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/build-array-from-permutation/discuss/2717909/Python-One-liner-solution. | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
n=len(nums)
for i in range(len(nums)):
nums.append(nums[nums[i]])
return nums[n:] | build-array-from-permutation | Python One liner solution. | guneet100 | 0 | 5 | build array from permutation | 1,920 | 0.912 | Easy | 27,100 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2706250/Easiest-Python-Brute-force-solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
nums2 = []
for i in nums:
nums2.append(i)
for i in range(0,len(nums)):
nums2[i] = nums[nums[i]]
return nums2 | build-array-from-permutation | Easiest Python - Brute force solution | jaiswaldevansh27 | 0 | 6 | build array from permutation | 1,920 | 0.912 | Easy | 27,101 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2666324/Python-or-list-comprehensions | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[num] for num in nums] | build-array-from-permutation | Python | list comprehensions | user9015KF | 0 | 3 | build array from permutation | 1,920 | 0.912 | Easy | 27,102 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2652590/Simple-Python-Solution-beats-98.1 | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
result = []
for i in range(len(nums)):
result.append(nums[nums[i]])
return result | build-array-from-permutation | Simple Python Solution beats 98.1% | shubhamshinde245 | 0 | 7 | build array from permutation | 1,920 | 0.912 | Easy | 27,103 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2651079/Python-3-Easy-Solution-Using-For-loop | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
ans = []
for i in range(len(nums)):
ans.append(nums[nums[i]])
return ans | build-array-from-permutation | Python 3 Easy Solution Using For loop | Maazkhattak | 0 | 2 | build array from permutation | 1,920 | 0.912 | Easy | 27,104 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2515118/Python-Solution-easy | class Solution(object):
def buildArray(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
ans = []
for i in range(len(nums)):
ans.append(nums[nums[i]])
return ans | build-array-from-permutation | Python Solution easy | savvy_phukan | 0 | 81 | build array from permutation | 1,920 | 0.912 | Easy | 27,105 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2481643/One-liner-faster-than-96 | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[nums[i]] for i in range(len(nums))] | build-array-from-permutation | One liner faster than 96% | rockdtaz | 0 | 76 | build array from permutation | 1,920 | 0.912 | Easy | 27,106 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2468784/Python-3-or-List-Comprehension-or-Easy-to-Understand-(With-Reference) | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
nums2 = [nums[i] for i in nums]
return nums2 | build-array-from-permutation | Python 3 | List Comprehension | Easy to Understand (With Reference) | danny42 | 0 | 62 | build array from permutation | 1,920 | 0.912 | Easy | 27,107 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2398508/Simple-python-code | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
#create an empty list
ans = []
#iterate through the list (nums)
for i in range(len(nums)):
#add the nums[nums[i]] to ans-> list
ans.append(nums[nums[i]])
#return ans-> list
return ans | build-array-from-permutation | Simple python code | thomanani | 0 | 154 | build array from permutation | 1,920 | 0.912 | Easy | 27,108 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2398508/Simple-python-code | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[i] for i in nums] | build-array-from-permutation | Simple python code | thomanani | 0 | 154 | build array from permutation | 1,920 | 0.912 | Easy | 27,109 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2373092/Easy-solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
a=[]
for i,v in enumerate(nums):
a.append(nums[nums[i]])
return a | build-array-from-permutation | Easy solution | ksn7 | 0 | 87 | build array from permutation | 1,920 | 0.912 | Easy | 27,110 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2275343/Python-Solution-using-List-Comprehension | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
ans=[nums[nums[i]] for i in range(len(nums))]
return ans | build-array-from-permutation | Python Solution using List Comprehension | diptaraj23 | 0 | 72 | build array from permutation | 1,920 | 0.912 | Easy | 27,111 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2259354/Python-one-liner-and-long-code | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[x] for x in nums]
def buildArrayLongCode(self, nums: List[int]) -> List[int]:
ans = []
for i in range(len(nums)):
ans.append(nums[nums[i]])
return ans | build-array-from-permutation | Python one liner and long code | user9702g | 0 | 49 | build array from permutation | 1,920 | 0.912 | Easy | 27,112 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2248431/simplest-python-solution-using-list.insert(indexelement)-and-enumerate | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
ans=[]
for index,value in enumerate(nums):
ans.insert(index,nums[value])
return ans | build-array-from-permutation | simplest python solution using list.insert(index,element) and enumerate | dalersingh13121998 | 0 | 89 | build array from permutation | 1,920 | 0.912 | Easy | 27,113 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2115969/Python-Easy-solutions-with-complexities | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
answer = []
for i in range(len(nums)):
answer.append(nums[nums[i]])
return answer
# space O(N)
# time O(N) | build-array-from-permutation | [Python] Easy solutions with complexities | mananiac | 0 | 167 | build array from permutation | 1,920 | 0.912 | Easy | 27,114 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2115969/Python-Easy-solutions-with-complexities | class Solution:
def buildArray(self,nums: List[int]) -> List[int]:
q = len(nums)
for i in range(len(nums)): #a = qb + r
r = nums[i]
b = nums[nums[i]] % q
nums[i] = q*b + r
for i in range(len(nums)):
nums[i] = nums[i] // q
return nums
# space O(1)
# time O(N) | build-array-from-permutation | [Python] Easy solutions with complexities | mananiac | 0 | 167 | build array from permutation | 1,920 | 0.912 | Easy | 27,115 |
https://leetcode.com/problems/build-array-from-permutation/discuss/2101241/easy-understanding-or-python-or-short-or-simply-ans-or-clean-code | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
l=[]
for i in range(len(nums)):
l.append(nums[nums[i]])
return l | build-array-from-permutation | easy understanding | python | short | simply ans | clean code | T1n1_B0x1 | 0 | 107 | build array from permutation | 1,920 | 0.912 | Easy | 27,116 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1959381/Python-3-Solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
ans = [0] * len(nums)
for i in range(len(nums)):
ans[i] = nums[nums[i]]
return ans | build-array-from-permutation | Python 3 Solution | AprDev2011 | 0 | 99 | build array from permutation | 1,920 | 0.912 | Easy | 27,117 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1959381/Python-3-Solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
n = len(nums)
for i in range(n):
nums[i] = n * (nums[nums[i]] % n) + nums[i]
for i in range(n):
nums[i] = nums[i] // n
return nums | build-array-from-permutation | Python 3 Solution | AprDev2011 | 0 | 99 | build array from permutation | 1,920 | 0.912 | Easy | 27,118 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1934290/Memory-Usage-less-than-91.65-of-Python3-submissions | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
new_arr = []
for x in nums:
new_arr.append(nums[x])
return new_arr | build-array-from-permutation | Memory Usage less than 91.65% of Python3 submissions | ruhiddin_uzb | 0 | 106 | build array from permutation | 1,920 | 0.912 | Easy | 27,119 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1928369/Python3-One-Line-Solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[nums[i]] for i in range(len(nums))] | build-array-from-permutation | [Python3] One-Line Solution | terrencetang | 0 | 73 | build array from permutation | 1,920 | 0.912 | Easy | 27,120 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1902326/Python-One-Liner!-x2 | class Solution:
def buildArray(self, nums):
return [nums[i] for i in nums] | build-array-from-permutation | Python - One Liner! x2 | domthedeveloper | 0 | 122 | build array from permutation | 1,920 | 0.912 | Easy | 27,121 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1902326/Python-One-Liner!-x2 | class Solution:
