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/is-graph-bipartite/discuss/2269192/Clear-simple-Python-solution-using-DFS-traversal | class Solution:
isBipartite = True;
colors = []
visited = []
def __init(self):
self.isBipartite = isBipartite;
self.colors = colors
self.visited = visited
def isBipartite(self, graph: List[List[int]]) -> bool:
n = len(graph)
self.isBipartite = True
s... | is-graph-bipartite | Clear simple Python solution using DFS traversal | leqinancy | 0 | 9 | is graph bipartite | 785 | 0.527 | Medium | 12,800 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1992330/Python3-Solution-with-using-dfs | class Solution:
def dfs(self, graph, color, source):
for neigb in graph[source]:
if neigb in color:
if color[neigb] == color[source]:
return False
else:
color[neigb] = 1 - color[source]
if not self.dfs(g... | is-graph-bipartite | [Python3] Solution with using dfs | maosipov11 | 0 | 8 | is graph bipartite | 785 | 0.527 | Medium | 12,801 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1992000/Python-3-Solution-or-BFS-or-Clean-Code | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
# -1 = no colour | 0 = first colour | 1 = second colour
colours = [-1] * len(graph)
q = deque()
def checkBipartite():
# BFS
while q:
node, currColour = q.popleft... | is-graph-bipartite | Python 3 Solution | BFS | Clean Code | Raja03 | 0 | 11 | is graph bipartite | 785 | 0.527 | Medium | 12,802 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1991032/Python-3-DFS | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
visited = [9 for i in range(len(graph))]
res = [True]
def dfs(node,color):
if visited[node]==9:
visited[node] = color
... | is-graph-bipartite | Python 3 DFS | Brillianttyagi | 0 | 23 | is graph bipartite | 785 | 0.527 | Medium | 12,803 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1990705/Python-3-BFS-Solution | class Solution:
def isBipartite(self, graph: list[list[int]]) -> bool:
vis = [False for n in range(0, len(graph))]
while sum(vis) != len(graph): # Since graph isn't required to be connected this process needs to be repeated
ind = vis.index(False) # Find the first entry in th... | is-graph-bipartite | [Python 3] BFS Solution | arsrbt | 0 | 8 | is graph bipartite | 785 | 0.527 | Medium | 12,804 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1990552/Python-Dictionary-Solution | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
dict_graph = collections.defaultdict(list)
for i, v in enumerate(graph) :
if v : dict_graph[i] = v
dict_tmp = dict_graph
while dict_tmp :
groupA = set()
... | is-graph-bipartite | [ Python ] Dictionary Solution | crazypuppy | 0 | 34 | is graph bipartite | 785 | 0.527 | Medium | 12,805 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1875232/Python3-DFS-solution | class Solution:
ok = True
colors = []
visited = []
def __init(self):
self.ok = ok
self.colors = colors
self.visited = visited
def isBipartite(self, graph: List[List[int]]) -> bool:
n = len(graph)
# bool array storing color(true/false) for each node
... | is-graph-bipartite | [Python3] DFS solution | leqinancy | 0 | 16 | is graph bipartite | 785 | 0.527 | Medium | 12,806 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1806072/Python-or-DFS-or-Notice-Separate-Nodes-Situation | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
def traverse(graph, s):
nonlocal visited, colored, res
# if the res has already been set to False, end backtracking
if not res:
return
# Mark this node as visit... | is-graph-bipartite | Python | DFS | Notice Separate Nodes Situation | Fayeyf | 0 | 37 | is graph bipartite | 785 | 0.527 | Medium | 12,807 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1743166/EASY-or-COLORING-or-DFS-or-PYTHON3 | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
# 1 red
# -1 blue
# 0 uncolored
N = len(graph)
color = [0]*N
def valid_color(node, color_code):
for neighbor in graph[node]:
if color[neighbor] == color_code:
... | is-graph-bipartite | EASY | COLORING | DFS | PYTHON3 | SN009006 | 0 | 49 | is graph bipartite | 785 | 0.527 | Medium | 12,808 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1696188/785-Bipartite-Graph-with-DFS | class Solution:
def isBipartite(self, graph):
visited = [False] * len(graph); color = visited[:]
bipartite = [True]
for u in range(len(graph)):
if not visited[u]:
self.dfs(graph, u, visited, color, bipartite)
return bipartite[0]
def dfs(self, graph, u, visited, color, bipartite):
if not biparti... | is-graph-bipartite | 785 Bipartite Graph with DFS | zwang198 | 0 | 61 | is graph bipartite | 785 | 0.527 | Medium | 12,809 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1694953/Python-DFS-coloring | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
def dfs(node, color):
if node in seen:
return seen[node] != color
color ^= 1
seen[node] = color
for neighbor in graph[node]:
... | is-graph-bipartite | Python DFS coloring | blue_sky5 | 0 | 42 | is graph bipartite | 785 | 0.527 | Medium | 12,810 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1544587/Python-simple-dfs-solution-with-coloring | class Solution:
def isBipartite(self, g: List[List[int]]) -> bool:
graph = defaultdict(set)
color = {}
n = 0
for i in range(len(g)):
v = g[i]
for x in v:
n = max(n, x)
graph[i].add(x)
unseen =... | is-graph-bipartite | Python simple dfs solution with coloring | byuns9334 | 0 | 133 | is graph bipartite | 785 | 0.527 | Medium | 12,811 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1168736/python-fastest-dfs-with-two-colors-set! | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
connections: dict[int, list[int]] = {index: nodelist for index, nodelist in enumerate(graph)}
result: list[bool] = [True]
two_colors: list[set] = [set(), set()]
visited: dict[int, bool] = {vertex: False for vertex in connections}
level: i... | is-graph-bipartite | python fastest dfs with two colors set! | rahul_sawhney | 0 | 90 | is graph bipartite | 785 | 0.527 | Medium | 12,812 |
https://leetcode.com/problems/is-graph-bipartite/discuss/1031608/Python-simple-DFS-solution-with-comments | class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
colors = [0]*len(graph) # 0: no color, 1: red, -1: green
def dfs(node: int, node_color: int) -> bool: # node: current node, node_color: current node's color
if colors[node] != 0: ... | is-graph-bipartite | Python simple DFS solution with comments | cj1989 | 0 | 259 | is graph bipartite | 785 | 0.527 | Medium | 12,813 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/2121258/Explained-Easiest-Python-Solution | class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
if len(arr) > 2:
res = [] # list for storing the list: [prime fraction of arr[i]/arr[j], arr[i], arr[j]]
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
# creating and adding the sublist to res
... | k-th-smallest-prime-fraction | [Explained] Easiest Python Solution | the_sky_high | 2 | 180 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,814 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/2744456/Binary-Search-%2B-Sliding-Window-(Beats-91.21) | class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
N = len(arr)
def count_less(v):
"""1. the number of fractions < v
2. the largest fraction l/r that is < v"""
li = 0
cnt, l, r = 0, arr[0], arr[-1]
for ... | k-th-smallest-prime-fraction | Binary Search + Sliding Window (Beats 91.21%) | GregHuang | 1 | 60 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,815 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/2848235/Python-Collect-(num1-num2-num1num2)-then-sort-on-fraction-ASC | class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
'''
I want to go from Input: arr = [1,2,3,5], k = 3 ; Output: [2,5]
Each [(num1, num2, fraction), ...]
