(self, isConnected, index_source, visited):
if index_source in visited:
return
visited.add(index_source)
for i in range(len(isConnected[index_source])):
if isConnected[index_source][i] == 1:
self.bfs(isConnected, i, visited)
def findCircleNum(self, is... | number-of-provinces | Simple Python BFS | LuraMisner | 0 | 9 | number of provinces | 547 | 0.634 | Medium | 9,616 |
https://leetcode.com/problems/number-of-provinces/discuss/2735841/Python-BFS-easy-understanding | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
adjList = {i:[] for i in range(1, n+1)}
for i in range(len(isConnected)):
for j in range(len(isConnected)):
if (isConnected[i][j] == 1 and (i != j)... | number-of-provinces | Python BFS easy understanding | logeshsrinivasans | 0 | 13 | number of provinces | 547 | 0.634 | Medium | 9,617 |
https://leetcode.com/problems/number-of-provinces/discuss/2700460/Python-Solution-or-DFS | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
row = len(isConnected)
col = len(isConnected[0])
ans = 0
seen = set()
def dfs(i):
for j in range(col):
if isConnected[i][j] == 1 and i != j and j not in seen:
... | number-of-provinces | Python Solution | DFS | maomao1010 | 0 | 12 | number of provinces | 547 | 0.634 | Medium | 9,618 |
https://leetcode.com/problems/number-of-provinces/discuss/2662696/Python3-Disjoint-Set-with-Path-Compression-and-Union-By-Rank | class Solution:
def find(self, x):
if x == self.root[x]:
return x
self.root[x] = self.find(self.root[x])
return self.root[x]
def union(self, x, y):
rootX = self.find(x)
rootY = self.find(y)
if rootX != rootY:
if self.rank[rootX] > se... | number-of-provinces | Python3 Disjoint Set with Path Compression and Union By Rank | PartialButton5 | 0 | 3 | number of provinces | 547 | 0.634 | Medium | 9,619 |
https://leetcode.com/problems/number-of-provinces/discuss/2615744/Simple-O(Nlog(N))-Union-Find-Python-Solution | class Solution(object):
def findCircleNum(self, isConnected):
N = len(isConnected)
parents = list(range(N))
rank = [1] * N
def find(x):
while x != parents[x]:
x = parents[x]
return x
def union(x, y):
x = fi... | number-of-provinces | Simple O(Nlog(N)) Union-Find Python Solution | tomascf | 0 | 57 | number of provinces | 547 | 0.634 | Medium | 9,620 |
https://leetcode.com/problems/number-of-provinces/discuss/2481080/Runtime%3A-194-ms-faster-than-95.28-Memory-Usage%3A-14.8-MB-less-than-21.29 | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
visited = set()
def getStart():
i = 0
while i < n:
if i not in visited:
return i
... | number-of-provinces | Runtime: 194 ms, faster than 95.28%; Memory Usage: 14.8 MB, less than 21.29% | GizDave | 0 | 18 | number of provinces | 547 | 0.634 | Medium | 9,621 |
https://leetcode.com/problems/number-of-provinces/discuss/2406406/Readable-broken-into-smaller-sub-logics-with-complexity-analysis-python3 | class Solution:
# O(n^2) time,
# O(n) space,
# Approach: DFS, hashset
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
unexplored = set([i for i in range(1, n+1)])
def findNeighbours(root):
row = isConnected[root-1]
... | number-of-provinces | Readable, broken into smaller sub logics with complexity analysis, python3 | destifo | 0 | 9 | number of provinces | 547 | 0.634 | Medium | 9,622 |
https://leetcode.com/problems/number-of-provinces/discuss/2402866/python-very-slow-solution-13..need-help-to-optimize | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
d={} #here is making dictionary holds city as key and nabours as a value
for i in range(len(isConnected)):
d[i]=[]
for j in range(len(isConnected)):
if i==j:
con... | number-of-provinces | python very slow solution 13%😔..need help to optimize | benon | 0 | 27 | number of provinces | 547 | 0.634 | Medium | 9,623 |
https://leetcode.com/problems/number-of-provinces/discuss/2360269/Python3-oror-2-Approaches-oror-BFS-oror-Union-Find | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected)
adj_list = collections.defaultdict(list)
for i in range(n):
for j in range(n):
if i != j:
if isConnected[i][j] == 1:
adj_... | number-of-provinces | Python3 || 2 Approaches || BFS || Union Find | s_m_d_29 | 0 | 104 | number of provinces | 547 | 0.634 | Medium | 9,624 |
https://leetcode.com/problems/number-of-provinces/discuss/2354556/Python-DFS-Beats-98-with-full-working-solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int: # Time: O(n*m) and Space: O(n)
seen = set()
def dfs(node):
# isConnected[node][nei]: nei is the index at node 0 index of the list, and adj is the value at that index
for nei, adj in enumerate(isConnected... | number-of-provinces | Python [DFS / Beats 98%] with full working solution | DanishKhanbx | 0 | 79 | number of provinces | 547 | 0.634 | Medium | 9,625 |