def buildArray(self, nums):
return map(lambda x:nums[x], nums) | build-array-from-permutation | Python - One Liner! x2 | domthedeveloper | 0 | 122 | build array from permutation | 1,920 | 0.912 | Easy | 27,122 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1885405/Python-one-liner | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[x] for x in nums] | build-array-from-permutation | Python one liner | khayaltech | 0 | 89 | build array from permutation | 1,920 | 0.912 | Easy | 27,123 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1861056/Python-One-Liner | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [ nums[number] for number in nums] | build-array-from-permutation | Python One Liner | hardik097 | 0 | 102 | build array from permutation | 1,920 | 0.912 | Easy | 27,124 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1821442/3-Lines-Python-Solution-oror-60-Faster-oror-Memory-less-than-96 | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
ans = [None]*len(nums)
for i in range(len(nums)): ans[i]=nums[nums[i]]
return ans | build-array-from-permutation | 3-Lines Python Solution || 60% Faster || Memory less than 96% | Taha-C | 0 | 82 | build array from permutation | 1,920 | 0.912 | Easy | 27,125 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1787138/Python3-recursive-NoForLoops-allowed | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
answer = []
def noForLoops(index):
if len(answer) >= len(nums):
return
answer.append(nums[nums[index]])
index += 1
noForLoops(index)
noForLoops(0)
return answer | build-array-from-permutation | Python3 recursive NoForLoops allowed | user5571e | 0 | 86 | build array from permutation | 1,920 | 0.912 | Easy | 27,126 |
https://leetcode.com/problems/build-array-from-permutation/discuss/1434015/Python-3-easy-solution | class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
x=[nums[nums[i]] for i in range(0,len(nums))]
return x; | build-array-from-permutation | Python 3 easy solution | Pranav447 | 0 | 331 | build array from permutation | 1,920 | 0.912 | Easy | 27,127 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1314370/Python3-3-line | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
for i, t in enumerate(sorted((d+s-1)//s for d, s in zip(dist, speed))):
if i == t: return i
return len(dist) | eliminate-maximum-number-of-monsters | [Python3] 3-line | ye15 | 4 | 312 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,128 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1314362/Python-oror-easy-comparator-sort | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
def help(a):
return a[0]/a[1]
temp = []
for i in range(len(dist)):
temp.append([dist[i],speed[i]])
temp = sorted(temp,key = help)
ans = 0
i = 0
while i<len(temp):
if (temp[i][0] - (temp[i][1])*i) > 0 :
ans += 1
i += 1
else:
break
return ans | eliminate-maximum-number-of-monsters | Python || easy comparator sort | harshhx | 3 | 243 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,129 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1314327/python-easy-solutionoror-sorting-arrival-time | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
if 0 in set(dist):
return 0
ans=[]
for i,el in enumerate(dist):
t=math.ceil(el/speed[i])
ans.append(t)
ans.sort()
count=0
prev=0
print(ans)
for i in range(len(ans)):
if prev==ans[i]:
return count
else :
count+=1
prev+=1
return count
``` | eliminate-maximum-number-of-monsters | python easy solution|| sorting arrival time🐍🐍 | aayush_chhabra | 3 | 260 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,130 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1366809/Sort-arrival-times-98-speed | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
for i, t in enumerate(sorted(d / s for d, s in zip(dist, speed))):
if t <= i:
return i
return len(dist) | eliminate-maximum-number-of-monsters | Sort arrival times, 98% speed | EvgenySH | 1 | 138 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,131 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/2662158/Python-3-simple-solution. | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
time = [math.ceil(x / y) for x, y in zip(dist, speed)]
time.sort()
res = 1
for i in range(1, len(time)):
if time[i] <= res:
return res
res += 1
return res | eliminate-maximum-number-of-monsters | Python 3 simple solution. | MaverickEyedea | 0 | 5 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,132 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/2513240/Heap-solution | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
from heapq import heappush, heappop
import math
if len(dist) == 1:
return 1
time = [math.ceil(dist[i]/speed[i]) for i in range(len(dist))]
heapify(time)
ans = 0
i = 0
while time:
i+=1
if heappop(time)-i>=0:
ans+=1
else:
break
return ans | eliminate-maximum-number-of-monsters | Heap solution | hacktheirlives | 0 | 20 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,133 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/2115996/python-3-oror-simple-greedy-solution | class Solution:
def eliminateMaximum(self, dists: List[int], speeds: List[int]) -> int:
times = sorted(dist / speed for dist, speed in zip(dists, speeds))
for i, time in enumerate(times):
if i >= time:
return i
return len(times) | eliminate-maximum-number-of-monsters | python 3 || simple greedy solution | dereky4 | 0 | 61 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,134 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1314969/calculate-times-and-sort-by-arrival | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
n=len(dist)
t=[]
for i in range(n):
t.append((dist[i]/speed[i]))
t.sort()
for i in range(n):
# current time is i and monster arrival time is t[i] if current time is greater than or equal to monster arrival time GAME OVER !!!
if(i>=t[i]):
i=i-1
break
return i+1 | eliminate-maximum-number-of-monsters | calculate times and sort by arrival | madhukar0011 | 0 | 23 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,135 |
https://leetcode.com/problems/eliminate-maximum-number-of-monsters/discuss/1314579/Well-Explained-ororSimple-Code-oror-4-Lines-oror-98-faster-oror-Easily-Understantable | class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
time = [d/s for d,s in zip(dist,speed)]
time.sort()
c=0
for i,t in enumerate(time):
if i<t:
c+=1
else:
break
return c | eliminate-maximum-number-of-monsters | 📌Well-Explained ||Simple Code || 4 Lines || 98% faster || Easily-Understantable 🐍 | abhi9Rai | 0 | 43 | eliminate maximum number of monsters | 1,921 | 0.379 | Medium | 27,136 |
https://leetcode.com/problems/count-good-numbers/discuss/1314484/Python3-Powermod-hack-3-lines | class Solution:
def countGoodNumbers(self, n: int) -> int:
'''
ans=1
MOD=int(10**9+7)
for i in range(n):
if i%2==0:
ans*=5
else:
ans*=4
ans%=MOD
return ans
'''
MOD=int(10**9+7)
fives,fours=n//2+n%2,n//2
# 5^fives*4^fours % MOD
# = 5^fives % MOD * 4^fours % MOD
return (pow(5,fives,MOD) * pow(4,fours,MOD)) % MOD | count-good-numbers | [Python3] Powermod hack, 3 lines | mikeyliu | 5 | 396 | count good numbers | 1,922 | 0.384 | Medium | 27,137 |
https://leetcode.com/problems/count-good-numbers/discuss/1314522/Well-explained-oror-4-Lines-oror-97-faster-oror-easily-understandable | class Solution:
def countGoodNumbers(self, n: int) -> int:
MOD = 10**9+7
# No. of even places
if n%2==0:
ne=n//2
else:
ne=(n+1)//2
# No. of odd places
no=n//2
te = pow(5,ne,MOD) #Total number of even places combinations.
tp = pow(4,no,MOD) #Total number of odd/prime combinations.
return (tp*te)%MOD | count-good-numbers | 📌 Well-explained || 4 Lines || 97% faster || easily-understandable 🐍 | abhi9Rai | 4 | 290 | count good numbers | 1,922 | 0.384 | Medium | 27,138 |
https://leetcode.com/problems/count-good-numbers/discuss/1314377/Python3-1-line | class Solution:
def countGoodNumbers(self, n: int) -> int:
return pow(5, (n+1)//2, 1_000_000_007) * pow(4, n//2, 1_000_000_007) % 1_000_000_007 | count-good-numbers | [Python3] 1-line | ye15 | 1 | 125 | count good numbers | 1,922 | 0.384 | Medium | 27,139 |
https://leetcode.com/problems/count-good-numbers/discuss/2781412/Fastest-Python-Solution | class Solution:
def myPow(self, x: float, n: int) -> float:
if n == 0:
return 1
if n==1:
return x
if n<0:
n = -1 * n
return 1/(self.myPow(x,n))
elif n%2 == 0:
halfvalue = self.myPow(x,n//2)
return halfvalue * halfvalue
else:
oddhalfvalue = self.myPow(x,(n-1)//2)
return oddhalfvalue * oddhalfvalue * x
def countGoodNumbers(self, n: int) -> int:
MOD = 10**9 + 7
primes = 4
evens = 5
ans = 1
if n%2 == 0:
odd_indexes = even_indexes = n//2