To: [(1, 5, 1/5), (1, 3, 1/3), (2, 5, 2/5), (1, 2, 1/2), (3, 5, 3/5), (2, 3, 2/3)]
... | k-th-smallest-prime-fraction | [Python] Collect (num1, num2, num1/num2), then sort on fraction ASC | graceiscoding | 0 | 1 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,816 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/2599568/Python3-or-Solved-Using-Sorting-and-Trying-Every-Possible-Pairings | class Solution:
#Time-Complexity: O(n^2 + n^2log(n^2)) -> O(n^2*log(n^2)) ->O(n^2*log(n))
#Space-Complexity: O(n^2)
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
array = []
for i in range(0, len(arr)-1):
numerator = arr[i]
for j in range(i+1... | k-th-smallest-prime-fraction | Python3 | Solved Using Sorting and Trying Every Possible Pairings | JOON1234 | 0 | 19 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,817 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/1669752/Python-SubOptimal-but-Easy-to-understand | class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
minHeap=[]
n = len(arr)
x=y=0
for i in range(n):
for j in range(i+1,n):
if arr[j] !=0:
heapq.heappush(minHeap, (arr[i]/arr[j], (arr[i],arr[j])))
... | k-th-smallest-prime-fraction | [Python] SubOptimal but Easy to understand | JimmyJammy1 | 0 | 89 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,818 |
https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/1305847/Python3-priority-queue | class Solution:
def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
pq = [(arr[i]/arr[-1], i, -1) for i in range(len(arr)-1)]
for _ in range(k):
_, i, j = heappop(pq)
if i - j + 1 < len(arr): heappush(pq, (arr[i]/arr[j-1], i, j-1))
return [arr[i]... | k-th-smallest-prime-fraction | [Python3] priority queue | ye15 | 0 | 90 | k th smallest prime fraction | 786 | 0.509 | Medium | 12,819 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2066105/Python-Easy-Solution-using-Dijkstra's-Algorithm | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
#Make graph
adj_list = {i:[] for i in range(n)}
for frm, to, price in flights:
adj_list[frm].append((to, price))
best_visited = [2**31]*n # Initializ... | cheapest-flights-within-k-stops | Python Easy Solution using Dijkstra's Algorithm | samirpaul1 | 6 | 445 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,820 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/790296/Readable-Python-(Djikstra) | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
graph = {}
for u in range(n):
graph[u] = []
for u,v,w in flights:
graph[u].append((v,w))
heap = [(0,-K,src)]
while heap:
(... | cheapest-flights-within-k-stops | Readable Python (Djikstra) | 2kvai777 | 6 | 911 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,821 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2831572/Pythonordpbfs | class Solution:
def dp(self, cur, price, step):
if cur == self.dst: return price
if step == 0: return inf
if self.memo[cur][step]: return self.memo[cur][step]
res = inf
for item in self.graph[cur]:
neighbor, cost = item[0], item[1]
re... | cheapest-flights-within-k-stops | Python|dp/bfs | lucy_sea | 0 | 8 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,822 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2831030/BFS | class Solution:
def findCheapestPrice(self, n: int, edges: List[List[int]], src: int, dst: int, k: int) -> int:
adj = [[] for _ in range(n)]
m = len(edges)
for i in range(m):
u, v, w = edges[i]
adj[u].append((v, w))
@lru_cache(None)
def dp(u, k):
... | cheapest-flights-within-k-stops | BFS | lillllllllly | 0 | 5 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,823 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2807007/Python-(Simple-Dynamic-Programming) | class Solution:
def findCheapestPrice(self, n, flights, src, dst, k):
dp = [[float("inf")]*(k+2) for _ in range(n)]
dp[src][0] = 0
for col in range(1,k+2):
for row in range(n):
dp[row][col] = dp[row][col-1]
for s,e,v in flights:
dp[e][... | cheapest-flights-within-k-stops | Python (Simple Dynamic Programming) | rnotappl | 0 | 11 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,824 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2740287/dictionary-solution-python | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
# Build the adjacency matrix
adj_matrix = [[0 for _ in range(n)] for _ in range(n)]
for s, d, w in flights:
adj_matrix[s][d] = w
# Short... | cheapest-flights-within-k-stops | dictionary solution python | yhu415 | 0 | 2 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,825 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2710124/Simple-Djikstra | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
graph = defaultdict(list)
for flight in flights:
s,d,price = flight
graph[s].append((d,price))
prices = [float("inf")]*n
prices[src] = 0
... | cheapest-flights-within-k-stops | Simple Djikstra | shriyansnaik | 0 | 12 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,826 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2672100/9-line-python-bellman-ford-beats-57 | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
dis = [float('inf')]*n
dis[src]=0
for i in range(k+1):
backup = dis[:]
for a,b,c in flights:
dis[b] = min(dis[b],backup[a]+c)
retu... | cheapest-flights-within-k-stops | 9 line python bellman-ford beats 57% | Ttanlog | 0 | 4 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,827 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2607312/Python3-DFS-w-Cache-or-O(k-*-(V-%2B-E)) | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
@cache
def dfs(cur, stops):
if stops > k:
return float('inf')
cheapest = float('inf')
for nxt, price in adj[cur]:
... | cheapest-flights-within-k-stops | Python3 DFS w/ Cache | O(k * (V + E)) | ryangrayson | 0 | 32 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,828 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2345657/python-easy-fast | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
prices = [float("inf")] * n
prices[src] = 0
for i in range(k + 1):
tempPrices = prices.copy()
for s,d,p in flights:
... | cheapest-flights-within-k-stops | python easy fast | soumyadexter7 | 0 | 142 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,829 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/2249651/HELP-NEEDED!-Why-does-DFS-does-not-work-in-this-solution | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
adj = [[] for _ in range(n)]
costs = {}
for flight_info in flights:
c1 = flight_info[0]
c2 = flight_info[1]
cost = flight_in... | cheapest-flights-within-k-stops | HELP NEEDED! Why does DFS does not work in this solution? | gabhinav001 | 0 | 93 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,830 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/1132908/Python-Simple-DP-O(n3) | class Solution:
def findCheapestPrice(self, n: int, edges: List[List[int]], src: int, dst: int, k: int) -> int:
adj = [[] for _ in range(n)]
m = len(edges)
for i in range(m):
u, v, w = edges[i]
adj[u].append((v, w))
@lru_cache(None)
def dp(u, k):
... | cheapest-flights-within-k-stops | [Python] Simple DP O(n^3) | carloscerlira | 0 | 109 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,831 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/687236/Python3-solution | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
graph = defaultdict(list) # adjacency list
for u, v, w in flights:
graph[u].append([v, w])
Q = deque([ [ src, 0] ]) # current node, cum cost
min_cost = float... | cheapest-flights-within-k-stops | Python3 solution | dalechoi | 0 | 66 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,832 |
https://leetcode.com/problems/cheapest-flights-within-k-stops/discuss/1042290/Bellman-Ford-Solution-in-C%2B%2B-and-Python3 | class Solution:
def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, K: int) -> int:
dist_price = [float('inf') for _ in range(n)]
dist_price[src]=0
for source,dest,cost in flights:
if src==source:
dist_price[dest] = cost
... | cheapest-flights-within-k-stops | Bellman Ford Solution in C++ and Python3 | aparna_g | -4 | 338 | cheapest flights within k stops | 787 | 0.359 | Medium | 12,833 |
https://leetcode.com/problems/rotated-digits/discuss/1205605/Python3-simple-solution-using-two-approaches | class Solution:
def rotatedDigits(self, N: int) -> int:
count = 0
for x in range(1, N+1):
x = str(x)
if '3' in x or '4' in x or '7' in x:
continue
if '2' in x or '5' in x or '6' in x or '9' in x:
count+=1
return count | rotated-digits | Python3 simple solution using two approaches | EklavyaJoshi | 4 | 207 | rotated digits | 788 | 0.568 | Medium | 12,834 |
https://leetcode.com/problems/rotated-digits/discuss/1205605/Python3-simple-solution-using-two-approaches | class Solution:
def rotatedDigits(self, N: int) -> int:
d = {'0':'0','1':'1','2':'5','5':'2','6':'9','8':'8','9':'6'}
count = 0
for i in range(1,N+1):
x = ''
flag = True
for j in str(i):
if j not in d.keys():
flag = Fals... | rotated-digits | Python3 simple solution using two approaches | EklavyaJoshi | 4 | 207 | rotated digits | 788 | 0.568 | Medium | 12,835 |
https://leetcode.com/problems/rotated-digits/discuss/551837/Python3-solution-using-a-string-conversion | class Solution:
def rotatedDigits(self, N: int) -> int:
quantity = 0
for num in range(1, N+1):
tally = str(num)
if any([True if x in '347' else False for x in tally]):
continue
if all([True if x in '018' else False for x in tally]):
... | rotated-digits | Python3 solution, using a string conversion | altareen | 2 | 191 | rotated digits | 788 | 0.568 | Medium | 12,836 |
https://leetcode.com/problems/rotated-digits/discuss/343982/Solution-in-Python-3-(beats-~100)-O(log-n)-(Combinatoric-Solution)-(Not-Brute-Force) | class Solution:
def rotatedDigits(self, N: int) -> int:
N, t, c = str(N), 0, 1
L, a, b = len(N) - 1, [1,2,3,3,3,4,5,5,6,7], [1,2,2,2,2,2,2,2,3,3]
for i in range(L):
if N[i] == '0': continue
t += a[int(N[i])-1]*7**(L-i) - c*b[int(N[i])-1]*3**(L-i)
if N[i] in '347': return t
... | rotated-digits | Solution in Python 3 (beats ~100%) O(log n) (Combinatoric Solution) (Not Brute Force) | junaidmansuri | 2 | 671 | rotated digits | 788 | 0.568 | Medium | 12,837 |
https://leetcode.com/problems/rotated-digits/discuss/2063497/Python-straightforward-solution | class Solution:
def rotatedDigits(self, n: int) -> int:
ans = 0
for i in range(1, n+1):
p = ''
if '3' in str(i) or '4' in str(i) or '7' in str(i):
continue
for j in str(i):
if j == '0':
p += '0'
e... | rotated-digits | Python straightforward solution | StikS32 | 1 | 161 | rotated digits | 788 | 0.568 | Medium | 12,838 |
https://leetcode.com/problems/rotated-digits/discuss/454860/Python3%3A-20ms-(99.86-faster)-12.7MB-(100-memory) | class Solution:
def rotatedDigits(self, N: int) -> int:
smallSet = {0,1,8}
bigSet = {2,5,6,9}
smallNum = [0,0,1,1,1,2,3,3,3,4][N % 10]
bigNum = [1,2,3,3,3,4,5,5,6,7][N % 10]
N = N // 10
smInc, bgInc = 4, 7
while N:
x = N % 10
N = N // 1... | rotated-digits | Python3: 20ms (99.86% faster) 12.7MB (100% memory) | andnik | 1 | 326 | rotated digits | 788 | 0.568 | Medium | 12,839 |
