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 |
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https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/2211868/python-3-or-simple-solution-or-O(log(max(a-b-c)))O(1) | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
res = 0
while a or b or c:
aBit, bBit, cBit = a % 2, b % 2, c % 2
if cBit:
res += not (aBit or bBit)
else:
res += aBit + bBit
a >>= 1
b >>= 1
c >>= 1
return res | minimum-flips-to-make-a-or-b-equal-to-c | python 3 | simple solution | O(log(max(a, b, c)))/O(1) | dereky4 | 0 | 28 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,700 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/1741099/1-Line-Python-with-Explanation | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
return (bin(~(c|(~a))) + bin(~(c|(~b))) + bin(c&(~(a|b)))).count('1') | minimum-flips-to-make-a-or-b-equal-to-c | 1 Line Python with Explanation | bbiuyy | 0 | 51 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,701 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/1200717/Python-fast-and-pythonic | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
result = 0
a, b, c = bin(a)[2:], bin(b)[2:], bin(c)[2:]
max_len = max(len(a), len(b), len(c))
a, b, c = a.zfill(max_len), b.zfill(max_len), c.zfill(max_len)
for i in range(len(c)):
if c[i] == '1':
result += a[i] == b[i] == '0'
else:
result += 2 if a[i] == b[i] == '1' else (a[i] == '1' or b[i] == '1')
return result | minimum-flips-to-make-a-or-b-equal-to-c | [Python] fast and pythonic | cruim | 0 | 71 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,702 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/979018/Beats-98 | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
ans = 0
while a or b or c:
if c % 2:
if a%2 or b%2:
pass
else:
ans += 1
else:
if a%2 and b%2:
ans += 2
elif (a%2 == 0 and b%2 == 1) or (a%2 == 1 and b%2 == 0):
ans += 1
else:
pass
a = a >> 1
b = b >> 1
c = c >> 1
return ans | minimum-flips-to-make-a-or-b-equal-to-c | Beats 98% | Aditya380 | 0 | 49 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,703 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/776962/Python3-Simple-with-comments | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
# convert numbers to binary representation
bin_a = bin(a)[2:]
bin_b = bin(b)[2:]
bin_c = bin(c)[2:]
# pad numbers with zeros on the left side (aka rjust /right align) to make sure all 3 numbers are the same length
max_len = max(len(bin_a), len(bin_b), len(bin_c))
bin_a = bin_a.rjust(max_len, '0')
bin_b = bin_b.rjust(max_len, '0')
bin_c = bin_c.rjust(max_len, '0')
flips = 0
for i in range(max_len):
if bin_c[i] == '1':
# if c bit is 1, then at least one of a or b has to be 1
if bin_a[i] == '0' and bin_b[i] == '0':
flips += 1
else:
# if c bit is 0, then a and/or b need to flipped if they are not 0
if bin_a[i] == '1':
flips += 1
if bin_b[i] == '1':
flips += 1
return flips | minimum-flips-to-make-a-or-b-equal-to-c | Python3 - Simple with comments | beyondcloudtech | 0 | 66 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,704 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/477724/Python3-Loop-through-bits | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
ans = 0
while a or b or c:
a, aa = divmod(a, 2)
b, bb = divmod(b, 2)
c, cc = divmod(c, 2)
if cc == 1: ans += (aa == bb == 0)
else: ans += (aa == 1) + (bb == 1)
return ans | minimum-flips-to-make-a-or-b-equal-to-c | [Python3] Loop through bits | ye15 | 0 | 44 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,705 |
https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/discuss/1326153/Python3-solution-single-pass | class Solution:
def minFlips(self, a: int, b: int, c: int) -> int:
a = bin(a)[2:]
b = bin(b)[2:]
c = bin(c)[2:]
x = max([len(a),len(b),len(c)])
a = a.zfill(x)
b = b.zfill(x)
c = c.zfill(x)
count = 0
for i in range(x-1,-1,-1):
if int(c[i]) != (int(a[i]) or int(b[i])):
if c[i] == '0':
if a[i] == '1':
count += 1
if b[i] == '1':
count += 1
else:
count += 1
return count | minimum-flips-to-make-a-or-b-equal-to-c | Python3 solution single-pass | EklavyaJoshi | -1 | 47 | minimum flips to make a or b equal to c | 1,318 | 0.66 | Medium | 19,706 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2420269/Operations-to-make-network-connected-oror-Python3-oror-Union-Find | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
self.components = n
# We need atleast n-1 connections to connect n networks
if(len(connections) < n-1):
return -1
# If we have n-1 connections, we only need to count to number of components
# Union-Find
parent = [i for i in range(0, n)]
rank = [0] * n
for x, y in connections:
self.union(x, y, parent, rank)
# we require no. of components - 1 edge to connect n components
return self.components - 1
def find(self, x, parent):
if(parent[x] != x):
parent[x] = self.find(parent[x], parent)
return parent[x]
def union(self, x, y, parent, rank):
parent_x = self.find(x, parent)
parent_y = self.find(y, parent)
if(parent_x == parent_y):
return
rank_x = rank[parent_x]
rank_y = rank[parent_y]
if(rank_x > rank_y):
parent[parent_y] = parent_x
elif(rank_x < rank_y):
parent[parent_x] = parent_y
else:
parent[parent_y] = parent_x
rank[parent_x] += 1
self.components -= 1 | number-of-operations-to-make-network-connected | Operations to make network connected || Python3 || Union-Find | vanshika_2507 | 1 | 30 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,707 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2299765/Python3-DFS-solution | class Solution:
def makeConnected(self, n: int, a: List[List[int]]) -> int:
length = len(a)
if length < n - 1:
return -1
nodes = set()
connections = defaultdict(list)
for src,dest in a:
connections[src].append(dest)
connections[dest].append(src)
def dfs(src):
if src in nodes:
return
nodes.add(src)
for node in connections[src]:
dfs(node)
result = 0
for i in range(n):
if i not in nodes:
dfs(i)
result+=1
return result - 1 | number-of-operations-to-make-network-connected | 📌 Python3 DFS solution | Dark_wolf_jss | 1 | 38 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,708 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2159089/Connected-Compnents-oror-DFS-oror-Fastest-Optimal-Solution-oror-TC-%3A-O(V2) | class Solution:
def dfs(self, comp, idx, visited, graph):
visited[idx] = True
comp.append(idx)
for neighbour in graph[idx]:
if visited[neighbour] is False:
comp = self.dfs(comp, neighbour, visited, graph)
return comp
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
graph = {}
for i in range(n):
graph[i] = []
for connection in connections:
graph[connection[0]].append(connection[1])
graph[connection[1]].append(connection[0])
totalEdges = 0
for key, value in graph.items():