else:
odd_indexes = n//2
even_indexes = odd_indexes + 1
ans1 = (pow(primes,odd_indexes,1_000_000_007)) % MOD
ans2 = (pow(evens,even_indexes,1_000_000_007)) % MOD
ans = (ans1 * ans2) % MOD
return int(ans) | count-good-numbers | Fastest Python Solution | aayushjain001 | 0 | 6 | count good numbers | 1,922 | 0.384 | Medium | 27,140 |
https://leetcode.com/problems/count-good-numbers/discuss/2748732/Python-or-Fast-Exponentation | class Solution:
def countGoodNumbers(self, n: int) -> int:
p = 10**9 + 7
if n % 2 == 0:
return pow(20, n // 2, p)
else:
return (5 * pow(20, (n-1) // 2, p)) % p | count-good-numbers | Python | Fast Exponentation | on_danse_encore_on_rit_encore | 0 | 8 | count good numbers | 1,922 | 0.384 | Medium | 27,141 |
https://leetcode.com/problems/count-good-numbers/discuss/1777792/Python3-Concise | class Solution:
def countGoodNumbers(self, n: int) -> int:
fours = n // 2
fives = (n + 1) // 2
mod = 10**9 + 7
ans = pow(4, fours, mod) * pow(5, fives, mod)
return ans % mod | count-good-numbers | Python3 Concise | shtanriverdi | 0 | 148 | count good numbers | 1,922 | 0.384 | Medium | 27,142 |
https://leetcode.com/problems/count-good-numbers/discuss/1408117/python-O(logn)-using-5n2-at-even-indexes-and-4n2-at-odd-indices-75-faster-runtime | class Solution:
def countGoodNumbers(self, n: int) -> int:
# Reusing the super power logic
# to calculate pow(x,n)%mod
def myPow(x: int, n: int, mod: int) -> int:
if n == 0:
return 1
elif n == 1:
return x
else:
half = myPow(x, n//2,mod)
if n % 2 == 0:
# Eg: n = 4 we can split into 2 and 2
# so we multiply two halves alone
return (half*half)%mod
# Eg: n = 5 we get n//2 = 2
# which we can split into 2,2,1
# so we multiply the halves twice and multiply the x for once
return ((half*half)%mod*x)%mod
# For an odd length string we get n//2+1 odd indices and n//2 even indices
# For an even length string we get n//2 odd indices and n//2 even indices
if n%2 == 0:
return ((myPow(4,n//2,10**9+7))*(myPow(5,n//2,10**9+7)))%(10**9+7)
return ((myPow(4,n//2,10**9+7))*(myPow(5,1+n//2,10**9+7)))%(10**9+7) | count-good-numbers | [python] O(logn) using 5^n//2 at even indexes and 4^n//2 at odd indices - 75 % faster runtime | Hariharan-SV | 0 | 238 | count good numbers | 1,922 | 0.384 | Medium | 27,143 |
https://leetcode.com/problems/count-good-numbers/discuss/1366913/Built-in-pow()-with-mod-as-argument-98-speed | class Solution:
def countGoodNumbers(self, n: int) -> int:
odds = n // 2
return (pow(5, n - odds, 1_000_000_007) * pow(4, odds, 1_000_000_007)
% 1_000_000_007) | count-good-numbers | Built-in pow() with mod as argument, 98% speed | EvgenySH | 0 | 164 | count good numbers | 1,922 | 0.384 | Medium | 27,144 |
https://leetcode.com/problems/count-square-sum-triples/discuss/2318104/Easy-Solution-oror-PYTHON | ```class Solution:
def countTriples(self, n: int) -> int:
count = 0
sqrt = 0
for i in range(1,n-1):
for j in range(i+1, n):
sqrt = ((i*i) + (j*j)) ** 0.5
if sqrt % 1 == 0 and sqrt <= n:
count += 2
return (count)
*Please Upvote if you like* | count-square-sum-triples | Easy Solution || PYTHON | Jonny69 | 2 | 134 | count square sum triples | 1,925 | 0.68 | Easy | 27,145 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1336127/Python3-enumeration | class Solution:
def countTriples(self, n: int) -> int:
ans = 0
for a in range(1, n):
for b in range(a+1, n):
c = int(sqrt(a*a + b*b))
if a*a + b*b == c*c and c <= n: ans += 2
return ans | count-square-sum-triples | [Python3] enumeration | ye15 | 1 | 125 | count square sum triples | 1,925 | 0.68 | Easy | 27,146 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1331860/Easy-Python3-faster-than-100 | class Solution:
def countTriples(self, n: int) -> int:
ans=0
for i in range(1,n):
for j in range(1,n):
c= (i**2+j**2)**0.5
if c.is_integer() and c<=n:
ans+=1
return ans | count-square-sum-triples | Easy Python3 faster than 100% | svr300 | 1 | 153 | count square sum triples | 1,925 | 0.68 | Easy | 27,147 |
https://leetcode.com/problems/count-square-sum-triples/discuss/2582971/Python3-oror-O(N*N)-optimized-solution | class Solution:
def countTriples(self, n: int) -> int:
dt = defaultdict(lambda:0)
res = 0
for i in range(n+1):
dt[i*i] = i
for i in range(1,n-1):
for j in range(i+1,n):
if dt[(i*i)+(j*j)]:
res+=2
return res | count-square-sum-triples | Python3 || O(N*N) optimized solution | shacid | 0 | 10 | count square sum triples | 1,925 | 0.68 | Easy | 27,148 |
https://leetcode.com/problems/count-square-sum-triples/discuss/2575802/solving-using-binary-search | class Solution:
def isSquare(self, target, n):
i, j = 1,n
while i <= j:
mid = (i+j)//2
s = mid*mid
# print(i,j, mid, s)
if s == target:
return True
elif s < target:
i = mid + 1
else:
j = mid - 1
return False
def countTriples(self, n: int) -> int:
ans = 0
for i in range(1, n+1):
for j in range(1, n+1):
cond = self.isSquare(i*i + j*j, n)
if cond:
ans += 1
# print(i,j)
return ans | count-square-sum-triples | solving using binary search | tejtharun625 | 0 | 23 | count square sum triples | 1,925 | 0.68 | Easy | 27,149 |
https://leetcode.com/problems/count-square-sum-triples/discuss/2188619/Python-simple-solution | class Solution:
def countTriples(self, n: int) -> int:
ans = 0
for i in range(1,n+1):
for j in range(1,n+1):
if ((i**2 + j**2)**0.5) == int((i**2 + j**2)**0.5) and int((i**2 + j**2)**0.5) <= n:
ans += 1
return ans | count-square-sum-triples | Python simple solution | StikS32 | 0 | 103 | count square sum triples | 1,925 | 0.68 | Easy | 27,150 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1983732/Python3-90-faster-with-explanation | class Solution:
def countTriples(self, n: int) -> int:
numlist, mapper, counter = [], {}, 0
for i in range(1, n + 1):
numlist.append(i * i)
mapper[i *i] = 1
for i in range(len(numlist)):
j = i + 1
while j < len(numlist):
if numlist[i] + numlist[j] in mapper:
counter += 1
# print(numlist[i], numlist[j])
j += 1
#print(mapper, numlist)
return counter * 2 | count-square-sum-triples | Python3, 90% faster with explanation | cvelazquez322 | 0 | 262 | count square sum triples | 1,925 | 0.68 | Easy | 27,151 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1905338/Python-easy-solution-for-beginners | class Solution:
def countTriples(self, n: int) -> int:
count = 0
for i in range(1, n+1):
for j in range(i+1, n+1):
if math.sqrt((i * i) + (j * j)) <= n and int(math.sqrt((i * i) + (j * j))) == math.sqrt((i * i) + (j * j)):
count += 2
return count | count-square-sum-triples | Python easy solution for beginners | alishak1999 | 0 | 142 | count square sum triples | 1,925 | 0.68 | Easy | 27,152 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1845506/6-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-100 | class Solution:
def countTriples(self, n: int) -> int:
ans=0
for i in range(1,n+1):
for j in range(1,i):
k=sqrt(i**2+j**2)
if int(k)==k and k<=n: ans+=2
return ans | count-square-sum-triples | 6-Lines Python Solution || 50% Faster || Memory less than 100% | Taha-C | 0 | 126 | count square sum triples | 1,925 | 0.68 | Easy | 27,153 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1845506/6-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-100 | class Solution:
def countTriples(self, n: int) -> int:
ans=0 ; S=set()
for i in range(1,n+1): S.add(i**2)
for i in range(1,n+1):
for j in range(i+1,n+1):
if i**2+j**2 in S: ans+=2
return ans | count-square-sum-triples | 6-Lines Python Solution || 50% Faster || Memory less than 100% | Taha-C | 0 | 126 | count square sum triples | 1,925 | 0.68 | Easy | 27,154 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1541994/Precompute-squares-87-speed | class Solution:
squares = [i * i for i in range(1, 251)]
def countTriples(self, n: int) -> int:
upper_idx = bisect_right(Solution.squares, n * n)
set_upper = set(Solution.squares[:upper_idx])
count = 0
for i in range(upper_idx - 1):
for j in range(i + 1, upper_idx):
if Solution.squares[i] + Solution.squares[j] in set_upper:
count += 2
return count | count-square-sum-triples | Precompute squares, 87% speed | EvgenySH | 0 | 116 | count square sum triples | 1,925 | 0.68 | Easy | 27,155 |
https://leetcode.com/problems/count-square-sum-triples/discuss/1329201/Python-solution-using-memoization-below-500-ms-runtime | class Solution:
cache = {1:0, 2:0, 3:0, 4:0}
def countTriples(self, n: int) -> int:
if n in self.cache.keys():
return self.cache[n]
else:
self.cache[n] = self.findTriples(n) + self.countTriples(n-1)
return self.cache[n]
def findTriples(self, n: int) -> int:
count = 0
for a in range(0, n):
for b in range(a, n):
if self.isSquareTriple(a, b, n):
count += 1
return count*2