https://leetcode.com/problems/rotated-digits/discuss/325045/Python-solution-using-dictionary | class Solution:
def rotatedDigits(self, N: int) -> int:
count=0
d={0:0,1:1,2:5,3:-1,4:-1,5:2,6:9,7:-1,8:8,9:6}
for i in range(1,N+1):
l=list(str(i))
res=[]
for j in l:
if d[int(j)]!=-1:
res.append(str(d[int(j)]))
... | rotated-digits | Python solution using dictionary | ketan35 | 1 | 189 | rotated digits | 788 | 0.568 | Medium | 12,840 |
https://leetcode.com/problems/rotated-digits/discuss/2810782/Python-(Simple-Dynamic-Programming) | class Solution:
def dfs(self,x):
dict1, x, str1 = {"0":"0","1":"1","8":"8","2":"5","5":"2","6":"9","9":"6"}, str(x), ""
for i in x:
if i not in dict1:
return False
else:
str1 += dict1[i]
return str1 != x
def rotatedDigits(self, n... | rotated-digits | Python (Simple Dynamic Programming) | rnotappl | 0 | 5 | rotated digits | 788 | 0.568 | Medium | 12,841 |
https://leetcode.com/problems/rotated-digits/discuss/1463538/Brute-force-and-walrus | class Solution:
not_allowed = {"3", "4", "7"}
mirrored = {"0", "1", "8"}
def rotatedDigits(self, n: int) -> int:
return n - sum(int(bool((s := set(str(i))) & Solution.not_allowed) or
s.issubset(Solution.mirrored)) for i in range(1, n + 1)) | rotated-digits | Brute force and walrus | EvgenySH | 0 | 42 | rotated digits | 788 | 0.568 | Medium | 12,842 |
https://leetcode.com/problems/rotated-digits/discuss/1328227/Python3-dollarolution | class Solution:
def rotatedDigits(self, n: int) -> int:
v, l = ['0','1','8','2','5','6','9'], []
c = 0
for i in range(2,n+1):
x = 1
l = []
y = str(i)
for j in y:
if j not in v:
x = 0
brea... | rotated-digits | Python3 $olution | AakRay | 0 | 171 | rotated digits | 788 | 0.568 | Medium | 12,843 |
https://leetcode.com/problems/rotated-digits/discuss/408795/Python-Simple-Solution | class Solution:
def rotatedDigits(self, N: int) -> int:
numb = set(['6','9','2','5', '1', '0', '8'])
cnt = 0
for i in range(1,N+1):
t = set(str(i))
if t-numb==set():
if t-set(['8', '0', '1'])==set():
pass
else:
cnt+=1
return cnt | rotated-digits | Python Simple Solution | saffi | 0 | 502 | rotated digits | 788 | 0.568 | Medium | 12,844 |
https://leetcode.com/problems/escape-the-ghosts/discuss/1477363/Python-3-or-Manhattan-Distance-Math-or-Explanation | class Solution:
def escapeGhosts(self, ghosts: List[List[int]], target: List[int]) -> bool:
t_x, t_y = target
m_x, m_y = abs(t_x), abs(t_y)
for x, y in ghosts:
manhattan = abs(t_x - x) + abs(t_y - y)
if manhattan <= m_x + m_y:
return False
retu... | escape-the-ghosts | Python 3 | Manhattan Distance, Math | Explanation | idontknoooo | 1 | 117 | escape the ghosts | 789 | 0.607 | Medium | 12,845 |
https://leetcode.com/problems/escape-the-ghosts/discuss/930127/Python3-brainteaser | class Solution:
def escapeGhosts(self, ghosts: List[List[int]], target: List[int]) -> bool:
xx, yy = target
return all(abs(x-xx) + abs(y-yy) > abs(xx) + abs(yy) for x, y in ghosts) | escape-the-ghosts | [Python3] brainteaser | ye15 | 1 | 86 | escape the ghosts | 789 | 0.607 | Medium | 12,846 |
https://leetcode.com/problems/escape-the-ghosts/discuss/1427939/Python3-simple-O(N)-time-and-O(1)-space-solution | class Solution:
def escapeGhosts(self, ghosts: List[List[int]], target: List[int]) -> bool:
for i in ghosts:
if abs(i[0]-target[0]) + abs(i[1]-target[1]) <= abs(target[0])+abs(target[1]):
return False
return True | escape-the-ghosts | Python3 simple O(N) time and O(1) space solution | EklavyaJoshi | 0 | 55 | escape the ghosts | 789 | 0.607 | Medium | 12,847 |
https://leetcode.com/problems/escape-the-ghosts/discuss/1085358/python-2-line-easy-faster-than-80 | class Solution:
def escapeGhosts(self, ghosts: List[List[int]], target: List[int]) -> bool:
l = [abs(ghost[0]-target[0])+abs(ghost[1]-target[1]) for ghost in ghosts]
return min(l)>abs(target[0])+abs(target[1]) | escape-the-ghosts | python 2-line easy faster than 80% | zzj8222090 | 0 | 105 | escape the ghosts | 789 | 0.607 | Medium | 12,848 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1620809/PythonJAVACC%2B%2B-DP-oror-Image-Visualized-Explanation-oror-100-Faster-oror-O(N) | class Solution(object):
def numTilings(self, n):
dp = [1, 2, 5] + [0] * n
for i in range(3, n):
dp[i] = (dp[i - 1] * 2 + dp[i - 3]) % 1000000007
return dp[n - 1] | domino-and-tromino-tiling | ✅ [Python/JAVA/C/C++] DP || Image Visualized Explanation || 100% Faster || O(N) | linfq | 84 | 2,800 | domino and tromino tiling | 790 | 0.484 | Medium | 12,849 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1620809/PythonJAVACC%2B%2B-DP-oror-Image-Visualized-Explanation-oror-100-Faster-oror-O(N) | class Solution(object):
def numTilings(self, n):
dp, dpa = [1, 2] + [0] * n, [1] * n
for i in range(2, n):
dp[i] = (dp[i - 1] + dp[i - 2] + dpa[i - 1] * 2) % 1000000007
dpa[i] = (dp[i - 2] + dpa[i - 1]) % 1000000007
return dp[n - 1] | domino-and-tromino-tiling | ✅ [Python/JAVA/C/C++] DP || Image Visualized Explanation || 100% Faster || O(N) | linfq | 84 | 2,800 | domino and tromino tiling | 790 | 0.484 | Medium | 12,850 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1620640/Python-dynamic-programming-in-4-lines-O(N)-time-and-O(1)-space | class Solution:
def numTilings(self, n: int) -> int:
full_0, full_1, incomp_1 = 1, 2, 2
for i in range(2, n):
full_0, full_1, incomp_1 = full_1, full_0 + full_1 + incomp_1, 2 * full_0 + incomp_1
return full_1 % (10 ** 9 + 7) if n >= 2 else 1 | domino-and-tromino-tiling | Python dynamic programming in 4 lines, O(N) time and O(1) space | kryuki | 8 | 567 | domino and tromino tiling | 790 | 0.484 | Medium | 12,851 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1620640/Python-dynamic-programming-in-4-lines-O(N)-time-and-O(1)-space | class Solution:
def numTilings(self, n: int) -> int:
#edge case
if n == 1:
return 1
mod = 10 ** 9 + 7
dp_full = [0 for _ in range(n)]
dp_incomp = [0 for _ in range(n)]
dp_full[0] = 1
dp_full[1] = 2
dp_incomp[1] = 2
... | domino-and-tromino-tiling | Python dynamic programming in 4 lines, O(N) time and O(1) space | kryuki | 8 | 567 | domino and tromino tiling | 790 | 0.484 | Medium | 12,852 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1621616/Python3-recursion-%2B-memo-(-dp-on-broken-profile-) | class Solution:
def __init__(self):
self.table = {(0,0):1, (0,3):1}
self.mod = 1000000007
def func(self, pos, state):
if pos < 0: return 0
elif (pos, state) not in self.table:
if state == 0:
self.table[(pos, state)] = self.func(pos-1, 3)
... | domino-and-tromino-tiling | Python3 recursion + memo ( dp on broken profile ) | aditya04848 | 0 | 19 | domino and tromino tiling | 790 | 0.484 | Medium | 12,853 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/1621456/Simple-Fast-Python-Solution | class Solution:
def numTilings(self, n: int) -> int:
a, b, c = 0, 1, 1
i = 1
while i < n:
a, b, c = a+b, c, a*2 + b + c
i += 1
return c % (10**9 + 7) | domino-and-tromino-tiling | Simple Fast Python Solution | VicV13 | 0 | 61 | domino and tromino tiling | 790 | 0.484 | Medium | 12,854 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/930186/Python3-top-down-and-bottom-up-dp | class Solution:
def numTilings(self, N: int) -> int:
@cache
def fn(n):
"""Return number of ways to tile board."""
if n < 0: return 0
if n <= 1: return 1
return (2*fn(n-1) + fn(n-3)) % 1_000_000_007
return fn(N) | domino-and-tromino-tiling | [Python3] top-down & bottom-up dp | ye15 | 0 | 130 | domino and tromino tiling | 790 | 0.484 | Medium | 12,855 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/930186/Python3-top-down-and-bottom-up-dp | class Solution:
def numTilings(self, N: int) -> int:
ans = [1]*(N+1)
prefix = 2
for i in range(2, N+1):
ans[i] = 2*prefix - ans[i-1] - ans[i-2]
prefix += ans[i]
return ans[-1] % 1_000_000_007 | domino-and-tromino-tiling | [Python3] top-down & bottom-up dp | ye15 | 0 | 130 | domino and tromino tiling | 790 | 0.484 | Medium | 12,856 |
https://leetcode.com/problems/domino-and-tromino-tiling/discuss/930186/Python3-top-down-and-bottom-up-dp | class Solution:
def numTilings(self, N: int) -> int:
f0, f1, f2 = 0, 1, 1
for i in range(N-1): f0, f1, f2 = f1, f2, (2*f2 + f0) % 1_000_000_007
return f2 | domino-and-tromino-tiling | [Python3] top-down & bottom-up dp | ye15 | 0 | 130 | domino and tromino tiling | 790 | 0.484 | Medium | 12,857 |
https://leetcode.com/problems/custom-sort-string/discuss/2237060/Simple-yet-interview-friendly-or-Faster-than-99.97-or-Custom-Sorting-in-Python | class Solution:
def customSortString(self, order: str, s: str) -> str:
rank = [26]*26
for i in range(len(order)):
rank[ord(order[i]) - ord('a')] = i
return "".join(sorted(list(s), key= lambda x: rank[ord(x) - ord('a')])) | custom-sort-string | ✅ Simple yet interview friendly | Faster than 99.97% | Custom Sorting in Python | reinkarnation | 2 | 62 | custom sort string | 791 | 0.693 | Medium | 12,858 |
https://leetcode.com/problems/custom-sort-string/discuss/2237060/Simple-yet-interview-friendly-or-Faster-than-99.97-or-Custom-Sorting-in-Python | class Solution:
def customSortString(self, order: str, s: str) -> str:
def serialOrder(x):
return rank[ord(x) - ord('a')]
rank = [26]*26
for i in range(len(order)):
rank[ord(order[i]) - ord('a')] = i
print(rank)
... | custom-sort-string | ✅ Simple yet interview friendly | Faster than 99.97% | Custom Sorting in Python | reinkarnation | 2 | 62 | custom sort string | 791 | 0.693 | Medium | 12,859 |
https://leetcode.com/problems/custom-sort-string/discuss/1813444/Python-easy-to-read-and-understand | class Solution:
def customSortString(self, order: str, s: str) -> str:
ans = ""
for ch in order:
cnt = s.count(ch)
for i in range(cnt):
ans += ch
for ch in s:
if ch not in order:
ans += ch
return an... | custom-sort-string | Python easy to read and understand | sanial2001 | 1 | 82 | custom sort string | 791 | 0.693 | Medium | 12,860 |
https://leetcode.com/problems/custom-sort-string/discuss/1735581/Python-99.8-Less-Memory-38.1-Faster-easy-to-understand | class Solution:
def customSortString(self, order: str, s: str) -> str:
#output string
sout = ''
count = 0
#iterate though the order
for sortletter in order:
#check if sortletter is in s, add to output var
count = s.count(sortletter)
... | custom-sort-string | Python 99.8% Less Memory, 38.1 Faster - easy to understand | ovidaure | 1 | 127 | custom sort string | 791 | 0.693 | Medium | 12,861 |
https://leetcode.com/problems/custom-sort-string/discuss/1191458/Python3-simple-solution-beats-99-users | class Solution:
def customSortString(self, S: str, T: str) -> str:
x = ''
t = {}
for i in T:
if i not in S:
x += i
t[i] = t.get(i,0) + 1
for i in S:
if i in T:
x += i*t[i]
return x | custom-sort-string | Python3 simple solution beats 99% users | EklavyaJoshi | 1 | 60 | custom sort string | 791 | 0.693 | Medium | 12,862 |
https://leetcode.com/problems/custom-sort-string/discuss/2808426/using-built-in-sorted-and-hashmap | class Solution:
def customSortString(self, order: str, s: str) -> str:
# Q does not say we cannot use built-in sort
# len(s) = n, len(order) = m
# Space Complexity: O(m+n)
# Time Complexity: O(nlogn) # average case for built in Python Tim Sort
order_map = {o: i for i... | custom-sort-string | using built in `sorted` and hashmap | curiosity_kids | 0 | 2 | custom sort string | 791 | 0.693 | Medium | 12,863 |
https://leetcode.com/problems/custom-sort-string/discuss/2796144/Beats-91.2-2-Liner | class Solution:
def customSortString(self, order: str, s: str) -> str:
cmap=collections.Counter(s)
return "".join(i*cmap[i] for i in order if i in s) + "".join(i for i in s if i not in order) | custom-sort-string | Beats 91.2% - 2 Liner | avinash_konduri | 0 | 2 | custom sort string | 791 | 0.693 | Medium | 12,864 |
https://leetcode.com/problems/custom-sort-string/discuss/2796143/Beats-91.2-2-Liner | class Solution:
def customSortString(self, order: str, s: str) -> str:
cmap=collections.Counter(s)
return "".join(i*cmap[i] for i in order if i in s) + "".join(i for i in s if i not in order) | custom-sort-string | Beats 91.2% - 2 Liner | avinash_konduri | 0 | 2 | custom sort string | 791 | 0.693 | Medium | 12,865 |
https://leetcode.com/problems/custom-sort-string/discuss/2778186/Clean-and-Concise-python-1-line-beats-69-memory | class Solution:
def customSortString(self, order: str, s: str) -> str:
return "".join(sorted(s, key=(lambda x: order.index(x) if x in order else 100))) | custom-sort-string | Clean and Concise python 1 line beats 69% memory | AryaDot | 0 | 2 | custom sort string | 791 | 0.693 | Medium | 12,866 |
https://leetcode.com/problems/custom-sort-string/discuss/2614831/python-easy-solution | class Solution:
def customSortString(self, order: str, s: str) -> str:
str1=""
for i in s:
if i not in order:
str1+=i
res=""
d = Counter(s)
for i in order:
if i in d:
res+=i*d[i]
return res+str1 | custom-sort-string | python easy solution | anshsharma17 | 0 | 13 | custom sort string | 791 | 0.693 | Medium | 12,867 |
https://leetcode.com/problems/custom-sort-string/discuss/2608411/Python-or-3-Liner-Solution-or-Easy-to-understand-or-Detailed-Solution | class Solution:
def customSortString(self, order: str, s: str) -> str:
# Assign each character a value in order and store it in hash map.
orderMap = {c: i for i, c in enumerate(order)}
# On the basis of order hash map, create the array of character map. If the char is not there in o... | custom-sort-string | Python | 3 Liner Solution | Easy to understand | Detailed Solution | Aexki | 0 | 16 | custom sort string | 791 | 0.693 | Medium | 12,868 |
https://leetcode.com/problems/custom-sort-string/discuss/2471050/easy-python-solution | class Solution:
def customSortString(self, order: str, s: str) -> str:
orderChar = [i for i in order]
sChar = [i for i in s]
notPresent = []
ans = ''
for i in orderChar :
if i in sChar :
for times in range(sChar.count(i)) :
an... | custom-sort-string | easy python solution | sghorai | 0 | 20 | custom sort string | 791 | 0.693 | Medium | 12,869 |
https://leetcode.com/problems/custom-sort-string/discuss/2293310/Python3-or-using-Counter | class Solution:
def customSortString(self, order: str, s: str) -> str:
count_s, res = Counter(s), ""
for char in order:
if char in count_s:
res += (char * count_s[char])
del count_s[char]
return res + "".join([(k*v) for k, v in count_s... | custom-sort-string | Python3 | using Counter | Ploypaphat | 0 | 36 | custom sort string | 791 | 0.693 | Medium | 12,870 |
https://leetcode.com/problems/custom-sort-string/discuss/2186078/python-solution-with-explanation | class Solution:
def customSortString(self, order: str, s: str) -> str:
"""
Example:
Input: order = "cba", s = "abcdeab"
Output: "cbbaade"
1. Create a map of string s
2. Iterate over the string order and add similar characters from the map to keep the ans strings characters in correct order and set... | custom-sort-string | python solution with explanation | yash921 | 0 | 20 | custom sort string | 791 | 0.693 | Medium | 12,871 |
https://leetcode.com/problems/custom-sort-string/discuss/1779999/Python-really-simple-using-hashmap | class Solution:
def customSortString(self, order: str, s: str) -> str:
o = collections.defaultdict(lambda: 0)
for i, c in enumerate(order): o[c] = i;
return ''.join(sorted(s, key = lambda c: o[c])) | custom-sort-string | Python really simple using hashmap | kaichamp101 | 0 | 80 | custom sort string | 791 | 0.693 | Medium | 12,872 |
https://leetcode.com/problems/custom-sort-string/discuss/1684699/Simple-Python3-beats-87.41 | class Solution:
def customSortString(self, order: str, s: str) -> str:
ranking = {v:i for i, v in enumerate(order)}
rest = ''
ans = ''
for x in s:
if x not in ranking:
rest += x
continue
ans += x
ans = sorted(an... | custom-sort-string | Simple Python3 beats 87.41% | mclovin286 | 0 | 53 | custom sort string | 791 | 0.693 | Medium | 12,873 |
https://leetcode.com/problems/custom-sort-string/discuss/1569065/Python-hashmap | class Solution:
def customSortString(self, order: str, s: str) -> str:
order_map = collections.defaultdict(lambda: -1)
for i, c in enumerate(order):
order_map[c] = i
return ''.join(sorted(s, key=lambda x: order_map[x])) | custom-sort-string | Python hashmap | dereky4 | 0 | 161 | custom sort string | 791 | 0.693 | Medium | 12,874 |
https://leetcode.com/problems/custom-sort-string/discuss/1559181/Python3-2-liners | class Solution:
def customSortString(self, order: str, s: str) -> str:
order_dict = {c: i + 1 for i, c in enumerate(order)}
return ''.join(sorted(s, key=lambda x: order_dict.get(x, 0))) | custom-sort-string | Python3 2 liners | needforspeed | 0 | 51 | custom sort string | 791 | 0.693 | Medium | 12,875 |