totalEdges += len(value)
totalEdges = totalEdges//2
visited = [False]*n
connectedComponents = []
for i in range(n):
if len(graph[i]) > 0 and visited[i] is False:
component = []
connectedComponents.append(self.dfs(component, i, visited, graph))
elif len(graph[i]) == 0:
connectedComponents.append([i])
totalComponents = len(connectedComponents)
redundantEdges = totalEdges - ((n - 1) - (totalComponents - 1))
requiredEdges = totalComponents - 1
if redundantEdges >= requiredEdges:
return requiredEdges
else:
return -1 | number-of-operations-to-make-network-connected | Connected Compnents || DFS || Fastest Optimal Solution || TC :- O(V^2) | Vaibhav7860 | 1 | 93 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,709 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2817975/Number-of-Operations-to-Make-Network-Connected | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
def solve(node, parent, vis, adj,check):
vis[node] = True
for i in adj[node]:
#print(parent, node, i)
if i == parent:
continue
if vis[i] and i != parent:
#print(i,node)
if check[str(sorted([i,node]))] == 0:
solve.ans += 1
check[str(sorted([i,node]))] += 1
elif not vis[i]:
solve(i, node, vis, adj, check)
adj = [[] for i in range(n)]
check = {}
for i in connections:
u = i[0]
v = i[1]
adj[u].append(v)
adj[v].append(u)
check[str(sorted([u,v]))] = 0
#print(check)
empty = -1
parent = -1
vis = [False for i in range(n)]
solve.ans = 0
for i in range(n):
if not vis[i]:
solve(i, parent, vis, adj, check)
empty += 1
if (solve.ans) >= empty:
return empty
return -1 | number-of-operations-to-make-network-connected | Number of Operations to Make Network Connected | sarthakchawande14 | 0 | 2 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,710 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2782904/Python-Union-Find-Method | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
#Check the numbers of cables
if n>len(connections)+1:
return -1
uf=UnionFind(n)
for [x,y] in connections:
uf.union(x,y)
return uf.groups-1
# UnionFind class
class UnionFind:
def __init__(self, size):
self.root = [i for i in range(size)]
self.groups=size
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:
self.root[rootY] = rootX
self.groups-=1 | number-of-operations-to-make-network-connected | [Python] Union Find Method | Hikari-Tsai | 0 | 3 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,711 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2748216/python3-oror-easy-oror-intuitive-disjoint-set-approach | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
parentArray=[i for i in range(n)]
sizeArray=[1 for i in range(n)]
extraEdges=0
components=0
def findUltimateParent(node):
if parentArray[node]==node:
return node
parentArray[node]=findUltimateParent(parentArray[node])
return parentArray[node]
def union(u,v):
parentU=findUltimateParent(u)
parentV=findUltimateParent(v)
if parentU==parentV:
return
if sizeArray[parentU]<sizeArray[parentV]:
parentArray[parentU]=parentV
sizeArray[parentV]+=sizeArray[parentU]
else:
parentArray[parentV]=parentU
sizeArray[parentU]+=sizeArray[parentV]
for i,j in connections:
if findUltimateParent(i)==findUltimateParent(j):
extraEdges+=1
else:
union(i,j)
#counting connected components
for i in range(len(parentArray)):
if parentArray[i]==i:
components+=1
if extraEdges>=(components-1):
return components-1
return -1 | number-of-operations-to-make-network-connected | python3 || easy || intuitive disjoint set approach | _soninirav | 0 | 4 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,712 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2745075/DFS-BFS-UnionFind-to-find-number-of-components | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
if len(connections) < n-1: return -1
graph = [[] for _ in range(n)]
for connection in connections:
graph[connection[0]].append(connection[1])
graph[connection[1]].append(connection[0])
visited = set()
def dfs(node):
visited.add(node)
for nei in graph[node]:
if nei not in visited:
dfs(nei)
return
components = 0
for node in range(n):
if node not in visited:
dfs(node)
components += 1
return components - 1 | number-of-operations-to-make-network-connected | DFS, BFS, UnionFind to find number of components | shriyansnaik | 0 | 4 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,713 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2745075/DFS-BFS-UnionFind-to-find-number-of-components | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
if len(connections) < n-1: return -1
graph = [[] for _ in range(n)]
for connection in connections:
graph[connection[0]].append(connection[1])
graph[connection[1]].append(connection[0])
visited = set()
def bfs(v):
queue = deque()
queue.append(v)
visited.add(v)
while queue:
node = queue.popleft()
for nei in graph[node]:
if nei not in visited:
queue.append(nei)
visited.add(nei)
return
components = 0
for node in range(n):
if node not in visited:
bfs(node)
components += 1
return components - 1 | number-of-operations-to-make-network-connected | DFS, BFS, UnionFind to find number of components | shriyansnaik | 0 | 4 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,714 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2363013/or-Python-3-or-DFS-Solution-or | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
if len(connections) < n - 1:
return -1
visited = [False for vis in range(n)]
graph = [[] for grp in range(n)]
for connection in connections:
graph[connection[0]].append(connection[1])
graph[connection[1]].append(connection[0])
def dfs(idx) -> None:
if visited[idx]:
return
visited[idx] = True
for destination in graph[idx]:
dfs(destination)
answer = -1
i = 0
while i < n:
if not visited[i]:
answer += 1
dfs(i)
i += 1
return answer | number-of-operations-to-make-network-connected | | Python 3 | DFS Solution | | YudoTLE | 0 | 22 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,715 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/2318137/Python3-Find-number-of-redundant-connection-and-connected-component | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
# if number of connected component - 1 == number of redundant wire
# use union find
par = [i for i in range(n)]
rank = [1 for i in range(n)]
connected = n
redundant_wire = 0
def find(n1):
p = par[n1]
while p != par[p]:
p = par[p]
return p
def union(n1, n2):
p1 = find(n1)
p2 = find(n2)
if p1 == p2: # redundant connetion
return False
if rank[p1] >= rank[p2]:
par[p2] = p1
rank[p1] += rank[p2]
else:
par[p1] = p2
rank[p2] += rank[p1]
return True
for connection in connections:
if union(connection[0], connection[1]):
connected -= 1
else:
redundant_wire += 1
# if we have more redundant wire, we just need connected - 1 wires
if connected - 1 <= redundant_wire:
return connected-1
return -1 | number-of-operations-to-make-network-connected | [Python3] Find number of redundant connection and connected component | Gp05 | 0 | 50 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,716 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/1899475/Python-99-Using-DFS | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
def dfs_(u):
#internal dfs for going into depth of node
# q = [u]
if vis[u]:
return
#print(u,vis)
vis[u] = True
for j in graph[u]:
if vis[j]: continue;
#vis[j] = True
dfs_(j)
def dfs():
#dfs for each node
c = 0
for i in range(n):
if vis[i]:
continue;
#vis[i] = Trueu
#print(i)
dfs_(i)
c+=1
return c-1
vis = [False]*n
graph = defaultdict(list)
graph = {i:[] for i in range(n)}
for u,v in connections:
graph[u].append(v)
graph[v].append(u)
#print(graph)
return dfs() if n-1<=len(connections) else -1 | number-of-operations-to-make-network-connected | [Python] 99% Using DFS | rstudy211 | 0 | 88 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,717 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/1818518/Python-Union-Find | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
parent = [i for i in range(n)]
rank = [1] * n
groups = [n]
def find(x: int) -> int:
p = parent[x]
while p != parent[p]:
p = parent[p]
parent[x] = p
return p
def union(x: int, y: int) -> None:
x, y = find(x), find(y)
if x != y:
if rank[y] > rank[x]:
x, y = y, x
parent[y] = x
rank[x] += rank[y]
groups[0] -= 1
for node, neigh in connections:
union(node, neigh)
# when we have n-1 or more connections it's always possible to connect all the components by moving the connections around
if len(connections) >= n-1:
return groups[0]-1 # for n amount of components, we only need n-1 connections to connect them all
return -1 | number-of-operations-to-make-network-connected | Python Union-Find | Rush_P | 0 | 67 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,718 |
https://leetcode.com/problems/number-of-operations-to-make-network-connected/discuss/478485/Python3-simple-solution-using-a-graph | class Solution:
def makeConnected(self, n: int, connections: List[List[int]]) -> int:
if len(connections)<n-1: return -1
self.graph,self.visited = {node:set() for node in range(n)},[0]*n
for n1,n2 in connections:
self.graph[n1].add(n2)
self.graph[n2].add(n1)
count = -1
for node in range(n):
count += self.dfs(node)
return count
def dfs(self,node):
if self.visited[node]: return 0
self.visited[node]=1
for n in self.graph[node]:
self.dfs(n)
return 1 | number-of-operations-to-make-network-connected | Python3 simple solution using a graph | jb07 | 0 | 55 | number of operations to make network connected | 1,319 | 0.585 | Medium | 19,719 |
https://leetcode.com/problems/minimum-distance-to-type-a-word-using-two-fingers/discuss/1509241/Well-Coded-oror-Clean-and-Concise-oror-93-faster | class Solution:
def minimumDistance(self, word: str) -> int:
def dist(pre,cur):
if pre==None:
return 0
x1,y1 = divmod(ord(pre)-ord('A'),6)
x2,y2 = divmod(ord(cur)-ord('A'),6)
return abs(x1-x2) + abs(y1-y2)
@lru_cache(None)
def fingers(i,l,r):
if i == len(word):
return 0
n1 = dist(l,word[i]) + fingers(i+1,word[i],r)
n2 = dist(r,word[i]) + fingers(i+1,l,word[i])
return min(n1,n2)
return fingers(0,None,None) | minimum-distance-to-type-a-word-using-two-fingers | 📌📌 Well-Coded || Clean & Concise || 93% faster 🐍 | abhi9Rai | 2 | 228 | minimum distance to type a word using two fingers | 1,320 | 0.597 | Hard | 19,720 |
https://leetcode.com/problems/minimum-distance-to-type-a-word-using-two-fingers/discuss/1485756/Python-or-81-speed-or-Intuitive-or-O(676-*-orwordor)-or-Recursion-%2B-Memo | class Solution:
def minimumDistance(self, word: str) -> int:
locations = {}
x, y = 0, 0
for i in range(65, 91):
locations[chr(i)] = (x, y)
y += 1
if y > 5:
x += 1
y = 0
#print(locations)
@lru_cache(None)
def func(idx, f1, f2):
if idx == len(word):
return 0
# use f1
if f1 == '':
x1, y1 = locations[word[idx]]
else:
x1, y1 = locations[f1]
x, y = locations[word[idx]]
dist1 = abs(x1 - x) + abs(y1 - y)
useF1 = dist1 + func(idx + 1, word[idx], f2)
if f2 == '':
x2, y2 = locations[word[idx]]
else:
x2, y2 = locations[f2]
dist2 = abs(x2 - x) + abs(y2 - y)
useF2 = dist2 + func(idx + 1, f1, word[idx])
return min(useF1, useF2)
return func(0, '', '') | minimum-distance-to-type-a-word-using-two-fingers | Python | 81% speed | Intuitive | O(676 * |word|) | Recursion + Memo | detective_dp | 1 | 179 | minimum distance to type a word using two fingers | 1,320 | 0.597 | Hard | 19,721 |
https://leetcode.com/problems/minimum-distance-to-type-a-word-using-two-fingers/discuss/478022/Python-3-(eight-lines)-(beats-100)-(376-ms)-(BFS-w-Memo) | class Solution:
def minimumDistance(self, W: str) -> int:
A, B, I, W = {(ord(W[0])-65,-1):0}, {}, math.inf, [ord(w)-65 for w in W]
def dist(u,v): return abs(u//6 - v//6) + abs(u % 6 - v % 6)
for w in W[1:]:
for a in A:
B[(w,a[1])] = min(B.get((w,a[1]),I), A[a] + dist(a[0],w))
B[(a[0],w)] = min(B.get((a[0],w),I), A[a] + (a[1] != -1)*dist(a[1],w))
A, B = B, {}
return min(A.values())
- Junaid Mansuri
- Chicago, IL | minimum-distance-to-type-a-word-using-two-fingers | Python 3 (eight lines) (beats 100%) (376 ms) (BFS w/ Memo) | junaidmansuri | 1 | 236 | minimum distance to type a word using two fingers | 1,320 | 0.597 | Hard | 19,722 |
https://leetcode.com/problems/minimum-distance-to-type-a-word-using-two-fingers/discuss/1106569/Python3-top-down-dp | class Solution:
def minimumDistance(self, word: str) -> int:
word = [ord(x)-65 for x in word]
dist = lambda x, y: 0 if -1 in (x, y) else abs(x//6 - y//6) + abs(x%6 - y%6) # l1 distance
@cache
def fn(i, f1=-1, f2=-1):
"""Return minimum distance of typing word[i:] with 2 fingers."""
if i == len(word): return 0
return min(dist(word[i], f1) + fn(i+1, word[i], f2), dist(word[i], f2) + fn(i+1, f1, word[i]))
return fn(0) | minimum-distance-to-type-a-word-using-two-fingers | [Python3] top-down dp | ye15 | 0 | 131 | minimum distance to type a word using two fingers | 1,320 | 0.597 | Hard | 19,723 |
https://leetcode.com/problems/maximum-69-number/discuss/2787125/PYTHON-oror-EASY-TO-UNDERSTAND-oror-WELL-EXPLAINED | class Solution:
def maximum69Number (self, nums: int) -> int:
nums = str(nums) #changing integer to string
j = 1 #maximum number you can change atmost
for i in range(len(nums)):
if nums[i] == "6" and (j == 1): #checking if the element is "6" and we are change only once
bef = nums[:i] #stored element before the element 6 using slicing
aft = nums[i+1:] #stored element after the element 6 using slicing
nums = bef + "9"+aft #adding element in place of 6
j-=1 # we will reduct 1. so that above condition cant satisfy again.