def isSquareTriple(self, a: int, b: int, c: int) -> bool:
if ((a*a) + (b*b) == (c*c)):
return True
else:
return False | count-square-sum-triples | Python solution using memoization, below 500 ms runtime | njain07 | 0 | 134 | count square sum triples | 1,925 | 0.68 | Easy | 27,156 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/1329534/Python-3-or-BFS-Deque-In-place-or-Explanation | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
q = collections.deque([(*entrance, 0)])
m, n = len(maze), len(maze[0])
maze[entrance[0]][entrance[1]] == '+'
while q:
x, y, c = q.popleft()
if (x == 0 or x == m-1 or y == 0 or y == n-1) and [x, y] != entrance:
return c
for i, j in [(x+_x, y+_y) for _x, _y in [(-1, 0), (1, 0), (0, -1), (0, 1)]]:
if 0 <= i < m and 0 <= j < n and maze[i][j] == '.':
maze[i][j] = '+'
q.append((i, j, c + 1))
return -1 | nearest-exit-from-entrance-in-maze | Python 3 | BFS, Deque, In-place | Explanation | idontknoooo | 5 | 197 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,157 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2713932/Fastest-Python-BFS-solution | class Solution:
def nearestExit(self, grid: List[List[str]], entrance: List[int]) -> int:
m=len(grid)
n=len(grid[0])
lst=[[entrance[0],entrance[1],0]]
visited=[[-1]*n for i in range(m)]
row=[-1,1,0,0]
col=[0,0,-1,1]
visited[entrance[0]][entrance[1]]=1
while lst:
x,y,d=lst.pop(0)
for i in range(4):
if x+row[i]>=0 and x+row[i]<m and y+col[i]>=0 and y+col[i]<n and visited[x+row[i]][y+col[i]]==-1 and grid[x+row[i]][y+col[i]]=='.':
if x+row[i]==0 or x+row[i]==m-1 or y+col[i]==0 or y+col[i]==n-1:
return d+1
lst.append([x+row[i],y+col[i],d+1])
visited[x+row[i]][y+col[i]]=1
return -1 | nearest-exit-from-entrance-in-maze | Fastest, Python BFS solution | beneath_ocean | 4 | 270 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,158 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/1343637/WEEB-DOES-BFS-PYTHON | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
row, col = len(maze), len(maze[0])
visited, steps = set([]), 0
queue = deque([entrance])
answer = set([(0,y) for y in range(0,col) if maze[0][y] == "."] + [(row-1,y) for y in range(0,col) if maze[row-1][y] == "."] + [(x,0) for x in range(1,row-1) if maze[x][0] == "."] + [(x,col-1) for x in range(1,row-1) if maze[x][col-1] == "."])
while queue:
for _ in range(len(queue)):
x, y = queue.popleft()
if (x,y) in visited: continue
visited.add((x,y))
for nx, ny in [[x-1,y],[x+1,y],[x,y+1],[x,y-1]]:
if 0<=nx<row and 0<=ny<col and maze[nx][ny] == ".":
if (nx,ny) in answer and (nx,ny) not in visited: return steps+1
queue.append((nx,ny))
steps+=1
return -1 | nearest-exit-from-entrance-in-maze | WEEB DOES BFS PYTHON | Skywalker5423 | 2 | 123 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,159 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2837306/Python-BFS-Easy-with-comments | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
# BFS
rows, cols = len(maze), len(maze[0])
# Set visited spaces to "+"
maze[entrance[0]][entrance[1]] = '+'
# Process stops in a queue
queue = collections.deque()
# Add first stop in queue
queue.appendleft([entrance[0],entrance[1],0])
# Iterate until queue empty or we reach an exit
while queue:
row, col, steps = queue.pop()
# Check each direction breadth first
for r, c in [[row+1, col], [row-1, col], [row, col+1], [row, col-1]]:
# Check in bounds and it not a wall
if 0 <= r < rows and 0 <= c < cols and maze[r][c] == '.':
# Check for exit
if (r == 0) or (c == 0) or (r == rows - 1) or (c == cols -1):
return steps+1
# Add stop to visited
maze[r][c] = '+'
# BFS, new stops get added at the end of the queue, not the front
queue.appendleft([r,c,steps+1])
# No exit found
return -1 | nearest-exit-from-entrance-in-maze | 😃 Python BFS Easy with comments | drblessing | 1 | 25 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,160 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835192/python-solution | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows=len(maze)
cols=len(maze[0])
sx,sy=entrance
maze[sx][sy]='s'
for x in range(rows):
if maze[x][0]=='.':
maze[x][0]='e'
if maze[x][-1]=='.':
maze[x][-1]='e'
for y in range(cols):
if maze[0][y]=='.':
maze[0][y]='e'
if maze[-1][y]=='.':
maze[-1][y]='e'
dir=[(0,1),(1,0),(-1,0),(0,-1)]
done=[[False]*cols for _ in range(rows)]
queue=collections.deque()
queue.append((0,sx,sy))
done[sx][sy]=True
while len(queue)>0:
d,x,y=queue.popleft()
for dx,dy in dir:
nx,ny=x+dx,y+dy
if 0<=nx<rows and 0<=ny < cols and not done[nx][ny]:
if maze[nx][ny]==".":
queue.append((d+1,nx,ny))
elif maze[nx][ny] =="e":
return d+1
return -1
# [6] BFS failed, there is no escape | nearest-exit-from-entrance-in-maze | python solution | ashishneo | 1 | 55 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,161 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2288054/Python3-BFS-similar-to-shortest-path-in-Binary-matrix | class Solution:
def nearestExit(self, grid: List[List[str]], entrance: List[int]) -> int:
m,n,visited,queue = len(grid), len(grid[0]), {}, [(entrance[0],entrance[1],0)]
while queue:
i,j,distance = queue.pop(0)
k = visited.get((i,j))
if(k == None or k > distance):
visited[(i,j)] = distance
for x,y in [(0,-1),(0,1),(-1,0),(1,0)]:
newI, newJ = i+x, j+y
if(newI >= 0 and newI < m and newJ >= 0 and newJ < n and grid[newI][newJ] != "+"):
queue.append((newI, newJ, distance + 1))
result = float("inf")
for i in [0,m-1]:
for j in range(n):
k = visited.get((i,j))
if k != None and [i,j] != entrance:
result = min(result, k)
for j in [0, n-1]:
for i in range(m):
k = visited.get((i,j))
if k != None and [i,j] != entrance:
result = min(result, k)
return -1 if result == float("inf") else result | nearest-exit-from-entrance-in-maze | 📌 Python3 BFS similar to shortest path in Binary matrix | Dark_wolf_jss | 1 | 40 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,162 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2845537/Clean-BFS-in-Python-3 | class Solution:
def nearestExit(self, M: List[List[str]], entrance: List[int]) -> int:
from math import inf
from collections import deque
m, n = len(M), len(M[0])
q = deque([(tuple(entrance), 0)])
d = ((0, 1), (0, -1), (1, 0), (-1, 0))
visited = set([tuple(entrance)])
ans = inf
while q:
cur, step = q.popleft()
x, y = cur
for dx, dy in d:
nx, ny = x + dx, y + dy
if not (0 <= nx < m and 0 <= ny < n) or (nx, ny) in visited or M[nx][ny] == "+":
continue
if (0 in (nx, ny) or nx == m - 1 or ny == n - 1):
ans = min(ans, step + 1)
else:
visited.add((nx, ny))
q += ((nx, ny), step + 1),
return (ans, -1)[ans == inf] | nearest-exit-from-entrance-in-maze | Clean BFS in Python 3 | mousun224 | 0 | 4 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,163 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2842616/Python-Solution-Using-BFS-oror-Beats-100 | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m,n = len(maze), len(maze[0])
que = collections.deque()
que.append([entrance[0],entrance[1],0])
maze[entrance[0]][entrance[1]] = '+'
while len(que)>0:
x,y,cost = que.popleft()
if [x,y] != entrance and (x==0 or y==0 or x==m-1 or y==n-1):
return cost
else:
if x>0 and maze[x-1][y] == '.':
maze[x-1][y] = '+'
que.append([x-1,y,cost+1])
if x<m-1 and maze[x+1][y] == '.':
maze[x+1][y] = '+'
que.append([x+1,y,cost+1])
if y>0 and maze[x][y-1] == '.':
maze[x][y-1] = '+'
que.append([x,y-1,cost+1])
if y<n-1 and maze[x][y+1] == '.':
maze[x][y+1] = '+'
que.append([x,y+1,cost+1])
return -1 | nearest-exit-from-entrance-in-maze | Python Solution, Using BFS || Beats 100% | aryan_codes_here | 0 | 2 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,164 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2839938/BFS-Modular-Solution-with-Input-Validation-The-Best | class Solution:
def __init__(self):
self.maze = None
self.entrance = None
self.rows, self.cols = 0, 0
def nearestExit(self, maze: list[list[str]], entrance: list[int]):
self.configure(maze, entrance)
self.is_empty(*entrance, 0)
moves, index = deque([entrance]), 0
while moves:
row, col = moves.popleft()
# print(row, col)
if self.is_exit(row, col):
return self.maze[row][col]
index += 1
moves += self.get_moves_and_mark(row, col)
return -1
def configure(self, maze, entrance):
self.maze = maze
self.entrance = entrance
self.rows = len(maze)
assert self.rows > 0
self.cols = len(maze[0])
assert all(len(row_length) == self.cols for row_length in maze)
assert all(all(cell in '.+' for cell in row) for row in maze)
assert len(entrance) == 2
assert self.is_empty(*entrance)
def is_empty(self, row, col, mark=None):
empty = 0 <= row < self.rows and 0 <= col < self.cols and self.maze[row][col] == '.'