https://leetcode.com/problems/number-of-provinces/discuss/2350013/547.-My-Python-Solution-with-comments | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
# typical union find algo
n = len(isConnected)
# initialize disjoint set, with parent of x is x itself
parent = {x:x for x in range(n)}
for i in range(n):
for j in range(i+1,n):
... | number-of-provinces | 547. My Python Solution with comments | JunyiLin | 0 | 18 | number of provinces | 547 | 0.634 | Medium | 9,626 |
https://leetcode.com/problems/number-of-provinces/discuss/2297949/python-dfs-solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
n = len(isConnected[0])
d={i:[] for i in range(n)}
for i in range(n):
for j in range(n):
if isConnected[i][j]!=0 and j!=i:
d[i].append(j)
visit=set()... | number-of-provinces | python dfs solution | sprasu21 | 0 | 35 | number of provinces | 547 | 0.634 | Medium | 9,627 |
https://leetcode.com/problems/number-of-provinces/discuss/2297281/Java-Python-Optimized-Space-O(n)-Time-Easy-to-Understand | class Solution:
def findCircleNum(self, mat: List[List[int]]) -> int:
parent = [x for x in range(len(mat) + 1)]
def find(x: int) -> int:
while parent[x] != x:
#Wire x to point to its grandparent, which may
#be identical to parent. Then move x up g... | number-of-provinces | [Java, Python] Optimized Space, O(n) Time, Easy to Understand | arceusx | 0 | 22 | number of provinces | 547 | 0.634 | Medium | 9,628 |
https://leetcode.com/problems/number-of-provinces/discuss/2171734/Solved-using-Connected-Components-oror-DFS | class Solution:
def dfs(self, comp, i, graph, visited):
visited[i] = True
comp.append(i)
for neighbour in graph[i]:
if visited[neighbour] == False:
comp = self.dfs(comp, neighbour, graph, visited)
return comp
def findCircleNum(self, isCon... | number-of-provinces | Solved using Connected Components || DFS | Vaibhav7860 | 0 | 38 | number of provinces | 547 | 0.634 | Medium | 9,629 |
https://leetcode.com/problems/number-of-provinces/discuss/2136147/Python-DFS-modifying-the-array | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def dfs(city):
isConnected[city][city] = 0
for c in range(n):
if isConnected[city][c]:
isConnected[city][c] = 0 ... | number-of-provinces | Python, DFS modifying the array | blue_sky5 | 0 | 50 | number of provinces | 547 | 0.634 | Medium | 9,630 |
https://leetcode.com/problems/number-of-provinces/discuss/2015486/Easy-DFS-Solution(90-faster-and-90-less-memory) | class Solution:
def dfs(self, isConnected, i):
self.visited[i] = True
# Visit all the neighbours of city i and there neighbours
for j in range(self.n):
if isConnected[i][j] == 1 and not self.visited[j]:
self.dfs(isConnected, j)
def findCircl... | number-of-provinces | Easy DFS Solution(90% faster and 90% less memory) | dbansal18 | 0 | 35 | number of provinces | 547 | 0.634 | Medium | 9,631 |
https://leetcode.com/problems/number-of-provinces/discuss/2006811/Union-Find-or-DSU-or-Easy-code | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
f = {}
def find(x):
f.setdefault(x, x)
if f[x] != x:
f[x] = find(f[x])
return f[x]
def union(x, y):
f[find(x)] = find(y)
for i in ran... | number-of-provinces | Union Find | DSU | Easy code | divyanshugairola | 0 | 47 | number of provinces | 547 | 0.634 | Medium | 9,632 |
https://leetcode.com/problems/number-of-provinces/discuss/1963382/dfs-and-union-find-python-solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
# In this questions first take count=len(isConnected) i.e. No. of different components.
#Than keeps on decrementing it if they have not same parents bcoz we have to find no. of #different componenets.
# Method ... | number-of-provinces | dfs and union find , python solution | Aniket_liar07 | 0 | 79 | number of provinces | 547 | 0.634 | Medium | 9,633 |
https://leetcode.com/problems/number-of-provinces/discuss/1815856/python3-DFS-defeat-80-with-explanation | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
cities = len(isConnected)
ans = 0
white, grey, black = set(range(cities)), set(), set()
def dfs(city, root = False):
nonlocal white, grey, black, ans
related = isConnected[city]
... | number-of-provinces | python3 DFS defeat 80% with explanation | 752937603 | 0 | 73 | number of provinces | 547 | 0.634 | Medium | 9,634 |
https://leetcode.com/problems/number-of-provinces/discuss/1769241/Python-DFS-Solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def dfs(node):
for i in range(0,len(isConnected[node])):
if i != node and isConnected[node][i] == 1 and i not in s:
s.add(i)
dfs(i)
prov... | number-of-provinces | Python DFS Solution | DietCoke777 | 0 | 82 | number of provinces | 547 | 0.634 | Medium | 9,635 |
https://leetcode.com/problems/number-of-provinces/discuss/1611936/This-Question-is-eerily-Similar-to-323-and-261-with-Solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