nums = int(nums)
return nums | maximum-69-number | PYTHON || EASY TO UNDERSTAND || WELL EXPLAINED | thesunnysinha | 5 | 372 | maximum 69 number | 1,323 | 0.821 | Easy | 19,724 |
https://leetcode.com/problems/maximum-69-number/discuss/2787127/Detailed-Explanation-of-the-Python-Solution-or-99-Faster | class Solution:
def maximum69Number (self, num: int) -> int:
num = list(str(num))
for i, n in enumerate(num):
if n == '6':
num[i] = '9'
break
return int(''.join(num)) | maximum-69-number | ✔️ Detailed Explanation of the Python Solution | 99% Faster 🔥 | pniraj657 | 4 | 144 | maximum 69 number | 1,323 | 0.821 | Easy | 19,725 |
https://leetcode.com/problems/maximum-69-number/discuss/486561/Python-beats-100-no-string-conversion-explained! | class Solution:
def maximum69Number(self, num):
i = j = 0
original_num = num
while num:
i += 1
d = num % 10
if d == 6: j = i
num //= 10
return original_num + 3*10**(j-1) if j > 0 else original_num | maximum-69-number | Python - beats 100%, no string conversion - explained! | domthedeveloper | 4 | 680 | maximum 69 number | 1,323 | 0.821 | Easy | 19,726 |
https://leetcode.com/problems/maximum-69-number/discuss/484317/Python-3-(one-line)-(beats-100) | class Solution:
def maximum69Number (self, n: int) -> int:
return int(str(n).replace('6','9',1))
- Junaid Mansuri
- Chicago, IL | maximum-69-number | Python 3 (one line) (beats 100%) | junaidmansuri | 4 | 399 | maximum 69 number | 1,323 | 0.821 | Easy | 19,727 |
https://leetcode.com/problems/maximum-69-number/discuss/1222931/python-95-fasteror-easy-solution | class Solution:
def maximum69Number (self, num: int) -> int:
string = str(num)
string = string.replace('6','9',1)
return int(string) | maximum-69-number | [python] 95% faster| easy solution | arkumari2000 | 3 | 353 | maximum 69 number | 1,323 | 0.821 | Easy | 19,728 |
https://leetcode.com/problems/maximum-69-number/discuss/2786667/Easy-Python-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
m=num
s=(str(num))
for i in range(len(s)):
if s[i]=="6":
temp=(int(s[:i]+"9"+s[i+1:]))
else:
temp=(int(s[:i]+"6"+s[i+1:]))
m=max(m,temp)
return m | maximum-69-number | Easy Python Solution | Vistrit | 2 | 156 | maximum 69 number | 1,323 | 0.821 | Easy | 19,729 |
https://leetcode.com/problems/maximum-69-number/discuss/683315/Python-Very-Easy-One-Liner. | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace('6', '9', 1)) | maximum-69-number | Python, Very Easy One-Liner. | Cavalier_Poet | 2 | 301 | maximum 69 number | 1,323 | 0.821 | Easy | 19,730 |
https://leetcode.com/problems/maximum-69-number/discuss/1324678/Python-solution-using-replace | class Solution:
def maximum69Number (self, num: int) -> int:
num = str(num)
if '6' in num:
num = num.replace('6','9',1)
return int(num) | maximum-69-number | Python solution using replace | _Mansiii_ | 1 | 76 | maximum 69 number | 1,323 | 0.821 | Easy | 19,731 |
https://leetcode.com/problems/maximum-69-number/discuss/522661/Python-Solution-using-Stack-O(log10(n)) | class Solution:
def maximum69Number (self, num: int) -> int:
numStack = []
while num!=0:
rem = num%10
num = num // 10
numStack.append(rem)
newNum = 0
found = False
while len(numStack)!=0:
dig = numStack.pop()
if dig == 6 and not found:
dig = 9
found = True
newNum *= 10
newNum += dig
return newNum | maximum-69-number | [Python] Solution using Stack O(log10(n)) | shreyashag | 1 | 216 | maximum 69 number | 1,323 | 0.821 | Easy | 19,732 |
https://leetcode.com/problems/maximum-69-number/discuss/492058/Python3-String-replace | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace("6", "9", 1)) | maximum-69-number | [Python3] String replace | ye15 | 1 | 107 | maximum 69 number | 1,323 | 0.821 | Easy | 19,733 |
https://leetcode.com/problems/maximum-69-number/discuss/2843599/Easy-to-understand-solution | class Solution:
def maximum69Number (self, num: int) -> int:
count = []
change = []
res=0
value=1
six=0
for i in str(num):
change.append(i)
for j in change:
if j =='9':
res += 9*10**int(len(change)-value)
value +=1
if j == '6':
if six ==0:
res += 9*10**int(len(change)-value)
value+=1
six+=1
else:
res += 6*10**int(len(change)-value)
value+=1
return res | maximum-69-number | Easy to understand solution | Warrior-Quant | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,734 |
https://leetcode.com/problems/maximum-69-number/discuss/2843599/Easy-to-understand-solution | class Solution:
def maximum69Number (self, num: int) -> int:
strNum = str(num)
found = False
for i in range(len(strNum)):
if strNum[i] == '6' and not found:
strNum = strNum[:i] + '9' + strNum[i+1:]
found = True
print(strNum)
return int(strNum) | maximum-69-number | Easy to understand solution | Warrior-Quant | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,735 |
https://leetcode.com/problems/maximum-69-number/discuss/2816414/Fast-and-Simply-Solution-Python | class Solution:
def maximum69Number (self, num: int) -> int:
num = str(num)
if "6" in num:
index = num.index("6")
number = ""
for i in range(0, len(num)):
if i == index:
number += "9"
else:
number += num[i]
return int(number)
else:
return int(num) | maximum-69-number | Fast and Simply Solution - Python | PranavBhatt | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,736 |
https://leetcode.com/problems/maximum-69-number/discuss/2805375/simple | class Solution:
def maximum69Number (self, n: int) -> int:
return int(str(n).replace('6','9',1)) | maximum-69-number | simple | siva00100 | 0 | 1 | maximum 69 number | 1,323 | 0.821 | Easy | 19,737 |
https://leetcode.com/problems/maximum-69-number/discuss/2798768/Beats-98 | class Solution:
def maximum69Number (self, num: int) -> int:
num_list = list(str(num))
for i in range(len(num_list)):
if num_list[i] == str(6):
num_list[i]=str(9)
break
print(num_list)
num=int("".join(num_list))
return num | maximum-69-number | Beats 98% | user5333xb | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,738 |
https://leetcode.com/problems/maximum-69-number/discuss/2795483/Python-easy-to-read-and-understand | class Solution:
def maximum69Number (self, num: int) -> int:
nums = list(str(num))
for i, num in enumerate(nums):
if num == '6':
nums[i] = '9'
break
return int(''.join(nums)) | maximum-69-number | Python easy to read and understand | sanial2001 | 0 | 4 | maximum 69 number | 1,323 | 0.821 | Easy | 19,739 |
https://leetcode.com/problems/maximum-69-number/discuss/2794393/Maximum-69-number-Python3-Easy | class Solution:
def maximum69Number (self, num: int) -> int:
num_arr=list(str(num))
for i in range(len(num_arr)):
if num_arr[i]=='6':
num_arr[i]='9'
break
return int(''.join(num_arr)) | maximum-69-number | Maximum 69 number-Python3-Easy | phanee16 | 0 | 1 | maximum 69 number | 1,323 | 0.821 | Easy | 19,740 |
https://leetcode.com/problems/maximum-69-number/discuss/2794241/Maximum69-Solution-or-Python-or-Enumerate | class Solution:
def maximum69Number (self, num: int) -> int:
numbers_list = [int(d) for d in str(num)]
for i,n in enumerate(numbers_list):
if n == 6:
numbers_list[i] = 9
break
new_num = int("".join(map(str, numbers_list)))
return new_num | maximum-69-number | Maximum69 Solution | Python | Enumerate | mitterbabu | 0 | 1 | maximum 69 number | 1,323 | 0.821 | Easy | 19,741 |
https://leetcode.com/problems/maximum-69-number/discuss/2793401/Python-oror-String-oror-Easy-to-Understand | class Solution:
def maximum69Number (self, num: int) -> int:
newNum=""
num=str(num)
i=0
while (i<len(num)):
if (num[i]=='6'):
newNum+='9'
break
newNum+=num[i]
i+=1
newNum+=num[i+1:]
return int(newNum) | maximum-69-number | Python || String || Easy to Understand | vishal_niet | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,742 |
https://leetcode.com/problems/maximum-69-number/discuss/2792251/98.46-FASTERPYTHONand-ONE-LINER-EXPLAINED | class Solution:
def maximum69Number (self, num: int) -> int:
s = str(num)
l = list(s)
if '6' in s:
six = s.index('6')
l[six] = '9'
return int("".join(l)) | maximum-69-number | ✅98.46% FASTER🐍PYTHON💥& ONE-LINER EXPLAINED | shubhamdraj | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,743 |
https://leetcode.com/problems/maximum-69-number/discuss/2792251/98.46-FASTERPYTHONand-ONE-LINER-EXPLAINED | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace("6", "9", 1)) | maximum-69-number | ✅98.46% FASTER🐍PYTHON💥& ONE-LINER EXPLAINED | shubhamdraj | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,744 |
https://leetcode.com/problems/maximum-69-number/discuss/2790640/Using-String-Python3 | class Solution:
def maximum69Number (self, num: int) -> int:
num= list(str(num))
for i in range(len(num)):
if num[i]=='6':
num[i]='9'
break
return ''.join(num) | maximum-69-number | Using String Python3 | SAI_KRISHNA_PRATHAPANENI | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,745 |
https://leetcode.com/problems/maximum-69-number/discuss/2790572/divmod-solution | class Solution:
def maximum69Number (self, num: int) -> int:
#k = int(log(num, 10))
k = 4
pos = 10**k
cur_num = num
while pos >= 1:
cur_digit, cur_num = divmod(cur_num, pos)
if cur_digit == 6:
return num + 3*pos
pos = pos // 10
return num | maximum-69-number | divmod solution | nonchalant-enthusiast | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,746 |
https://leetcode.com/problems/maximum-69-number/discuss/2790461/Python-3-solution | class Solution:
def maximum69Number (self, num: int) -> int:
num = list(str(num))
for i in range(0,len(num)):
if num[i]=='6':
num[i]='9'
break
return ''.join(num) | maximum-69-number | Python 3 solution | mj1602 | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,747 |
https://leetcode.com/problems/maximum-69-number/discuss/2790423/Faster-than-82.37 | class Solution:
def maximum69Number (self, num: int) -> int:
num_str = list(str(num))
change = False
for i in range(len(num_str)):
if num_str[i] == "6":
change = True
break
if not change:
return int("".join(num_str))
else:
for i in range(len(num_str)):
if num_str[i] == "6":
num_str[i] = "9"
break
return int("".join(num_str)) | maximum-69-number | Faster than 82.37% 🔥 | starlingvibes | 0 | 1 | maximum 69 number | 1,323 | 0.821 | Easy | 19,748 |
https://leetcode.com/problems/maximum-69-number/discuss/2790221/Use-math.log10-without-string-conversion | class Solution:
def maximum69Number (self, num: int) -> int:
ans = 0
while num:
k = 10 ** int(math.log10(num))
first, num = divmod(num, k)
ans += 9 * k
if first == 6:
return ans + num
return ans | maximum-69-number | Use math.log10 without string conversion | kcostya | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,749 |
https://leetcode.com/problems/maximum-69-number/discuss/2790120/Python3-Solution-one-liner | class Solution:
def maximum69Number (self, num: int) -> int:
return num if str(num).find('6') == -1 else int(str(num).replace('6', '9', 1)) | maximum-69-number | Python3 Solution one-liner | sipi09 | 0 | 4 | maximum 69 number | 1,323 | 0.821 | Easy | 19,750 |
https://leetcode.com/problems/maximum-69-number/discuss/2789974/Maximum-69-number-oror-PYTHON | class Solution:
def maximum69Number (self, num: int) -> int:
arr = [i for i in str(num)]
ans = [ ]
op=""
for i in range(len(arr)):
if arr[i]!="9":
temp = op
op = op + "9"+ "".join(arr[i+1:])
ans.append(int(op))
op = temp + arr[i]
else:
op+=arr[i]
m=0
for i in ans: m=max(i,m)
return max(m,num) | maximum-69-number | Maximum 69 number || PYTHON | cheems_ds_side | 0 | 6 | maximum 69 number | 1,323 | 0.821 | Easy | 19,751 |
https://leetcode.com/problems/maximum-69-number/discuss/2789950/Intuition-Explained | class Solution:
def maximum69Number (self, num: int) -> int:
# convert integer to string
num = str(num)
# convert this string into array of chars, cuz python string is immutable
num = list(num)
# since 9 bigger than 6, best way to maximize value would be to find the 6 with highest position...
# then swap it for a 9...
# if number has no 6, no scope of increasing value..
for i in range(len(num)):
if num[i] == '6':
# found the first 6...make it 9 and break out of the loop
num[i] = '9'
break
# if we got no 6, we will go through full loop without changing anything
# convert our list back into an integer
# list to string
num = ''.join(num)
# string to num
num = int(num)
return num | maximum-69-number | Intuition Explained | g_aswin | 0 | 4 | maximum 69 number | 1,323 | 0.821 | Easy | 19,752 |
https://leetcode.com/problems/maximum-69-number/discuss/2789808/Maximum-69-Number-or-Python-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace('6', '9', 1)) | maximum-69-number | Maximum 69 Number | Python Solution | nishanrahman1994 | 0 | 6 | maximum 69 number | 1,323 | 0.821 | Easy | 19,753 |
https://leetcode.com/problems/maximum-69-number/discuss/2789799/Easy-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
m=num
t=list(str(num))
for i in range(len(t)):
t=list(str(num))
t[i]='6' if(t[i]=='9')else '9'
m=max(m,int("".join(t)))
return m | maximum-69-number | Easy Solution | soumya262003 | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,754 |
https://leetcode.com/problems/maximum-69-number/discuss/2789709/Python3-Solution-with-using-remainder | class Solution:
def maximum69Number (self, num: int) -> int:
shift = 1
num_copy = num
add_num = 0
while num_copy:
digit = num_copy % 10
if digit == 6:
add_num = shift * 3
shift *= 10
num_copy //= 10
return num + add_num | maximum-69-number | [Python3] Solution with using remainder | maosipov11 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,755 |
https://leetcode.com/problems/maximum-69-number/discuss/2789643/Simple-Python-Solution-or-Using-Strings | class Solution:
def maximum69Number (self, num: int) -> int:
temp = str(num)
for i in range(len(temp)):
if temp[i] == '6':
return int(temp[:i] + '9' + temp[i+1:])
return num | maximum-69-number | Simple Python Solution | Using Strings | Depender | 0 | 5 | maximum 69 number | 1,323 | 0.821 | Easy | 19,756 |
https://leetcode.com/problems/maximum-69-number/discuss/2789642/Python-easy-to-understand-solution-(Faster-than-94) | class Solution:
def maximum69Number (self, num: int) -> int:
numCopy = num
i = 0
sixIdx = -1
while numCopy:
r = numCopy % 10
numCopy //= 10
i += 1
if r == 6:
sixIdx = i
if sixIdx != -1:
return num + 3 * (10 ** (sixIdx - 1))
return num | maximum-69-number | Python easy to understand solution (Faster than 94%) | KevinJM17 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,757 |
https://leetcode.com/problems/maximum-69-number/discuss/2789642/Python-easy-to-understand-solution-(Faster-than-94) | class Solution:
def maximum69Number (self, num: int) -> int:
numdigit = list(str(num))
for i in range(len(numdigit)):
if numdigit[i] == '6':
numdigit[i] = '9'
return int(''.join(numdigit))