if empty and mark is not None:
self.maze[row][col] = mark
return empty
def is_exit(self, row, col):
return (row in {0, self.rows-1} or col in {0, self.cols-1}) and [row, col] != self.entrance
def get_moves_and_mark(self, row, col):
return [[row + dy, col + dx] for (dy, dx) in ((-1, 0), (1, 0), (0, -1), (0, 1)) if
self.is_empty(row + dy, col + dx, self.maze[row][col] + 1)] | nearest-exit-from-entrance-in-maze | BFS, Modular Solution with Input Validation, The Best | Triquetra | 0 | 5 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,165 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2837724/Python-Solution | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m = len(maze)
n = len(maze[0])
queue = deque()
seen = set()
moves = [[0, 1], [0, -1], [1, 0], [-1, 0]]
steps = 0
queue.append(entrance)
while queue:
count = len(queue)
steps += 1
for _ in range(count):
cell = queue.popleft()
for move in moves:
x = cell[0] + move[0]
y = cell[1] + move[1]
if x < 0 or x >= m:
continue
if y < 0 or y >= n:
continue
if maze[x][y] == '+':
continue
if x == entrance[0] and y == entrance[1]:
continue
if x == 0 or x == m - 1:
return steps
if y == 0 or y == n - 1:
return steps
if (x, y) in seen:
continue
queue.append((x, y))
seen.add((x, y))
return -1 | nearest-exit-from-entrance-in-maze | Python Solution | mansoorafzal | 0 | 12 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,166 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2837334/simple-and-efficient | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
paths = collections.deque([entrance])
maze[entrance[0]][entrance[1]] = "+"
steps = 0
while paths:
for _ in range(len(paths)):
row, col = paths.pop()
if steps > 0 and (row in (0, len(maze)-1) or col in (0, len(maze[0])-1)):
return steps
if row > 0 and maze[row-1][col] == ".":
paths.appendleft((row-1, col))
maze[row-1][col] = "+"
if row < len(maze)-1 and maze[row+1][col] == ".":
paths.appendleft((row+1, col))
maze[row+1][col] = "+"
if col > 0 and maze[row][col-1] == ".":
paths.appendleft((row, col-1))
maze[row][col-1] = "+"
if col < len(maze[0])-1 and maze[row][col+1] == ".":
paths.appendleft((row, col+1))
maze[row][col+1] = "+"
steps += 1
return -1 | nearest-exit-from-entrance-in-maze | simple and efficient | TrickyUnicorn | 0 | 5 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,167 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836955/BFS-and-Python | class Solution:
def isItExit(self, curr_x, curr_y, n,m):
return curr_x == 0 or curr_y == 0 or curr_x == n-1 or curr_y == m-1
def isPossibleToProcess(self, new_x, new_y, maze, n,m):
return 0<= new_x < n and 0<= new_y < m and maze[new_x][new_y] == "."
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
dirxn = ((-1,0), (1,0), (0, -1), (0,1))
n,m = len(maze), len(maze[0])
x, y = entrance
maze[x][y] = "+"
queue = collections.deque()
queue.append((x,y,0))
result = -1
while queue:
curr_x, curr_y, curr_result = queue.popleft()
for dx, dy in dirxn:
new_x = curr_x + dx
new_y = curr_y + dy
if(self.isPossibleToProcess(new_x, new_y, maze, n,m)):
if(self.isItExit(new_x, new_y, n,m)):
return curr_result + 1
# we will block the current tiles
maze[new_x][new_y] = "+"
queue.append((new_x, new_y, curr_result+1))
return result | nearest-exit-from-entrance-in-maze | BFS and Python | Sanjaychandak95 | 0 | 6 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,168 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836652/BFS-solution-python3-with-explanation | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
ROWS, COLS = len(maze), len(maze[0])
#for finding all the available exits
exits = set()
for r in range(ROWS):
for c in range(COLS):
if (maze[r][c] == '.' and
((r == 0 or r == ROWS - 1) or
(c == 0 or c == COLS - 1))):
exits.add((r, c))
#removing the entrance, if it is in exits
if tuple(entrance) in exits:
exits.remove(tuple(entrance))
directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
#normal BFS
q = deque()
q.append([entrance[0], entrance[1], 0])
while q:
r, c, steps = q.popleft()
if (r, c) in exits:
return steps
for dr, dc in directions:
nr, nc = r + dr, c + dc
if 0 <= nr < ROWS and 0 <= nc < COLS and maze[nr][nc] == '.':
#changing the state of the visited place in the maze
maze[nr][nc] = 0
q.append([nr, nc, steps + 1])
return -1 | nearest-exit-from-entrance-in-maze | BFS solution python3 with explanation | mohitkumar36 | 0 | 3 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,169 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836556/1926.-Nearest-Exit-from-Entrance-in-Maze-or-Python3 | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows, cols = len(maze), len(maze[0])
#create list of directions we can travel in
dirs = ((1, 0), (-1, 0), (0, 1), (0,-1))
#create values for the entrance and mark it as visited
sr, sc = entrance
maze[sr][sc] = "+"
#create a queue and add the entrance
queue = collections.deque()
queue.append([sr, sc, 0])
while queue:
#set current row, current column, current distance to the next cell in the queue
cr, cc, cd = queue.popleft()
#check every direction
for d in dirs:
#next row/column equals the current row/column plus the direction
nr = cr + d[0]
nc = cc + d[1]
#check to see if the next cell is in the maze and open
if (0 <= nr < rows) and (0 <= nc < cols) and maze[nr][nc] == ".":
#check to see if its on the edge meaning an exit
if 0 == nr or nr == rows-1 or 0 == nc or nc == cols-1:
#return the current distance + 1
return cd + 1
#if not an exit then mark it as visited and add it to queue
maze[nr][nc] = "+"
queue.append([nr, nc, cd+1])
#return -1 if nothing is found
return -1 | nearest-exit-from-entrance-in-maze | 1926. Nearest Exit from Entrance in Maze | Python3 | AndrewMitchell25 | 0 | 6 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,170 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836261/Python-Solution-or-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
EMPTY, WALL = '.', '+'
directions = {(0, 1), (0, -1), (1, 0), (-1, 0)}
m, n = len(maze), len(maze[0])
q, visited = deque(), set()
def in_bounds(r, c):
return 0 <= r < m and 0 <= c < n
def is_exit(r, c):
return maze[r][c] == EMPTY and (r == 0 or r == m-1 or c == 0 or c == n-1)
def is_wall(r, c):
return maze[r][c] == WALL
ans = 0
q.append((entrance[0], entrance[1]))
visited.add((entrance[0], entrance[1]))
while q:
ans += 1
for i in range(len(q)):
r, c = q.popleft()
for Δr, Δc in directions:
rr, cc = r + Δr, c + Δc
if not in_bounds(rr, cc) or (rr, cc) in visited or is_wall(rr, cc):
continue
if is_exit(rr, cc):
return ans
q.append((rr, cc))
visited.add((rr, cc))
return -1 | nearest-exit-from-entrance-in-maze | Python Solution | BFS | on_danse_encore_on_rit_encore | 0 | 7 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,171 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836224/Python3-or-BFS-%2B-Heap-or-Shortest-path-or-Dijkstra | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
dirList = [[0, 1], [0, -1], [1, 0], [-1, 0]]
bfs = [(0, entrance[0], entrance[1])]
seen = {}
n, m = len(maze), len(maze[0])
def isExit(x, y):
if x == entrance[0] and y == entrance[1]:
return False
return x == 0 or y == 0 or x == n-1 or y == m-1
while bfs:
curd, curx, cury = heapq.heappop(bfs)
if maze[curx][cury] == '+':
continue
maze[curx][cury] = '+'
if isExit(curx, cury):
return curd
for dx, dy in dirList:
x, y = curx + dx, cury + dy
if x < 0 or x >= n or y < 0 or y >= m:
continue
heapq.heappush(bfs, (curd+1, x, y))
return -1 | nearest-exit-from-entrance-in-maze | Python3 | BFS + Heap | Shortest path | Dijkstra | vikinam97 | 0 | 4 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,172 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836211/Clean-BFS-Solution-oror-Python-3-oror-Easy-Approach | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m,n=len(maze),len(maze[0])