# Graph initialization
graph = {x:[] for x in range(len(isConnected))}
# Here graph is given in Adjacency matrix form, we would build accordingly.
for i in range(len(isConnected)):
for j in range(len(isCon... | number-of-provinces | This Question is eerily Similar to 323 and 261 [with Solution] | pratushah | 0 | 66 | number of provinces | 547 | 0.634 | Medium | 9,636 |
https://leetcode.com/problems/number-of-provinces/discuss/1481461/Python3-DFS-solution | class Solution:
def __init__(self):
self.seen = set()
def dfs(self, isConnected, i):
self.seen.add(i)
for j in range(len(isConnected[i])):
if isConnected[i][j] == 1 and j not in self.seen:
self.dfs(isConnected, j)
def ... | number-of-provinces | [Python3] DFS solution | maosipov11 | 0 | 66 | number of provinces | 547 | 0.634 | Medium | 9,637 |
https://leetcode.com/problems/number-of-provinces/discuss/1460323/Python-using-Union-Find-base-methods | class Solution:
'''
cities belong to one province if they're
drectly or indirectly connected => we need
to count only distinct provincies.
And due to construction of Disjoint Set,
`self.root` will contain roots of provincies.
Ex: [1, 2, 3, 4] with [[1,1,0,0], [1,1,0,1],
[0,0,1,0], [... | number-of-provinces | Python using Union Find base methods | SleeplessChallenger | 0 | 97 | number of provinces | 547 | 0.634 | Medium | 9,638 |
https://leetcode.com/problems/number-of-provinces/discuss/1422691/Short-Python-DFS-beats-99.24 | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def dfs(node):
self.visited[node] = True
for i in range(n):
if isConnected[node][i] and not self.visited[i]:
dfs(i)
ret = 0
n = len(isConnected)
s... | number-of-provinces | Short Python DFS beats 99.24% | Charlesl0129 | 0 | 246 | number of provinces | 547 | 0.634 | Medium | 9,639 |
https://leetcode.com/problems/number-of-provinces/discuss/1418547/Just-use-DFS.-That's-it.-Simple-readable-solution. | class Solution:
def dfs(self, edges: List[set], is_visited: set, node: int):
if node not in is_visited:
is_visited.add(node)
for adj_node in edges[node]:
self.dfs(edges, is_visited, adj_node)
return
def makeGraph(self, isConnected: List[List[int]... | number-of-provinces | Just use DFS. That's it. Simple readable solution. | ssshukla26 | 0 | 81 | number of provinces | 547 | 0.634 | Medium | 9,640 |
https://leetcode.com/problems/number-of-provinces/discuss/1414849/Python3-dfs-solution | class Solution:
def findCircleNum(self, isConnected: List[List[int]]) -> int:
def dfs(i,isConnected,visited):
st = []
for j,k in enumerate(isConnected[i]):
if j != i and j not in visited and k == 1:
visited.append(j)
st.append(j... | number-of-provinces | Python3 dfs solution | EklavyaJoshi | 0 | 29 | number of provinces | 547 | 0.634 | Medium | 9,641 |
https://leetcode.com/problems/number-of-provinces/discuss/1394460/Python-or-DFS | class Solution(object):
def findCircleNum(self, isConnected):
"""
:type isConnected: List[List[int]]
:rtype: int
"""
g={}
for i in range(len(isConnected)):
for j in range(len(isConnected[i])):
if isConnected[i][j]==1 and i!=j:
... | number-of-provinces | Python | DFS | nmk0462 | 0 | 161 | number of provinces | 547 | 0.634 | Medium | 9,642 |
https://leetcode.com/problems/number-of-provinces/discuss/1049167/Python-Find-union-with-path-compression-and-union-by-rank | class Solution:
def find(self, x: int) -> int:
"""
C = cities
Time: O(1) amortized time
"""
if self.parents[x] != x:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x: int, y: int) -> bool:
"""... | number-of-provinces | Python, Find union with path compression and union by rank | arauter | 0 | 96 | number of provinces | 547 | 0.634 | Medium | 9,643 |
https://leetcode.com/problems/student-attendance-record-i/discuss/356636/Solution-in-Python-3-(one-line) | class Solution:
def checkRecord(self, s: str) -> bool:
return (s.count('A') < 2) and ('LLL' not in s)
- Junaid Mansuri
(LeetCode ID)@hotmail.com | student-attendance-record-i | Solution in Python 3 (one line) | junaidmansuri | 16 | 940 | student attendance record i | 551 | 0.481 | Easy | 9,644 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1407236/Python-or-Faster-Than-100-or-Runtime-%3A-12-ms | class Solution:
def checkRecord(self, s: str) -> bool:
d=0
for i in range(len(s)):
if s[i]=='A':
d+=1
if d==2:
return False
if i>=2 and s[i]==s[i-1]==s[i-2]=='L':
return False
return True | student-attendance-record-i | Python | Faster Than 100% | Runtime : 12 ms | satyamshrma | 4 | 224 | student attendance record i | 551 | 0.481 | Easy | 9,645 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1290958/Python3-greedy | class Solution:
def checkRecord(self, s: str) -> bool:
absent = late = 0
for i, ch in enumerate(s):
if ch == "A": absent += 1
elif ch == "L":
if i == 0 or s[i-1] != "L": cnt = 0
cnt += 1
late = max(late, cnt)
return ... | student-attendance-record-i | [Python3] greedy | ye15 | 1 | 95 | student attendance record i | 551 | 0.481 | Easy | 9,646 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1290958/Python3-greedy | class Solution:
def checkRecord(self, s: str) -> bool:
return s.count("A") < 2 and "LLL" not in s | student-attendance-record-i | [Python3] greedy | ye15 | 1 | 95 | student attendance record i | 551 | 0.481 | Easy | 9,647 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1236515/Python3-simple-solution-%22One-liner%22-beats-95-users | class Solution:
def checkRecord(self, s: str) -> bool:
return s.count('A') < 2 and 'LLL' not in s | student-attendance-record-i | Python3 simple solution "One-liner" beats 95% users | EklavyaJoshi | 1 | 37 | student attendance record i | 551 | 0.481 | Easy | 9,648 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1017572/Simple-Solution-faster-than-99.83 | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count("A")>1:
return False
for i in range(len(s)-2):
if s[i]==s[i+1]==s[i+2]=="L":
return False
return True | student-attendance-record-i | Simple Solution- faster than 99.83% | thisisakshat | 1 | 107 | student attendance record i | 551 | 0.481 | Easy | 9,649 |
https://leetcode.com/problems/student-attendance-record-i/discuss/429080/Python-one-line | class Solution:
def checkRecord(self, s):
return not ('LLL' in s or s.count('A') > 1) | student-attendance-record-i | Python one-line | domthedeveloper | 1 | 86 | student attendance record i | 551 | 0.481 | Easy | 9,650 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2798011/2-lines-O(n) | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count('A')>=2 or "LLL" in s:
return False
return True | student-attendance-record-i | 2 lines, O(n) | albararamli | 0 | 2 | student attendance record i | 551 | 0.481 | Easy | 9,651 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2696561/Python-One-Pass-Solution | class Solution:
def checkRecord(self, s: str) -> bool:
lates = absences = 0
for status in s:
if status == 'L':
lates += 1
if lates >= 3:
return False
continue
if status == 'A':
absences += 1
... | student-attendance-record-i | Python One Pass Solution | kcstar | 0 | 6 | student attendance record i | 551 | 0.481 | Easy | 9,652 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2647251/python-solution | class Solution:
def checkRecord(self, s: str) -> bool:
p=s.count("A")
r="LLL"
if p>=2 or r in s:
return False
return True | student-attendance-record-i | python solution | kissi616 | 0 | 6 | student attendance record i | 551 | 0.481 | Easy | 9,653 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2584184/Python-3-One-liner-%2B-descriptive-Solution | class Solution:
def checkRecord(self, s: str) -> bool:
#return s.count("A") < 2 and s.count("LLL") < 1
absent = late = 0
for i in s:
if i == "L":
late+=1
elif i== "A":
... | student-attendance-record-i | Python 3 One liner + descriptive Solution | abhisheksanwal745 | 0 | 16 | student attendance record i | 551 | 0.481 | Easy | 9,654 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2491612/Python3-faster-than-88 | class Solution(object):
def checkRecord(self, s):
"""
:type s: str
:rtype: bool
"""
count_abscence = 0
previous_pos = -1
count_late = 0
for pos, i in enumerate(s):
if i == 'A':
count_abscence += 1
if count_a... | student-attendance-record-i | Python3 faster than 88% | gourav14051992 | 0 | 18 | student attendance record i | 551 | 0.481 | Easy | 9,655 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2410191/Python3-easy-solution | class Solution:
def checkRecord(self, s: str) -> bool:
absent = 0
for i in range(len(s)):
if i+2<len(s) and s[i]=='L' and s[i+1]=="L" and s[i+2]=="L":
return False
if s[i]=='A':
absent+=1
if absent >=2:
return False
... | student-attendance-record-i | Python3 easy solution | shacid | 0 | 19 | student attendance record i | 551 | 0.481 | Easy | 9,656 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2081200/Python-2-solutions%3A-Basic-and-Oneliner | class Solution:
def checkRecord(self, s: str) -> bool:
A,L = 0,0
for i in s:
if i =='A':
A+=1
elif i=='L':
L+=1
if i!='L':
L=0
if A>=2 or L>=3:
return False
return True | student-attendance-record-i | [Python] 2 solutions: Basic and Oneliner | rtyagi1 | 0 | 59 | student attendance record i | 551 | 0.481 | Easy | 9,657 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2081200/Python-2-solutions%3A-Basic-and-Oneliner | class Solution:
def checkRecord2(self, s:str) ->bool:
return s.count('A') <= 1 and s.count('LLL') == 0 | student-attendance-record-i | [Python] 2 solutions: Basic and Oneliner | rtyagi1 | 0 | 59 | student attendance record i | 551 | 0.481 | Easy | 9,658 |
https://leetcode.com/problems/student-attendance-record-i/discuss/2042359/Python-oneliner | class Solution:
def checkRecord(self, s: str) -> bool:
return False if 'LLL' in s or s.count('A') >= 2 else True | student-attendance-record-i | Python oneliner | StikS32 | 0 | 44 | student attendance record i | 551 | 0.481 | Easy | 9,659 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1903039/Easiest-and-Simplest-Python3-Solution-oror-Beginner-Friendly-Code-oror-Very-Easy | class Solution:
def checkRecord(self, s: str) -> bool:
cL=0
cA=0
l=0
a=0
for i in s:
if i=='L':
cL=cL+1
l=cL
elif i=='A':
cA=cA+1
l=cL
a=cA
cL=0
... | student-attendance-record-i | Easiest & Simplest Python3 Solution || Beginner Friendly Code || Very Easy | RatnaPriya | 0 | 32 | student attendance record i | 551 | 0.481 | Easy | 9,660 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1719206/Python3-accepted-solution | class Solution:
def checkRecord(self, s: str) -> bool:
return True if(s.count("A")<2 and s.count("LLL")==0)else False | student-attendance-record-i | Python3 accepted solution | sreeleetcode19 | 0 | 39 | student attendance record i | 551 | 0.481 | Easy | 9,661 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1687314/Python3-100-faster-with-explanation | class Solution:
def checkRecord(self, s: str) -> bool:
absent, late = 0,0
for x in s:
if x == 'A':
absent += 1
if absent >= 2:
return False
if x == 'L':
late += 1
if late >= 3:
... | student-attendance-record-i | Python3, 100% faster with explanation | cvelazquez322 | 0 | 90 | student attendance record i | 551 | 0.481 | Easy | 9,662 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1411326/Python-Solution-oror-Easy-to-understand | class Solution:
def checkRecord(self, s: str) -> bool:
x = s.count('A')
for i in range (len(s)-2):
if (s[i]==s[i+1]==s[i+2] == 'L'):
return False
if x<2:
return True
return False | student-attendance-record-i | Python Solution || Easy to understand | adityarichhariya7879 | 0 | 102 | student attendance record i | 551 | 0.481 | Easy | 9,663 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1273534/Easy-Python-Solution(28ms) | class Solution:
def checkRecord(self, s: str) -> bool:
c=Counter(s)
if(c['A']>=2):
return False
for i in range(len(s)):
if(s[i]=='L' and i<len(s)-2):
if(s[i+1]=='L' and i<len(s)-1):
if(s[i+2]=='L' and i<len(s)):
... | student-attendance-record-i | Easy Python Solution(28ms) | Sneh17029 | 0 | 80 | student attendance record i | 551 | 0.481 | Easy | 9,664 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1238245/Python-oror-99.24-faster | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count('A')<2 and s.count('L')<3:
return True
elif s.count('A')<2 and s.count('L')>2:
for c in range(3,s.count('L')+1):
temp ='L'*c
if temp in s:
return False
... | student-attendance-record-i | Python || 99.24% faster | dhrumilg699 | 0 | 61 | student attendance record i | 551 | 0.481 | Easy | 9,665 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1230393/PYTHON-3-FASTER-THAN-99.8-OF-CODE | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count('A') < 2 and s.count('LLL') < 1:
return True
return False | student-attendance-record-i | PYTHON 3 FASTER THAN 99.8% OF CODE | ehtesham22 | 0 | 27 | student attendance record i | 551 | 0.481 | Easy | 9,666 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1133191/Easy-and-simple-python | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count("A") > 1 or "LLL" in s:
return False
return True | student-attendance-record-i | Easy and simple python | pheobhe | 0 | 34 | student attendance record i | 551 | 0.481 | Easy | 9,667 |
https://leetcode.com/problems/student-attendance-record-i/discuss/794955/Python3-98-faster | class Solution:
def checkRecord(self, s: str) -> bool:
if 'LLL' not in s:
c=0
for i in s:
if i=='A':
c+=1
if c==2:
return False
return True
return False | student-attendance-record-i | Python3 98% faster | SurajJadhav7 | 0 | 68 | student attendance record i | 551 | 0.481 | Easy | 9,668 |
https://leetcode.com/problems/student-attendance-record-i/discuss/645907/Python3-simple-solution-using-counter | class Solution:
def checkRecord(self, s: str) -> bool:
checks = Counter(s)
if 'LLL' in s or checks['A'] > 1:
return False