return num | maximum-69-number | Python easy to understand solution (Faster than 94%) | KevinJM17 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,758 |
https://leetcode.com/problems/maximum-69-number/discuss/2789639/1323-%3A-Python-Easy-Solution-with-comment | class Solution(object):
def maximum69Number(self,num):
num_arr = list(str(num) #first we will create a variable and convert string to list : eg 9995 will be ['9','9','9','5']
for i in range(len(num_arr): #we are going to iterate through the list
if num_arr[i] == '6': #if index of num_arr is equal to 6 then
num_arr[i] ='9' # for getting the maximum number
break # to come out of the if statement
print(''.join('num_sum')) # combine the list | maximum-69-number | ✅✔1323 : Python Easy Solution with comment | nitin_207013 | 0 | 4 | maximum 69 number | 1,323 | 0.821 | Easy | 19,759 |
https://leetcode.com/problems/maximum-69-number/discuss/2789575/Without-converting-to-string-One-Liner-or-Python | class Solution:
def maximum69Number (self, num: int) -> int:
ind = 0
power = -1
var = num
while var > 0:
if var % 10 == 6:
power = ind
var = var //10
ind += 1
return (num + 3*(10**power)) if power != -1 else num | maximum-69-number | Without converting to string , One Liner | Python | Abhi_-_- | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,760 |
https://leetcode.com/problems/maximum-69-number/discuss/2789575/Without-converting-to-string-One-Liner-or-Python | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace('6', '9', 1)) | maximum-69-number | Without converting to string , One Liner | Python | Abhi_-_- | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,761 |
https://leetcode.com/problems/maximum-69-number/discuss/2789575/Without-converting-to-string-One-Liner-or-Python | class Solution:
def maximum69Number (self, num: int) -> int:
dig = list(str(num))
if '6' in dig:
dig[dig.index('6')] = '9'
return int(''.join(dig))
return num | maximum-69-number | Without converting to string , One Liner | Python | Abhi_-_- | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,762 |
https://leetcode.com/problems/maximum-69-number/discuss/2789459/Python-or-LeetCode-or-1323.-Maximum-69-Number | class Solution:
def maximum69Number (self, num: int) -> int:
list_num = list(str(num))
for i, e in enumerate(list_num):
if e == '6':
list_num[i] = '9'
break
return int(''.join(list_num)) | maximum-69-number | Python | LeetCode | 1323. Maximum 69 Number | UzbekDasturchisiman | 0 | 7 | maximum 69 number | 1,323 | 0.821 | Easy | 19,763 |
https://leetcode.com/problems/maximum-69-number/discuss/2789459/Python-or-LeetCode-or-1323.-Maximum-69-Number | class Solution:
def maximum69Number (self, num):
exp = len(str(num)) - 1
n = num
while exp >= 0:
x = num // (10 ** exp)
y = num % (10 ** exp)
if x == 6:
return n + 3 * (10 ** exp)
else:
exp -= 1
num = y
return n | maximum-69-number | Python | LeetCode | 1323. Maximum 69 Number | UzbekDasturchisiman | 0 | 7 | maximum 69 number | 1,323 | 0.821 | Easy | 19,764 |
https://leetcode.com/problems/maximum-69-number/discuss/2789459/Python-or-LeetCode-or-1323.-Maximum-69-Number | class Solution:
def maximum69Number (self, num: int) -> int:
return int(str(num).replace('6', '9', 1)) | maximum-69-number | Python | LeetCode | 1323. Maximum 69 Number | UzbekDasturchisiman | 0 | 7 | maximum 69 number | 1,323 | 0.821 | Easy | 19,765 |
https://leetcode.com/problems/maximum-69-number/discuss/2789459/Python-or-LeetCode-or-1323.-Maximum-69-Number | class Solution:
def maximum69Number (self, num: int) -> int:
list_num = list(str(num))
for i, e in enumerate(list_num):
if e == "6":
list_num[i] = "9"
break
ret_str = ""
for j in list_num:
ret_str += j
return int(ret_str) | maximum-69-number | Python | LeetCode | 1323. Maximum 69 Number | UzbekDasturchisiman | 0 | 7 | maximum 69 number | 1,323 | 0.821 | Easy | 19,766 |
https://leetcode.com/problems/maximum-69-number/discuss/2789416/Simple-Python-Solution-using-String | class Solution:
def maximum69Number (self, num: int) -> int:
x = str(num)
if '6' in x:
return int(x.replace('6','9',1))
return num | maximum-69-number | Simple Python Solution using String | jps194 | 0 | 4 | maximum 69 number | 1,323 | 0.821 | Easy | 19,767 |
https://leetcode.com/problems/maximum-69-number/discuss/2789369/Python-simple-to-understand | class Solution:
def maximum69Number (self, num: int) -> int:
num_str:str = str(num)
for i,v in enumerate(num_str):
if v != "9":
num_str=num_str[:i] + "9" + num_str[i+1:]
return int(num_str)
return num | maximum-69-number | Python simple to understand | MuftawoOmar | 0 | 5 | maximum 69 number | 1,323 | 0.821 | Easy | 19,768 |
https://leetcode.com/problems/maximum-69-number/discuss/2789214/Simple-Python-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
s=str(num)
r=''
flag=1
for i in s:
if i=='6' and flag:
r+='9'
flag=0
else:
r+=i
return int(r) | maximum-69-number | Simple Python Solution | Siddharth_singh | 0 | 5 | maximum 69 number | 1,323 | 0.821 | Easy | 19,769 |
https://leetcode.com/problems/maximum-69-number/discuss/2789188/optimised-one-in-python3 | class Solution:
def maximum69Number (self, num: int) -> int:
x=str(num)
z=""
n=1
for i in x:
if(int(i)==6)and(n==1):z=z+"9";n=2;continue;
if(n==2)or(int(i)==9):z=z+i;
return int(z) | maximum-69-number | optimised one in python3 | pavankumar19992208 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,770 |
https://leetcode.com/problems/maximum-69-number/discuss/2789185/Python-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
# create result array
res = []
num_list = str(num)
n = len(num_list)
# if num has no 6 int, then return num
if '6' not in num_list:
return num
# Linearly iterate and subst 6 to 9 and 9 to 6
for i in range(len(num_list) -1):
if num_list[i] == '6':
res.append(int(num_list[:i] +'9' + num_list[i+1:]))
else:
res.append(int(num_list[:i] +'6' + num_list[i+1:]))
# Check and update last element of string
if num_list.endswith('6'):
res.append(int(num_list[:n-1]+'9'))
# return max number
return max(res) | maximum-69-number | Python Solution | ratva0717 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,771 |
https://leetcode.com/problems/maximum-69-number/discuss/2789127/Python-3-or-Simple-Math | class Solution:
def maximum69Number (self, num: int) -> int:
temp = str(num)
if '6' not in temp:
return num
left = 0
n = len(temp)
while temp[left] != '6':
if temp[left] != '6':
left += 1
num += 3 * (10 ** (n-left -1))
return num | maximum-69-number | Python 3 | Simple Math | sweetkimchi | 0 | 1 | maximum 69 number | 1,323 | 0.821 | Easy | 19,772 |
https://leetcode.com/problems/maximum-69-number/discuss/2789082/Python3-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
num=str(num)
num=list(num)
for i in range(0,len(num)):
if(num[i]=='6'):
num[i]='9'
break
num = ''.join(num)
return int(num) | maximum-69-number | Python3 Solution | jayesh_s_patil | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,773 |
https://leetcode.com/problems/maximum-69-number/discuss/2789081/Python3-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
num=str(num)
num=list(num)
for i in range(0,len(num)):
if(num[i]=='6'):
num[i]='9'
break
num = ''.join(num)
return int(num) | maximum-69-number | Python3 Solution | jayesh_s_patil | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,774 |