exit=set()
for i in range(m):
if maze[i][0]=="." and [i,0]!=entrance:
exit.add((i,0))
if maze[i][n-1]=="." and [i,n-1]!=entrance:
exit.add((i,n-1))
for i in range(n):
if maze[0][i]=="." and [0,i]!=entrance:
exit.add((0,i))
if maze[m-1][i]=="." and [m-1,i]!=entrance:
exit.add((m-1,i))
def bfs(maze,i,j):
nonlocal exit,m,n
seen=set()
seen.add((i,j))
q=[(i,j)]
ans=0
while q:
for i in range(len(q)):
x,y=q.pop(0)
if (x,y) in exit:
return ans
for cx,cy in [(1,0),(0,1),(-1,0),(0,-1)]:
nx,ny=x+cx,y+cy
if 0<=nx<m and 0<=ny<n and maze[nx][ny]=="." and (nx,ny) not in seen:
q.append((nx,ny))
seen.add((nx,ny))
ans+=1
return -1
return bfs(maze,entrance[0],entrance[1]) | nearest-exit-from-entrance-in-maze | Clean BFS Solution || Python 3 || Easy Approach | aditya1292 | 0 | 7 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,173 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836158/Python3-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m = len(maze)
n = len(maze[0])
startX, startY = entrance
DQ = deque([[startX, startY, 0]])
visited = {(startX, startY)}
while DQ:
X, Y, dist = DQ.popleft()
for x, y in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
currX, currY = X + x, Y + y
if 0 <= currX < m and 0 <= currY < n and (currX, currY) not in visited and maze[currX][currY] == '.':
if currX == 0 or currX == m - 1 or currY == 0 or currY == n - 1:
return dist + 1
DQ.append([currX, currY, dist + 1])
visited.add((currX, currY))
return -1 | nearest-exit-from-entrance-in-maze | Python3 BFS | mediocre-coder | 0 | 4 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,174 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2836152/Python3-using-BFS-with-List-%2B-Set | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
vis = set([tuple(entrance)])
bfs = [[*entrance, 0]]
for r, c, steps in bfs:
for nr, nc in [(r+1, c), (r-1, c), (r, c+1), (r, c-1)]:
if not 0 <= nr < len(maze) or not 0 <= nc < len(maze[0]):
continue
if maze[nr][nc] == '+':
continue
if (nr, nc) in vis:
continue
if nr == 0 or nc == 0 or nr+1 == len(maze) or nc+1 == len(maze[0]):
return steps + 1
vis.add((nr, nc))
bfs.append([nr, nc, steps+1])
return -1 | nearest-exit-from-entrance-in-maze | Python3 using BFS with List + Set | Nesop | 0 | 8 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,175 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835963/Python3-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m, n = len(maze), len(maze[0])
entrance = tuple(entrance)
lv = 1
seen = {entrance}
curLv = [entrance]
while curLv:
nxtLv = []
for r,c in curLv:
for nr,nc in [(r+1,c),(r-1,c),(r,c+1),(r,c-1)]:
if 0<=nr<m and 0<=nc<n and maze[nr][nc]=="." \
and (nr,nc) not in seen:
if nr in (0,m-1) or nc in (0,n-1):
return lv
nxtLv.append((nr,nc))
seen.add((nr,nc))
curLv = nxtLv
lv += 1
return -1 | nearest-exit-from-entrance-in-maze | [Python3] BFS | ruosengao | 0 | 6 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,176 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835737/Python3-BFS-simple-way | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
# BFS
q = [[entrance[0],entrance[1]]]
step = 0
rows = len(maze)
cols = len(maze[0])
# clockwise
directions = [[0,1],[1,0],[0,-1],[-1,0]]
maze[entrance[0]][entrance[1]] = '+'
while q:
for _ in range(len(q)):
r,c = q.pop(0)
# q.append so can not check this is '+'
# entrance can not be exit
if (0 == r or 0 == c or r == rows-1 or c ==cols-1) and [r,c] != [entrance[0], entrance[1]]:
return step
for x, y in directions:
new_x, new_y = r+x, c+y
if 0 <= new_x < rows and 0 <= new_y < cols and maze[new_x][new_y]=='.':
# Cuz change '.' to '+'
maze[new_x][new_y] = '+'
q.append([new_x,new_y])
step += 1
return -1 | nearest-exit-from-entrance-in-maze | Python3 BFS simple way | dad88htc816 | 0 | 7 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,177 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835734/Python-or-Java-or-Easy-BFS-Solution | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
queue = []
directions = [[0, -1], [0, 1], [1, 0], [-1, 0]]
rows = len(maze)
columns = len(maze[0])
ans = 0
# offer and mark visited
queue.append(entrance)
maze[entrance[0]][entrance[1]] = '+'
while queue:
size = len(queue)
ans += 1
while size > 0:
size -= 1
position = queue.pop(0)
x = position[0]
y = position[1]
for dir in directions:
newX = x + dir[0]
newY = y + dir[1]
if newX < 0 or newY < 0 or newX > rows - 1 or newY > columns - 1 or maze[newX][newY] == '+':
continue
elif newX == 0 or newY == 0 or newX == rows - 1 or newY == columns - 1:
return ans
# offer and mark visited
queue.append([newX, newY])
maze[newX][newY] = '+'
return -1 | nearest-exit-from-entrance-in-maze | Python | Java | Easy BFS Solution | rahul_mishra_ | 0 | 8 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,178 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835603/Python-or-BFS-solution-using-queue | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows, cols = len(maze), len(maze[0])
q = deque([entrance])
visited = set()
visited.add((entrance[0], entrance[1]))
directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
steps = 0
while q:
for i in range(len(q)):
r, c = q.popleft()
if [r, c] != entrance and (r == 0 or r == rows - 1 or c == 0 or c == cols - 1):
return steps
for dr, dc in directions:
nr, nc = r + dr, c + dc
if nr in range(rows) and nc in range(cols) and (nr, nc) not in visited and maze[nr][nc] == '.':
visited.add((nr, nc))
q.append([nr, nc])
steps += 1
return -1 | nearest-exit-from-entrance-in-maze | Python | BFS solution using queue | KevinJM17 | 0 | 4 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,179 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835566/Python-soln-using-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
n=len(maze)
m=len(maze[0])
q=deque()
q.append([entrance[0],entrance[1],0])
ret=[]
visit=set()
while q:
[x,y,s]=q.popleft()
if not (0<=x<n and 0<=y<m):
continue
if (x,y) in visit or maze[x][y]=='+':
continue
if x==n-1 or x==0 or y==m-1 or y==0:
if s:
return s
visit.add((x,y))
q.append([x+1,y,s+1])
q.append([x-1,y,s+1])
q.append([x,y+1,s+1])
q.append([x,y-1,s+1])
return -1 | nearest-exit-from-entrance-in-maze | Python soln using BFS | DhruvBagrecha | 0 | 4 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,180 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835526/Python-or-BFS-solution | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
q = deque([entrance])
m, n = len(maze) - 1, len(maze[0]) - 1
used, ans, dist = {tuple(entrance)}, 1e18, defaultdict(int)
dist[tuple(entrance)] = 0
while q:
nx = q.popleft()
if (nx[0] == m or nx[1] == n or nx[0] == 0 or nx[1] == 0) and nx != entrance:
ans = min(ans, dist[tuple(nx)])
for f, s in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
x, y = f + nx[0], s + nx[1]
if 0 <= x <= m and 0 <= y <= n and (x, y) not in used and maze[x][y] == '.':
dist[(x, y)] = dist[tuple(nx)] + 1
q.append((x, y))
used.add((x, y))
return -1 if ans == 1e18 else ans | nearest-exit-from-entrance-in-maze | Python | BFS solution | LordVader1 | 0 | 8 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,181 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835421/BFS-Easy-Solution-in-O(N) | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows = len(maze)
columns = len(maze[0])
q = deque([(entrance[0], entrance[1], 0)])
visited = set()
while q:
i, j, steps = q.popleft()
if (i == rows - 1 or i == 0 or j == columns - 1 or j == 0) and steps != 0:
return steps
visited.add((i, j))
for di, dj in [(1, 0), (0, 1), (-1, 0), (0, -1)]:
ni, nj = (di + i), (dj + j)