https://leetcode.com/problems/custom-sort-string/discuss/1559181/Python3-2-liners | class Solution:
def customSortString(self, order: str, s: str) -> str:
order_dict = {c: i + 1 for i, c in enumerate(order)}
return ''.join(sorted(s, key=lambda x: order_dict.get(x, len(order_dict)))) | custom-sort-string | Python3 2 liners | needforspeed | 0 | 51 | custom sort string | 791 | 0.693 | Medium | 12,876 |
https://leetcode.com/problems/custom-sort-string/discuss/1531759/Python3-Time%3A-O(s%2Bo)-and-Space%3A-O(s) | class Solution:
def customSortString(self, order: str, s: str) -> str:
# "cba", "abcd" => cbad
# "cbafg", "abcd" => "cbad"
# create dict for s
# iterate thru order and append it to new string if it exists in dict
# append left over characters from dict
# Time: O(s+o)
... | custom-sort-string | [Python3] Time: O(s+o) & Space: O(s) | jae2021 | 0 | 67 | custom sort string | 791 | 0.693 | Medium | 12,877 |
https://leetcode.com/problems/custom-sort-string/discuss/1257901/Python3-Straight-Forward-Method-easy-to-understand | class Solution:
def customSortString(self, order: str, strs: str) -> str:
res = [''] * len(strs)
not_appear = -1
for s in strs:
if s in order:
res[order.index(s)] += s
else:
res[not_appear] = s
not_appear -= 1
re... | custom-sort-string | Python3 Straight Forward Method, easy to understand | georgeqz | 0 | 89 | custom sort string | 791 | 0.693 | Medium | 12,878 |
https://leetcode.com/problems/custom-sort-string/discuss/1087673/Python-One-Liner-Sort-and-Lambda | class Solution:
def customSortString(self, S: str, T: str) -> str:
return ''.join(sorted(T,key=lambda k:[S.index(c) if c in S else len(S) for c in k])) | custom-sort-string | Python One Liner - Sort & Lambda | ashishpawar517 | 0 | 48 | custom sort string | 791 | 0.693 | Medium | 12,879 |
https://leetcode.com/problems/custom-sort-string/discuss/930194/Python3-custom-sorting | class Solution:
def customSortString(self, S: str, T: str) -> str:
mp = {c: i for i, c in enumerate(S)}
return "".join(sorted(T, key=lambda x: mp.get(x, 26))) | custom-sort-string | [Python3] custom sorting | ye15 | 0 | 62 | custom sort string | 791 | 0.693 | Medium | 12,880 |
https://leetcode.com/problems/custom-sort-string/discuss/930194/Python3-custom-sorting | class Solution:
def customSortString(self, order: str, str: str) -> str:
freq = {}
for c in str: freq[c] = 1 + freq.get(c, 0)
ans = []
for c in order:
if c in freq: ans.append(c * freq.pop(c))
return "".join(ans) + "".join(k*v for k, v in freq.items()) | custom-sort-string | [Python3] custom sorting | ye15 | 0 | 62 | custom sort string | 791 | 0.693 | Medium | 12,881 |
https://leetcode.com/problems/custom-sort-string/discuss/594065/Python-Super-Easy-Runtime-24ms-Complexity-O(n) | class Solution:
def customSortString(self, S: str, T: str) -> str:
m=len(S)
n=len(T)
sl=S.split()
l=[""]*n
left=[]
d={}
for i in range(n):
d[T[i]]=0
for i in range(n):
d[T[i]]+=1
... | custom-sort-string | Python Super Easy Runtime-24ms Complexity- O(n) | Ayu-99 | 0 | 47 | custom sort string | 791 | 0.693 | Medium | 12,882 |
https://leetcode.com/problems/custom-sort-string/discuss/455460/Python-3-(one-line) | class Solution:
def customSortString(self, S: str, T: str) -> str:
return ''.join(sorted(T, key = lambda x: {c:i for i,c in enumerate(S)}.get(x,0)))
- Junaid Mansuri
- Chicago, IL | custom-sort-string | Python 3 (one line) | junaidmansuri | 0 | 117 | custom sort string | 791 | 0.693 | Medium | 12,883 |
https://leetcode.com/problems/custom-sort-string/discuss/403513/Python-one-liner | class Solution:
def customSortString(self, S: str, T: str) -> str:
return "".join([x*(T.count(x)) for x in list(S)]+[x for x in T if x not in S]) | custom-sort-string | Python one liner | saffi | 0 | 112 | custom sort string | 791 | 0.693 | Medium | 12,884 |
https://leetcode.com/problems/custom-sort-string/discuss/315615/Python3-straightforward-and-concise-solution-beats-80 | class Solution:
def customSortString(self, S: str, T: str) -> str:
m={}
for i in S:
m.setdefault(i,len(m))
tem=[]
tem2=[]
for i in T:
if i in S:
tem.append(i)
else:
tem2.append(i)
tem=sorted(tem,key=lambda x:m[x])
return ''.join(tem)+''.join(tem2) | custom-sort-string | Python3 straightforward and concise solution beats 80% | jasperjoe | 0 | 50 | custom sort string | 791 | 0.693 | Medium | 12,885 |
https://leetcode.com/problems/custom-sort-string/discuss/304618/Python-O(S%2BT)-solution | class Solution:
def customSortString(self, S: str, T: str) -> str:
letter_count = collections.Counter(T)
other_letters = set(string.ascii_lowercase) - set(S)
order = S + ''.join(other_letters)
result = []
for letter in order:
result.extend([letter * letter_count[l... | custom-sort-string | Python O(S+T) solution | FooBarFooBarFooBar | 0 | 126 | custom sort string | 791 | 0.693 | Medium | 12,886 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1289476/Easy-Approach-oror-Well-explained-oror-95-faster | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
def is_sub(word):
index=-1
for ch in word:
index=s.find(ch,index+1)
if index==-1:
return False
return True
c=0
for word in words:
if is_sub(word):
... | number-of-matching-subsequences | 📌 Easy-Approach || Well-explained || 95% faster 🐍 | abhi9Rai | 27 | 1,200 | number of matching subsequences | 792 | 0.519 | Medium | 12,887 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/2077197/Python3-oror-cache-w-explanation-oror-TM%3A-99.995 | class Solution: # The plan is to iterate through the words and, for each word w, move