https://leetcode.com/problems/maximum-69-number/discuss/2789027/Python3-oror-Easy-way | class Solution:
def maximum69Number (self, num: int) -> int:
n = list(str(num))
for i in range(len(n)):
if n[i] == '6':
n[i] = '9'
break
return int(''.join(n)) | maximum-69-number | ✅ Python3 || Easy way | PabloVE2001 | 0 | 5 | maximum 69 number | 1,323 | 0.821 | Easy | 19,775 |
https://leetcode.com/problems/maximum-69-number/discuss/2789026/Simple-Python-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
s = list(str(num))
n = len(s)
for i in range(n):
if s[i] == '6':
s[i] = '9'
break;
return int(''.join(s)) | maximum-69-number | Simple Python Solution | mansoorafzal | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,776 |
https://leetcode.com/problems/maximum-69-number/discuss/2789005/Python | class Solution:
def maximum69Number (self, num: int) -> int:
res = [str(i) for i in list(str(num))]
for i in range(len(res)):
if res[i] != '9':
res[i] = '9'
break
return int(''.join(res)) | maximum-69-number | Python | JSTM2022 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,777 |
https://leetcode.com/problems/maximum-69-number/discuss/2788958/Well-Explained-or-100-True-Solution | class Solution:
def maximum69Number (self, num: int) -> int:
num=str(num)
l=[i for i in num]
for i in range(len(l)):
if l[i]=='6':
l[i]='9'
break
return ''.join(map(str,l)) | maximum-69-number | Well Explained | 100% True Solution | wht_hats_ani56 | 0 | 2 | maximum 69 number | 1,323 | 0.821 | Easy | 19,778 |
https://leetcode.com/problems/maximum-69-number/discuss/2788924/Easy-to-understand-python-solution-O(n)-worst-case-time-complexity | class Solution:
def maximum69Number (self, num: int) -> int:
if set(str(num)) == {'9'}:
return num
list_num = list(str(num))
for i in range(len(list_num)):
if list_num[i] == '6':
list_num[i] = '9'
break
return int(''.join(list_num)) | maximum-69-number | Easy to understand python solution, O(n) worst case time complexity | sheetalpatne | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,779 |
https://leetcode.com/problems/maximum-69-number/discuss/2788920/Solution-in-Python. | class Solution:
def maximum69Number (self, num: int) -> int:
lst_num = list(str(num))
for i in range(len(lst_num)):
if lst_num[i] == '6':
lst_num[i] = '9'
break
return int(''.join(lst_num)) | maximum-69-number | Solution in Python. | ahti1405 | 0 | 3 | maximum 69 number | 1,323 | 0.821 | Easy | 19,780 |
https://leetcode.com/problems/print-words-vertically/discuss/1277233/python-94.44-or-easy-or-with-comments | class Solution:
def printVertically(self, s: str) -> List[str]:
st=0 # track of index to take element from each word
s=s.split()
ans=[]
y=0
for i in s:
y=max(y,len(i))
while st<y:
u=[]
for i in s:
if st<len(i):
u.append(i[st])
else:
u.append(' ')# adding spaces if word length is less
while u[-1]==' ': # using stack operation to remove trailing spaces
u.pop()
ans.append(''.join(u))
st+=1# increasing index at each iteration
return ans | print-words-vertically | python 94.44% | easy | with comments | chikushen99 | 2 | 140 | print words vertically | 1,324 | 0.604 | Medium | 19,781 |
https://leetcode.com/problems/print-words-vertically/discuss/1184093/Python-Solution-oror-93.38-faster | class Solution:
def printVertically(self, s: str) -> List[str]:
t = s.split(' ')
l = 0
for i in t:
if l < len(i):
l = len(i)
final = []
i = 0
for j in range(l):
st = ''
for word in t:
if i < len(word) and word[i]:
st += word[i]
else:
st = st + ' '
while len(st) >= 0:
if st[-1] == ' ':
st = st[:-1]
else:
break
i += 1
final.append(st)
return final | print-words-vertically | Python Solution || 93.38% faster | KiranUpase | 1 | 104 | print words vertically | 1,324 | 0.604 | Medium | 19,782 |
https://leetcode.com/problems/print-words-vertically/discuss/2845275/Stripping-and-processing-words | class Solution:
def printVertically(self, s: str) -> List[str]:
# split the words by spaces
# find the longest word in s (how how elements in result array)
# iterate through the words and add the characters to each row (to get column words)
# use a pointer that indicates the position of the char in the word and the position to add the char into
# the result array. If it's in bounds add the char otherwise add the space and increment
# once done, iterate again and trim all leading spaces in the rows
# time O(m * n) m = num of words n = word size O(n)
words = s.split()
maxlen = len(max(words, key=len))
res = [""] * maxlen
for word in words:
i = 0
while i < maxlen:
if i < len(word):
res[i] += word[i]
else:
res[i] += " "
i += 1
for i in range(maxlen):
res[i] = res[i].rstrip()
return res | print-words-vertically | Stripping and processing words | andrewnerdimo | 0 | 1 | print words vertically | 1,324 | 0.604 | Medium | 19,783 |
https://leetcode.com/problems/print-words-vertically/discuss/2792933/Print-Vertically | class Solution:
def printVertically(self, s: str) -> List[str]:
# s="TO BE OR NOT TO BE"
l=s.split(' ')
# print(l)
maxlen=max(len(ele)for ele in l)
# print(maxlen)
m=[]
for i in range(maxlen):
str=''
for word in l:
if i>=len(word):
str+=' '
else:
str+=word[i]
m.append(str.rstrip())
return m | print-words-vertically | Print Vertically | shivansh2001sri | 0 | 1 | print words vertically | 1,324 | 0.604 | Medium | 19,784 |
https://leetcode.com/problems/print-words-vertically/discuss/2792859/Efficient-Solution-in-Python-oror-Easy-Understanding-oror-Optimized-Code | class Solution:
def printVertically(self, s: str) -> List[str]:
s = s.split(); maxlen = 0; j = 0;
for i in s:
maxlen = max(maxlen, len(i));
row = maxlen; col = len(s);
matrix = [[" "]*col for i in range(row)];
for strr in s:
n = len(strr);
for i in range(n):
matrix[i][j] = strr[i];
j += 1;
finalstr = ["".join(substr).rstrip() for substr in matrix];
return finalstr | print-words-vertically | Efficient Solution in Python || Easy Understanding || Optimized Code | avinashdoddi2001 | 0 | 2 | print words vertically | 1,324 | 0.604 | Medium | 19,785 |
https://leetcode.com/problems/print-words-vertically/discuss/2790006/Python3-Solution-99.80-faster | class Solution:
def printVertically(self, s: str) -> List[str]:
words = s.split()
lenWords = len(words)
maxChars = 0
for i, val in enumerate(words):
maxChars = max(len(val), maxChars)
res = [[] for _ in range(maxChars)]
for i in range(maxChars):
for word in words:
try:
res[i].append(word[i])
except:
res[i].append(' ')
return [''.join(val).rstrip() for val in res] | print-words-vertically | [Python3] Solution, 99.80% faster | user8780Q | 0 | 4 | print words vertically | 1,324 | 0.604 | Medium | 19,786 |
https://leetcode.com/problems/print-words-vertically/discuss/2788237/Simple-python-solution | class Solution:
def printVertically(self, s: str) -> List[str]:
ls = s.split()
maxlen = -1
for a in ls:
if len(a) > maxlen:
maxlen = len(a)
arr2D = []
for a in ls:
innerarr = list(a)
if len(a) < maxlen:
for i in range(maxlen-len(a)):
innerarr.append(' ')
arr2D.append(innerarr)
returnarr = []
for i in range(len(arr2D[0])):
string = ''
for j in range(len(arr2D)):
string += arr2D[j][i]
returnarr.append(string.rstrip())
return returnarr | print-words-vertically | Simple python solution | Tonmoy-saha18 | 0 | 3 | print words vertically | 1,324 | 0.604 | Medium | 19,787 |