if 0 <= ni < rows and 0 <= nj < columns and (ni, nj) not in visited and maze[ni][nj] == '.':
visited.add((ni, nj))
q.append((ni, nj, steps + 1))
return -1 | nearest-exit-from-entrance-in-maze | BFS - Easy Solution in O(N) | user6770yv | 0 | 5 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,182 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835283/Python-Simple-BFS-(TLE-explanation) | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
count = 1
queue = [[entrance[0], entrance[1]]]
row = len(maze)
col = len(maze[0])
maze[entrance[0]][entrance[1]] = '+'
while(queue):
for _ in range(len(queue)):
cur = queue.pop(0)
#if mark "+" in there we will got the TLE because we will count duplicate
#ex:queue[[1,2], [2,1]] if [1,1] is '.', then we will add [1,1] to queue two times
#maze[cur[0]][cur[1]] = '+'
for dx, dy in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
if(0 <= cur[0] + dx < row and 0 <= cur[1] + dy < col and maze[cur[0] + dx][cur[1] + dy] == '.'):
#mark "+" in there so we don't count duplicate
maze[cur[0] + dx][cur[1] + dy] = '+'
if(cur[0] + dx == 0 or cur[0] + dx == row - 1 or cur[1] + dy == 0 or cur[1] + dy == col - 1):
return count
queue.append([cur[0] + dx, cur[1] + dy])
count += 1
return -1 | nearest-exit-from-entrance-in-maze | Python Simple BFS (TLE explanation) | 55337123kk3 | 0 | 5 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,183 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835093/Python-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
Q = [entrance]
Q_next = []
maze[entrance[0]][entrance[1]] = '+'
near = [[1, 0], [-1, 0], [0, 1], [0, -1]]
row, col = len(maze) - 1, len(maze[0]) - 1
step = 0
while len(Q) > 0:
curr = Q.pop(0)
for n in near:
next = [curr[0] + n[0], curr[1] + n[1]]
if (next[0] >= 0) and (next[1] >= 0) and (next[0] <= row) and (next[1] <= col):
if maze[next[0]][next[1]] == '.':
if (next[0] == 0) or (next[1] == 0) or (next[0] == row) or (next[1] == col):
return step + 1
Q_next.append(next)
maze[next[0]][next[1]] = '+'
if len(Q) == 0:
Q, Q_next = Q_next, Q
step += 1
return -1 | nearest-exit-from-entrance-in-maze | [Python] BFS | mingyang-tu | 0 | 10 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,184 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2835041/Easy-for-understand | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows = len(maze)
cols = len(maze[0])
queue = collections.deque()
queue.append((entrance[0], entrance[1], 0))
seen = set()
mn = -1
while queue:
row, col, step = queue.popleft()
if mn != - 1 and mn < step:
continue
if row >= rows or row < 0 or col >= cols or col < 0:
continue
if maze[row][col] == '+' or (row, col) in seen:
continue
seen.add((row, col))
if not (row == entrance[0] and col == entrance[1]) and (row in [0, rows - 1] or col in [0, cols - 1]):
return step
queue.append((row - 1, col, step + 1))
queue.append((row + 1, col, step + 1))
queue.append((row, col - 1, step + 1))
queue.append((row, col + 1, step + 1))
return -1 | nearest-exit-from-entrance-in-maze | Easy for understand | lllymka | 0 | 16 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,185 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834999/Python-oror-Easy-Solution-oror-O(N)-oror-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
n, m = len(maze), len(maze[0])
visit = []
for i in range(n):
visit.append([False]*m)
def isValid(i, j):
return i >= 0 and j >= 0 and i < n and j < m and not visit[i][j] and maze[i][j] == "."
def isExit(i, j):
return (i == 0 or i == n - 1 or j == 0 or j == m - 1) and not (entrance[0] == i and entrance[1] == j)
ans = [float('inf')]
def get_ans(i, j, step):
queue = [[i, j, step]]
while len(queue) != 0:
r_queue = []
while len(queue) != 0:
i, j, step = queue.pop()
if isExit(i, j):
ans[0] = min(ans[0], step)
d = [[1, 0], [-1, 0], [0, 1], [0, -1]]
for dx, dy in d:
n_i, n_j = i + dx, j + dy
if isValid(n_i, n_j):
visit[n_i][n_j] = True
r_queue.append([n_i, n_j, step + 1])
queue = r_queue
get_ans(entrance[0], entrance[1], 0)
return ans[0] if ans[0] != float('inf') else -1 | nearest-exit-from-entrance-in-maze | Python || Easy Solution || O(N) || BFS | Rahul_Kantwa | 0 | 8 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,186 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834946/Typical-BFS-in-Python3 | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m, n = len(maze), len(maze[0])
exits = set()
# add all the exits from the matrix
for i in range(m):
for j in range(n):
if ((i == 0 or i == m - 1) or (j == 0 or j == n - 1)) and [i, j] != entrance and maze[i][j] == ".":
exits.add((i, j))
if not exits:
return -1
q = []
visited = set()
q.append(entrance + [0])
while q:
i, j, distance = q.pop(0)
if (i, j) in exits:
return distance
for di, dj in ((0, 1), (0, -1), (1, 0), (-1, 0)):
ni, nj = i + di, j + dj
if 0 <= ni < m and 0 <= nj < n and maze[ni][nj] != "+" and (ni, nj) not in visited:
q.append([ni, nj, distance + 1])
visited.add((ni, nj))
return -1 | nearest-exit-from-entrance-in-maze | Typical BFS in Python3 | hahashabi | 0 | 9 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,187 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834811/python3-Shortest-path-Dijkstra-sol-for-reference | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
R = len(maze)
C = len(maze[0])
dp = [[float('inf') for _ in range(C)] for _ in range(R)]
st = [(0, tuple(entrance))]
visited = defaultdict(bool)
visited[tuple(entrance)] = True
while st:
dist, point = heapq.heappop(st)
x,y = point
for dx,dy in [(-1,0),(1,0),(0,-1),(0,1)]:
nx = x + dx
ny = y + dy
if nx >= 0 and ny >= 0 and nx < R and ny < C and maze[nx][ny] == "." and not visited[(nx,ny)]:
visited[(nx,ny)] = True
dp[nx][ny] = min(dp[nx][ny], dist + 1)
heapq.heappush(st, (dp[nx][ny], (nx, ny)))
ans = min(min(dp[0]), min(dp[-1]), min([dp[i][0] for i in range(R)]), min([dp[i][-1] for i in range(R)]))
return ans if ans != float('inf') else -1 | nearest-exit-from-entrance-in-maze | [python3] Shortest path Dijkstra sol for reference | vadhri_venkat | 0 | 6 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,188 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834805/Simple-bfs-using-Python3.-Very-easy-to-understand. | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
def helper(r,c):
queue = collections.deque()
queue.append((r,c,0))
visited = set([(r,c)])
while queue:
r, c, dist = queue.popleft()
if ([r, c] != entrance) and (r == 0 or r == len(maze)-1 or c == 0 or c == len(maze[0]) - 1): #found exit on the edge
return dist
else:
for r1, c1 in [(r, c+1), (r, c-1), (r-1, c), (r+1, c)]:
#for all valid neighbors that are in bounds and don't have a wall, add them to the queue
if r1 < len(maze) and r1 >= 0 and c1 < len(maze[0]) and c1 >= 0 and (r1, c1) not in visited and maze[r1][c1] == '.':
queue.append((r1,c1,dist+1))
visited.add((r1,c1))
return -1
d = helper(entrance[0],entrance[1])
return d | nearest-exit-from-entrance-in-maze | Simple bfs using Python3. Very easy to understand. | dkashi | 0 | 7 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,189 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834770/Python3-Solution-with-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows=len(maze)
cols=len(maze[0])
sx,sy=entrance
maze[sx][sy]="s"
for x in range(rows):
if maze[x][0]==".":
maze[x][0]="e"
if maze[x][-1]==".":
maze[x][-1]="e"
for y in range(cols):
if maze[0][y]==".":
maze[0][y]="e"
if maze[-1][y]==".":
maze[-1][y]="e"
directions=[(0,1),(1,0),(0,-1),(-1,0)]
done=[[False]*cols for _ in range(rows)]
queue=collections.deque()
queue.append((0,sx,sy))
done[sx][sy]=True
while len(queue)>0:
d,x,y=queue.popleft()
for dx,dy in directions:
nx,ny=x+dx,y+dy