return True | student-attendance-record-i | Python3 simple solution using counter | pythongo | 0 | 77 | student attendance record i | 551 | 0.481 | Easy | 9,669 |
https://leetcode.com/problems/student-attendance-record-i/discuss/583978/Simple-Python-Solution | class Solution:
def checkRecord(self, s: str) -> bool:
if s.count('A')>1:
return False
l=[]
for i in range(len(s)):
if s[i]=='L':
l.append(i)
for i in range(len(l)-2):
if l[i]+1==l[i+1] an... | student-attendance-record-i | Simple Python Solution | Ayu-99 | 0 | 51 | student attendance record i | 551 | 0.481 | Easy | 9,670 |
https://leetcode.com/problems/student-attendance-record-i/discuss/528631/Python-One-Line-(-Ternary-) | class Solution:
def checkRecord(self, s: str) -> bool:
return False if s.count('A')>=2 or 'LLL' in s else True | student-attendance-record-i | Python One Line ( Ternary ) | bachana3435 | 0 | 61 | student attendance record i | 551 | 0.481 | Easy | 9,671 |
https://leetcode.com/problems/student-attendance-record-i/discuss/395804/Python%3A-32-ms-faster-than-92.22-of-submissions | class Solution:
def __init__(self):
self.absent = False
self.late = 0
def checkRecord(self, s: str) -> bool:
for c in s:
if c == 'P':
self.late = 0
elif c == 'A':
if self.absent:
return False
... | student-attendance-record-i | Python: 32 ms, faster than 92.22% of submissions | btjd | 0 | 107 | student attendance record i | 551 | 0.481 | Easy | 9,672 |
https://leetcode.com/problems/student-attendance-record-i/discuss/1299074/Python3-dollarolution | class Solution:
def checkRecord(self, s: str) -> bool:
a, l = 0, 0
for i in range(len(s)):
if s[i] == 'A':
a += 1
if l < 3:
l = 0
elif s[i] == 'L':
if l == 0:
l += 1
elif l... | student-attendance-record-i | Python3 $olution | AakRay | -1 | 70 | student attendance record i | 551 | 0.481 | Easy | 9,673 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/356750/Solution-in-Python-3-(five-lines)-(with-explanation) | class Solution:
def checkRecord(self, n: int) -> int:
C, m = [1,1,0,1,0,0], 10**9 + 7
for i in range(n-1):
a, b = sum(C[:3]) % m, sum(C[3:]) % m
C = [a, C[0], C[1], a + b, C[3], C[4]]
return (sum(C) % m) | student-attendance-record-ii | Solution in Python 3 (five lines) (with explanation) | junaidmansuri | 12 | 1,900 | student attendance record ii | 552 | 0.412 | Hard | 9,674 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/827561/Python-solution-with-explanation | class Solution:
def checkRecord(self, n: int) -> int:
"""
Suppose dp[i] is the number of all the rewarded sequences without 'A'
having their length equals to i, then we have:
1. Number of sequence ends with 'P': dp[i - 1]
2. Number of sequence ends with 'L':
... | student-attendance-record-ii | Python solution with explanation | eroneko | 9 | 458 | student attendance record ii | 552 | 0.412 | Hard | 9,675 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1290973/Python3-top-down-dp | class Solution:
def checkRecord(self, n: int) -> int:
@lru_cache(100)
def fn(i, absent, late):
"""Return number of attendance eligible for award."""
if i == n: return 1
ans = fn(i+1, absent, 0) # present
if absent == 0: ans += fn(i+1, 1, 0) ... | student-attendance-record-ii | [Python3] top-down dp | ye15 | 5 | 357 | student attendance record ii | 552 | 0.412 | Hard | 9,676 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1290973/Python3-top-down-dp | class Solution:
def checkRecord(self, n: int) -> int:
dp = [1, 2, 4]
for i in range(3, n+1):
dp.append((dp[i-3] + dp[i-2] + dp[i-1]) % 1_000_000_007)
ans = dp[n]
for i in range(n):
ans = (ans + dp[i] * dp[n-1-i]) % 1_000_000_007
return ans | student-attendance-record-ii | [Python3] top-down dp | ye15 | 5 | 357 | student attendance record ii | 552 | 0.412 | Hard | 9,677 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1290973/Python3-top-down-dp | class Solution:
def checkRecord(self, n: int) -> int:
f0, f1, f2 = 1, 1, 0
g0, g1, g2 = 1, 0, 0
for _ in range(n-1):
f0, f1, f2, g0, g1, g2 = (f0+f1+f2) % 1_000_000_007, f0, f1, (f0+f1+f2+g0+g1+g2) % 1_000_000_007, g0, g1
return (f0+f1+f2+g0+g1+g2) % 1_000_000_007 | student-attendance-record-ii | [Python3] top-down dp | ye15 | 5 | 357 | student attendance record ii | 552 | 0.412 | Hard | 9,678 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1962121/Python-or-DFS-%2B-Memo-(TLE)-or-DP-(success)-or-Simple-comments | class Solution:
def checkRecord(self, n: int) -> int:
hsh = {}
if n==1:
return 3
md = (10**9)+7
def helper(rem,countA,endL):
if countA>1 or endL>=3:
return 0
if (rem,countA,endL) in hsh:
return hsh[(rem,... | student-attendance-record-ii | Python | DFS + Memo (TLE) | DP (success) | Simple comments | vishyarjun1991 | 3 | 304 | student attendance record ii | 552 | 0.412 | Hard | 9,679 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1962121/Python-or-DFS-%2B-Memo-(TLE)-or-DP-(success)-or-Simple-comments | class Solution:
def checkRecord(self, n: int) -> int:
dp = [1,1,0,1,0,0]
md = (10**9)+7
def calculate_dp():
nonlocal dp
# carefully look at how case 1 to 5 is handled here and try to justify the reason, pls comment incase if this is not clear
# make sure to % dp at every st... | student-attendance-record-ii | Python | DFS + Memo (TLE) | DP (success) | Simple comments | vishyarjun1991 | 3 | 304 | student attendance record ii | 552 | 0.412 | Hard | 9,680 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1091774/Python-DP-solution-O(n)-time-with-detailed-comments | class Solution:
def checkRecord(self, n: int) -> int:
# dptotal[i] the number of rewardable records without A whose lenghth is i
dptotal = [0] * (n + 1)
dp1,dp2,dp3 = 1,1,0
# dp1: the number of rewardable records that end with one L and without A
# dp2: the number of reward... | student-attendance-record-ii | Python DP solution O(n) time with detailed comments | yuhaogogo123 | 2 | 467 | student attendance record ii | 552 | 0.412 | Hard | 9,681 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1914899/Python-Clean-and-Simple!-(DP) | class Solution:
def checkRecord(self, n):
return sum(map(self.check,product(["A","L","P"],repeat=n)))
def check(self, s):
return not ('LLL' in s or s.count('A') > 1) | student-attendance-record-ii | Python - Clean and Simple! (DP) | domthedeveloper | 1 | 255 | student attendance record ii | 552 | 0.412 | Hard | 9,682 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1914899/Python-Clean-and-Simple!-(DP) | class Solution:
def checkRecord(self, n):
mod = lambda x : x % (10**9+7)
dp = [1, 1, 0, 1, 0, 0]
for i in range(2, n+1):
dp = [mod(sum(dp[:3])), dp[0], dp[1], mod(sum(dp)), dp[3], dp[4]]
return mod(sum(dp)) | student-attendance-record-ii | Python - Clean and Simple! (DP) | domthedeveloper | 1 | 255 | student attendance record ii | 552 | 0.412 | Hard | 9,683 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/1914899/Python-Clean-and-Simple!-(DP) | class Solution:
def checkRecord(self, n):
mod = lambda x : x % (10**9+7)
absentZero = [1, 1, 0]
absentOnce = [1, 0, 0]
for i in range(2, n+1):
absentZero, absentOnce = \
[mod(sum(absentZero)), absentZero[0], absentZero[1]], \
[mod(sum(absentZero+... | student-attendance-record-ii | Python - Clean and Simple! (DP) | domthedeveloper | 1 | 255 | student attendance record ii | 552 | 0.412 | Hard | 9,684 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/2792165/Python-oror-Recursion-%2B-Memoization-oror-DP | class Solution:
mod = 1000000007
dp = {}
def checkRecord(self, n: int) -> int:
def helper(curr_n, leave_taken, late_count):
if (curr_n, leave_taken, late_count) in self.dp:
return self.dp[(curr_n, leave_taken, late_count)]
#Student cannot take more than 2 leav... | student-attendance-record-ii | Python || Recursion + Memoization || DP | buggybot | 0 | 10 | student attendance record ii | 552 | 0.412 | Hard | 9,685 |
https://leetcode.com/problems/student-attendance-record-ii/discuss/2721748/Top-Down-DP-Python-Memoization-O(n) | class Solution:
def checkRecord(self, n: int) -> int:
self.dp = defaultdict(int)
self.dp[0] = 1
self.dp[1] = 2
self.dp[2] = 4
self.dp[3] = 7
self.helper(n)
total = self.dp[n]
for i in range(1, n + 1):
total += self.dp[i - 1] * self.dp[n ... | student-attendance-record-ii | Top-Down DP - Python - Memoization - O(n) | sehi05 | 0 | 18 | student attendance record ii | 552 | 0.412 | Hard | 9,686 |
https://leetcode.com/problems/optimal-division/discuss/1265206/Python3-string-concatenation | class Solution:
def optimalDivision(self, nums: List[int]) -> str:
if len(nums) <= 2: return "/".join(map(str, nums))
return f'{nums[0]}/({"/".join(map(str, nums[1:]))})' | optimal-division | [Python3] string concatenation | ye15 | 3 | 86 | optimal division | 553 | 0.597 | Medium | 9,687 |
https://leetcode.com/problems/optimal-division/discuss/1265206/Python3-string-concatenation | class Solution:
def optimalDivision(self, nums: List[int]) -> str:
return "/".join(map(str, nums)) if len(nums) <= 2 else f'{nums[0]}/({"/".join(map(str, nums[1:]))})' | optimal-division | [Python3] string concatenation | ye15 | 3 | 86 | optimal division | 553 | 0.597 | Medium | 9,688 |
https://leetcode.com/problems/optimal-division/discuss/1265206/Python3-string-concatenation | class Solution:
def optimalDivision(self, nums: List[int]) -> str:
@cache
def fn(lo, hi):
"""Return max division of nums[lo:hi]."""