# letter by letter of w though the string s if possible to determine
# whether w is a subsequence of s. If so, we add to ans.
#
# We use a functio... | number-of-matching-subsequences | Python3 || cache, w explanation || T/M: 99.9%/95% | warrenruud | 4 | 134 | number of matching subsequences | 792 | 0.519 | Medium | 12,888 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1734042/Python-or-HashMap-or-Counter-or-faster-that-90 | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
alpha = defaultdict(list)
for w in words:
alpha[w[0]].append(w)
counter = 0
for c in s:
old_bucket = alpha[c]
alpha[c] = []
for w in old_bucket:
... | number-of-matching-subsequences | Python | HashMap | Counter | faster that 90% | holdenkold | 4 | 340 | number of matching subsequences | 792 | 0.519 | Medium | 12,889 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1289470/Number-of-Matching-Subsequences-Python-Easy | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
def issub(x, y):
it = iter(y)
return all(c in it for c in x)
c=0
wordsset = set(words)
for i in wordsset:
if issub(i,s):
c = c+words.count(i)
retu... | number-of-matching-subsequences | Number of Matching Subsequences Python Easy | user8744WJ | 4 | 342 | number of matching subsequences | 792 | 0.519 | Medium | 12,890 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1712999/Python-solution-Faster-than-99.32-of-python-Submissions | class Solution:
def check(self,original,new,index):
for ch in new:
index= original.find(ch,index)
if index==-1:return False
index+=1
return True
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
ans=0
for w in words:ans+=self.check(... | number-of-matching-subsequences | Python solution Faster than 99.32% of python Submissions | reaper_27 | 2 | 186 | number of matching subsequences | 792 | 0.519 | Medium | 12,891 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/2310050/Python3-oror-Fast-97-3-Approaches-oror-simple-oror-Explained | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
"""
# OPTIMAL Approach
# create wordMap = {a: [a, acd, ace], b: [bb] ...}
# on each iter it becomes {a:[], b: [b], c: [cd, ce] ...} and small
# Time Complexity: O(n) + O(m)
"""
... | number-of-matching-subsequences | Python3 || Fast 97% 3 Approaches || simple || Explained | Dewang_Patil | 1 | 102 | number of matching subsequences | 792 | 0.519 | Medium | 12,892 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/2308688/Python-Solution | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
result = len(words)
for word in words:
index = -1
for w in word:
index = s.find(w, index + 1)
if index == -1:
result -= 1
brea... | number-of-matching-subsequences | Python Solution | hgalytoby | 1 | 81 | number of matching subsequences | 792 | 0.519 | Medium | 12,893 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/2307644/Python3-Preprocessing-Next-Characters | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
# radix
R = 26
def char_to_int(ch: chr) -> int:
return ord(ch) - ord('a')
# preprocessng
recent_ind = [len(s)] * R
next_char = [()] * len(s) # next_char[i][j] g... | number-of-matching-subsequences | [Python3] Preprocessing Next Characters | jeffreyhu8 | 1 | 11 | number of matching subsequences | 792 | 0.519 | Medium | 12,894 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/2306885/Python-T%3A1001-ms-oror-Mem%3A17.4MB-oror-Commented-oror-Easy-to-understand-oror-HashMap | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
dict = {}
# fill dictionary for all letters with empty list
for c in 'abcdefghijklmnopqrstuvwxyz':
dict[c] = []
# fill lists occurance-indices in super string
for i, c in enu... | number-of-matching-subsequences | [Python] T:1001 ms || Mem:17.4MB || Commented || Easy to understand || HashMap | Buntynara | 1 | 115 | number of matching subsequences | 792 | 0.519 | Medium | 12,895 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1428238/PythonPython3-Simple-readable-solution-using-find-method | class Solution:
def mactchChars(self, s: str, word: str):
# For each char in a word
for char in word:
# Find the current char in the string
index = s.find(char)
# If char not found return false
if index == -1:
... | number-of-matching-subsequences | [Python/Python3] Simple readable solution using find method | ssshukla26 | 1 | 212 | number of matching subsequences | 792 | 0.519 | Medium | 12,896 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/1103585/PythonPython3-Number-of-Matching-Subsequences | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
cnt = 0
final_count = Counter(words)
for word in set(words):
it = iter(s)
if all(letter in it for letter in word):
cnt += final_count[word]
... | number-of-matching-subsequences | [Python/Python3] Number of Matching Subsequences | newborncoder | 1 | 395 | number of matching subsequences | 792 | 0.519 | Medium | 12,897 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/932263/Python3-two-approaches | class Solution:
def numMatchingSubseq(self, S: str, words: List[str]) -> int:
mp = {}
for i, w in enumerate(words): mp.setdefault(w[0], []).append((i, 0))
ans = 0
for c in S:
for i, k in mp.pop(c, []):
if k+1 == len(words[i]): ans += 1
... | number-of-matching-subsequences | [Python3] two approaches | ye15 | 1 | 198 | number of matching subsequences | 792 | 0.519 | Medium | 12,898 |
https://leetcode.com/problems/number-of-matching-subsequences/discuss/932263/Python3-two-approaches | class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
loc = {}
for i, ch in enumerate(s):
loc.setdefault(ch, []).append(i)
ans = 0
for word in words:
x = 0
for ch in word:
i = bisect_left(loc.get(c... | number-of-matching-subsequences | [Python3] two approaches | ye15 | 1 | 198 | number of matching subsequences | 792 | 0.519 | Medium | 12,899 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.