https://leetcode.com/problems/print-words-vertically/discuss/2787923/Python-or-33ms-or-Beats-92-or-Single-For-Loop-or-Only-Built-In-Libraries | class Solution:
def printVertically(self, s: str) -> List[str]:
mat = []
i = 0
j = 0
for ch in s:
if ch == ' ':
i = 0
j += 1
continue
if i == len(mat):
mat.append([' ' for _ in range(j)])
for _ in range(j - len(mat[i])):
mat[i].append(' ')
mat[i].append(ch)
i += 1
return [''.join(i) for i in mat] | print-words-vertically | Python | 33ms | Beats 92% | Single For Loop | Only Built-In Libraries | Aayush65 | 0 | 2 | print words vertically | 1,324 | 0.604 | Medium | 19,788 |
https://leetcode.com/problems/print-words-vertically/discuss/2787882/Python | class Solution:
def printVertically(self, s: str) -> List[str]:
x=s.split()
n=0
for i in x:
a=len(i)
if a>n:
n=a
m=len(x)
matrix=[]
#print(m,n)
for i in range(n):
matrix.append([' ']*m)
k=0
for i in range(m):
for j in range(len(x[k])):
matrix[j][i]=x[k][j]
k+=1
res=[]
for i in range(n):
strr=''.join(matrix[i])
strr=strr.rstrip()
res.append(strr)
return res | print-words-vertically | Python | Chetan_007 | 0 | 3 | print words vertically | 1,324 | 0.604 | Medium | 19,789 |
https://leetcode.com/problems/print-words-vertically/discuss/2787854/python-solution | class Solution:
def printVertically(self, s: str) -> List[str]:
array = s.split(" ")
maxLen = 0
for a in array:
if len(a) > maxLen:
maxLen = len(a)
ans=[""]*maxLen
for i in range(maxLen):
for j in array:
if(i>=len(j)):
ans[i]+=' '
else:
ans[i]+=j[i]
for i in range(len(ans)):
ans[i]=ans[i].rstrip()
return ans | print-words-vertically | python solution | Nileshsingh08 | 0 | 1 | print words vertically | 1,324 | 0.604 | Medium | 19,790 |
https://leetcode.com/problems/print-words-vertically/discuss/2712058/Runtime-O(total-words-*-max-word-length) | class Solution:
def printVertically(self, s: str) -> List[str]:
arr = s.split(" ");
length = len(max(arr, key=len))
ans = [""] * length;
for i in range(0, len(arr)):
word = arr[i];
for j in range(0, length):
if j >= len(arr[i]):
ans[j] += " ";
continue;
ans[j] += word[j];
return [ans[i].rstrip() for i in range(length)]; | print-words-vertically | Runtime O(total words * max word length) | Jaykant111 | 0 | 1 | print words vertically | 1,324 | 0.604 | Medium | 19,791 |
https://leetcode.com/problems/print-words-vertically/discuss/2508011/easy-python-solution | class Solution:
def printVertically(self, s: str) -> List[str]:
max_len = max([len(word) for word in s.split(' ')])
words = s.split(' ')
output = []
for i in range(max_len) :
val, flag = "", True
for word in words :
if i >= len(word) :
val += " "
else :
if flag :
space_count = len(val)
flag = False
val += word[i]
if val[-1] == ' ' :
val = ' '*space_count + val.strip()
output.append(val)
return output | print-words-vertically | easy python solution | sghorai | 0 | 16 | print words vertically | 1,324 | 0.604 | Medium | 19,792 |
https://leetcode.com/problems/print-words-vertically/discuss/2110968/Python-Solution | class Solution:
def printVertically(self, s: str) -> List[str]:
#s = "TO BE OR NOT TO BE"
#Output: ["TBONTB","OEROOE"," T"]
s = s.split(" ")
length = max([len(item) for item in s])
temp = []
for i in range(0,length):
current = ""
for j in range(0,len(s)):
try:
current += s[j][i]
except:
current += " "
current = self.reduce(current)
temp.append(current)
return temp
def reduce(self,current):
while current[-1] == " ":
current = current[0:-1]
return current | print-words-vertically | Python Solution | xevb | 0 | 29 | print words vertically | 1,324 | 0.604 | Medium | 19,793 |
https://leetcode.com/problems/print-words-vertically/discuss/2103265/python-3-oror-simple-solution | class Solution:
def printVertically(self, s: str) -> List[str]:
words = s.split()
maxLen = max(map(len, words))
verticals = []
for i in range(maxLen):
vertical = []
for word in words:
vertical.append(word[i] if i < len(word) else ' ')
verticals.append(''.join(vertical).rstrip())
return verticals | print-words-vertically | python 3 || simple solution | dereky4 | 0 | 35 | print words vertically | 1,324 | 0.604 | Medium | 19,794 |
https://leetcode.com/problems/print-words-vertically/discuss/1701890/Python-simple-solution | class Solution:
def printVertically(self, s: str) -> List[str]:
words = s.split(' ')
n = len(words)
m = 0
for w in words:
m = max(m, len(w))
res = [[" " for _ in range(n)] for _ in range(m)]
for j in range(n):
word = words[j]
for i in range(len(word)):
res[i][j] = word[i]
return ["".join(x).rstrip() for x in res] | print-words-vertically | Python simple solution | byuns9334 | 0 | 144 | print words vertically | 1,324 | 0.604 | Medium | 19,795 |
https://leetcode.com/problems/print-words-vertically/discuss/1413347/nearly-1-liner-in-Python-3-using-itertools.zip_longest | class Solution:
def printVertically(self, s: str) -> List[str]:
from itertools import zip_longest
return list(''.join(k).rstrip() for k in zip_longest(*s.split(), fillvalue=' ')) | print-words-vertically | nearly 1-liner in Python 3, using itertools.zip_longest | mousun224 | 0 | 79 | print words vertically | 1,324 | 0.604 | Medium | 19,796 |
https://leetcode.com/problems/print-words-vertically/discuss/1325235/Python3-beats-99-comments-in-code | class Solution:
def printVertically(self, s: str) -> List[str]:
words = s.split() # split into words for convenience
ans = []
for i in range(len(words)):
wordlen = len(words[i]) # length of word in question
if wordlen > len(ans): # if the length of this word is longer than the length of ans, then:
# we need to add the appropriate number of strings to ans so that we can fit this word vertically
for k in range(wordlen-len(ans)):
ans.append(' '*i) # the strings we add has to have an appropriate number of blank spaces
for j in range(len(words[i])): # loop to add the characters to the strings in ans
ans[j] += (' '*(i-len(ans[j]))) + words[i][j] # ' '*(i-len(ans[j])) is to add an appropriate number of blanks in case the word before was shorter than the current word
return ans | print-words-vertically | Python3 beats 99%, comments in code | knguyen06 | 0 | 121 | print words vertically | 1,324 | 0.604 | Medium | 19,797 |
https://leetcode.com/problems/print-words-vertically/discuss/1302937/Python3-Simple-solution-faster-than-94-submissions | class Solution:
def printVertically(self, s: str) -> list:
splits = s.split(' ')
max_index = len(max(s.split(' '), key=lambda s:len(s)))
vertical = []
vertical_str = ''
for index in range(0, max_index):
for split in splits:
if index >= len(split):
vertical_str+=' '
else:
char = split[index]
vertical_str += char
vertical_str = vertical_str.rstrip()
vertical.append(vertical_str)
vertical_str = ''
return vertical | print-words-vertically | [Python3] Simple solution, faster than 94% submissions | GauravKK08 | 0 | 54 | print words vertically | 1,324 | 0.604 | Medium | 19,798 |
https://leetcode.com/problems/print-words-vertically/discuss/1113477/Python-clear-short-pythonic | class Solution:
def printVertically(self, s: str) -> List[str]:
s = s.split()
output = []
for key in range(len(max(s, key=len))):
output.append(''.join(s[x][key] if len(s[x]) > key else ' ' for x in range(len(s))).rstrip())
return output | print-words-vertically | [Python], clear, short, pythonic | cruim | 0 | 95 | print words vertically | 1,324 | 0.604 | Medium | 19,799 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.