if lo + 1 == hi: return str(nums[lo])
ans = "-inf"
for mid in range(lo+1, hi):
cand = fn(lo,... | optimal-division | [Python3] string concatenation | ye15 | 3 | 86 | optimal division | 553 | 0.597 | Medium | 9,689 |
https://leetcode.com/problems/optimal-division/discuss/362944/Solution-in-Python-3-(beats-~93)-(one-line) | class Solution:
def optimalDivision(self, n: List[int]) -> str:
return f'{n[0]}/({"/".join(map(str,n[1:]))})' if len(n)>2 else "/".join(map(str,n))
- Junaid Mansuri
(LeetCode ID)@hotmail.com | optimal-division | Solution in Python 3 (beats ~93%) (one line) | junaidmansuri | 2 | 243 | optimal division | 553 | 0.597 | Medium | 9,690 |
https://leetcode.com/problems/optimal-division/discuss/1347507/oror-python-oror-clean-and-short-solution-oror-explained | class Solution(object):
def optimalDivision(self, nums):
A = list(map(str, nums))
if len(A) <= 2:
return '/'.join(A)
return A[0] + '/(' + '/'.join(A[1:]) + ')' | optimal-division | ✅ || python || clean and short solution || explained | chikushen99 | 1 | 162 | optimal division | 553 | 0.597 | Medium | 9,691 |
https://leetcode.com/problems/optimal-division/discuss/404017/Python-3-.-Make-a-division-of-1st-element-by-2nd-element-to-the-last-one-.-.-Return | class Solution(object):
def optimalDivision(self, nums):
"""
:type nums: List[int]
:rtype: str
"""
list_length=len(nums)
if list_length==1:
return str(nums[0])
elif list_length==2:
return str(nums[0])+'/'+str(nums[1])
else:
... | optimal-division | Python 3 . Make a division of 1st element by 2nd element to the last one . . Return | mathewjose09 | 1 | 182 | optimal division | 553 | 0.597 | Medium | 9,692 |
https://leetcode.com/problems/optimal-division/discuss/2830571/Python-(Simple-Maths) | class Solution:
def optimalDivision(self, nums):
n, k = len(nums), str(nums[0]) + "/("
if n == 1:
return str(nums[0])
elif n == 2:
return str(nums[0]) + "/" + str(nums[1])
else:
for i in nums[1:]:
k += str(i) + "/"
return ... | optimal-division | Python (Simple Maths) | rnotappl | 0 | 2 | optimal division | 553 | 0.597 | Medium | 9,693 |
https://leetcode.com/problems/optimal-division/discuss/1223726/Simple-8-lines-solution-97-fast | class Solution:
def optimalDivision(self, nums: List[int]) -> str:
s1,s=str(nums[0]),''
if(len(nums)==2):
return str(nums[0])+'/'+str(nums[1])
for i in range(1,len(nums)):
s=s+str(nums[i])+'/'
if(len(s)>0):
s1=s1+'/'+'('+s[:-1]+')'
... | optimal-division | Simple 8 lines solution 97% fast | Rajashekar_Booreddy | 0 | 90 | optimal division | 553 | 0.597 | Medium | 9,694 |
https://leetcode.com/problems/optimal-division/discuss/1201982/slow-sol-but-dp-approach | class Solution:
def optimalDivision(self, nums: List[int]) -> str:
n=len(nums)
new=[[(0,"") for i in range(0,n)] for i in range(0,n)]
for i in range(0,n):
new[i][i]=(nums[i],str(nums[i]))
for i in range(0,n-1):
for j in range(i+1,n):
new[i][j]=... | optimal-division | slow sol but dp approach | heisenbarg | 0 | 63 | optimal division | 553 | 0.597 | Medium | 9,695 |
https://leetcode.com/problems/brick-wall/discuss/1736767/python-easy-hashmap-solution | class Solution:
def leastBricks(self, wall: List[List[int]]) -> int:
count = defaultdict(int)
tot = len(wall)
if tot == 1 and len(wall[0]) > 1:
return 0
elif tot == 1 and len(wall[0]) == 1:
return 1
for w in wall:
s = 0
... | brick-wall | python easy hashmap solution | byuns9334 | 1 | 62 | brick wall | 554 | 0.532 | Medium | 9,696 |
https://leetcode.com/problems/brick-wall/discuss/1172325/python-easy-understanding-omn-solution | class Solution:
def leastBricks(self, wall: List[List[int]]) -> int:
table = dict()
for row in wall:
tmp_sum = 0
for item in row:
tmp_sum += item
if tmp_sum not in table:
table[tmp_sum] = 1
else:
... | brick-wall | python easy understanding omn solution | yingziqing123 | 1 | 68 | brick wall | 554 | 0.532 | Medium | 9,697 |
https://leetcode.com/problems/brick-wall/discuss/393969/Solution-in-Python-3-(three-lines) | class Solution:
def leastBricks(self, W: List[List[int]]) -> int:
L, C = len(W), collections.defaultdict(int)
for w in W:
s = 0
for b in w: s += b; C[s] += 1
C[s] = 0
return L - max(C.values()) | brick-wall | Solution in Python 3 (three lines) | junaidmansuri | 1 | 581 | brick wall | 554 | 0.532 | Medium | 9,698 |
https://leetcode.com/problems/brick-wall/discuss/393969/Solution-in-Python-3-(three-lines) | class Solution:
def leastBricks(self, W: List[List[int]]) -> int:
L, C = len(W), collections.Counter(sum([list(itertools.accumulate(w)) for w in W],[]))
C[sum(W[0])] = 0
return L - max(C.values())
- Junaid Mansuri
(LeetCode ID)@hotmail.com | brick-wall | Solution in Python 3 (three lines) | junaidmansuri | 1 | 581 | brick wall | 554 | 0.532 | Medium | 9,699 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.