if 0<=nx<rows and 0<=ny<cols and not done[nx][ny]:
if maze[nx][ny]==".":
done[nx][ny]=True
queue.append((d+1,nx,ny))
elif maze[nx][ny]=="e":
return d+1
return -1 | nearest-exit-from-entrance-in-maze | Python3 Solution with BFS | Motaharozzaman1996 | 0 | 6 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,190 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834761/BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m, n = len(maze), len(maze[0])
visited = set() # 记录去过的格子
def next_neighbor(x, y): # 邻居生成器
for xx, yy in ((x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)):
if 0 <= xx < m and 0 <= yy < n:
yield xx, yy
maze[entrance[0]][entrance[1]] = '+' # 把entrance标记成墙壁避免一开始就从entrance溜出去
queue = [tuple(entrance)] # 当前边界就是entrance一格
visited.add(tuple(entrance))
count = 0
while queue: # 只要还有边界可探就继续
temp_queue = [] # 记录新边界
for x, y in queue:
if (x == 0 or x == m - 1 or y == 0 or y == n - 1) and maze[x][y] == '.': # 判断出口
return count
for xx, yy in next_neighbor(x, y): # 遍历边界格子的所有邻居
if maze[xx][yy] == '.' and (xx, yy) not in visited: # 可以走而且没走过的格子就是新边界
temp_queue.append((xx, yy))
visited.add((xx, yy))
queue = temp_queue # 更新边界
count += 1 # 记录回合数
return -1 # 走投无路GG | nearest-exit-from-entrance-in-maze | BFS 广度优先搜索习题 | nathentasty | 0 | 3 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,191 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834664/Simple-Python-BFS | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
queue = [(entrance, 0),]
maze[entrance[0]][entrance[1]] = '+'
while(not len(queue) == 0):
(current, length) = queue.pop(0)
if(current[1]+1 < len(maze[0])):
if(maze[current[0]][current[1]+1] == '.'):
queue.append(([current[0],current[1]+1], length+1))
maze[current[0]][current[1]+1] = '+'
else:
if(length > 0):
return length
if(current[1]-1 >= 0):
if(maze[current[0]][current[1]-1] == '.'):
queue.append(([current[0],current[1]-1], length+1))
maze[current[0]][current[1]-1] = '+'
else:
if(length > 0):
return length
if(current[0]+1 < len(maze)):
if(maze[current[0]+1][current[1]] == '.'):
queue.append(([current[0]+1,current[1]], length+1))
maze[current[0]+1][current[1]] = '+'
else:
if(length > 0):
return length
if(current[0]-1 >= 0):
if(maze[current[0]-1][current[1]] == '.'):
queue.append(([current[0]-1,current[1]], length+1))
maze[current[0]-1][current[1]] = '+'
else:
if(length > 0):
return length
return -1 | nearest-exit-from-entrance-in-maze | Simple Python BFS | denfedex | 0 | 3 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,192 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834652/Python-BFS-approach-O(n)-time-O(n)-memory | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
steps = 0
n = len(maze)
m = len(maze[0])
i0, j0 = entrance[0], entrance[1]
maze[i0][j0] = '+'
level = [[i0, j0]]
neighbors = [[1, 0], [-1, 0], [0, 1], [0, -1]]
while level:
next_level = []
steps += 1
for i, j in level:
for di, dj in neighbors:
i1, j1 = i + di, j + dj
if 0 <= i1 < n and 0 <= j1 < m:
if maze[i1][j1] == '.':
if i1 == 0 or i1 == n-1 or j1 ==0 or j1 == m-1:
return steps
maze[i1][j1] = '+'
next_level.append([i1, j1])
level = next_level
return -1 | nearest-exit-from-entrance-in-maze | [Python] BFS approach O(n) time, O(n) memory | wtain | 0 | 5 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,193 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834623/Python3-BFS-%2B-Visted-Set-With-Helper-Function | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
visited, visit = set([(entrance[0], entrance[1])]), [(0,entrance[0], entrance[1])]
def if_good_add(c, i, j):
if (i,j) in visited or i<0 or j<0 or i>=len(maze) or j>=len(maze[0]) or maze[i][j] == '+':
return False
visit.append((c, i, j))
visited.add((i,j))
if i == 0 or i == len(maze) - 1 or j == 0 or j == len(maze[0]) - 1:
return True
while visit:
c, i, j = visit.pop(0)
for [id, jd] in [[1,0], [-1,0], [0,1], [0,-1]]:
if if_good_add(c+1, i + id, j + jd):
return c+1
return -1 | nearest-exit-from-entrance-in-maze | Python3 BFS + Visted Set With Helper Function | godshiva | 0 | 1 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,194 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2834598/Python-or-BFS-search | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows, cols = len(maze), len(maze[0])
bfsQue = [[entrance, 0]]
maze[entrance[0]][entrance[1]] = '+'
movements = [[-1, 0], [1, 0], [0, -1], [0, 1]]
while(bfsQue):
top, step = bfsQue[0]
s1 = step + 1
for m in movements:
nr, nc = top[0] + m[0], top[1] + m[1]
if(nr < 0 or nc < 0 or nr >= rows or nc >= cols or maze[nr][nc] != '.'):
continue
if(nr == rows - 1 or nr == 0 or nc == cols - 1 or nc == 0):
return s1
maze[nr][nc] = '+'
bfsQue.append([[nr, nc], s1])
bfsQue.pop(0)
return -1 | nearest-exit-from-entrance-in-maze | Python | BFS search | CosmosYu | 0 | 2 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,195 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2614850/Python-Readable-BFS-(and-explanation-why-not-DFS) | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
# make bfs as we want to search shortest path
queue = collections.deque()
queue.append((entrance[0], entrance[1], 0))
# get the matrix sizes
m = len(maze) - 1
n = len(maze[0]) - 1
# check for empty queue
while queue:
# pop the position
rx, cx, steps = queue.popleft()
# check if we hit a wall (or we already visited)
if maze[rx][cx] == '+':
continue
# tell that we visited (build a wall!)
# as we do bfs we can be shure we are on the
# shortest path
maze[rx][cx] = "+"
# check if it is exit
if rx == 0 or rx == m or cx == 0 or cx == n:
# check that it is something else than entrace
if not (rx == entrance[0] and cx == entrance[1]):
# return our shortest path
return steps
# go further in all directions (and check that we stay in bounds)
if rx < m:
queue.append((rx+1, cx, steps+1))
if rx > 0:
queue.append((rx-1, cx, steps+1))
if cx < n:
queue.append((rx, cx+1, steps+1))
if cx > 0:
queue.append((rx, cx-1, steps+1))
# return the minimum
return -1 | nearest-exit-from-entrance-in-maze | [Python] - Readable BFS (and explanation why not DFS) | Lucew | 0 | 19 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,196 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/2034044/Python-simple-and-efficient-BFS-solution | class Solution:
def nearestExit(self, grid: List[List[str]], entrance: List[int]) -> int:
q = deque()
q.append((entrance[0], entrance[1], 0))
seen = set()
while q :
i, j, d = q.popleft()
if (i == 0 or j == 0 or i == len(grid) - 1 or j == len(grid[0]) - 1) and [i, j] != entrance :
return d
for x, y in ((i-1, j), (i, j-1), (i+1, j), (i, j+1)) :
if 0 <= x < len(grid) and 0 <= y < len(grid[0]) :
if (x, y) not in seen and grid[x][y] == "." :
seen.add((x, y))
q.append((x, y, d+1))
# print(q)
return -1 | nearest-exit-from-entrance-in-maze | Python simple and efficient BFS solution | runtime-terror | 0 | 35 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,197 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/1888635/Python-BFS-solution | class Solution:
def nearestExit(self, A: List[List[str]], st: List[int]) -> int:
bounds = {(st[0], st[1])} # We will store our latest checked cells here
A[st[0]][st[1]] = '1' # We are going to change values of checked cells
m, n = len(A), len(A[0])
step = 1
while bounds:
new = []
for i,j in bounds:
for d in [(1,0),(-1,0),(0,1),(0,-1)]: # Check all the adjacent cells
x,y = i+d[0], j+d[1]
if 0 <= x < m and 0 <= y < n: # If the current cell is inside the borders
if A[x][y] == '.': # ... and if it is not the wall
if x in [0, m-1] or y in [0, n-1]: # check if it's an exit
return step
A[x][y] = '1' # ...if no, add it to list of our latest checked cells
new.append((x,y))
step += 1
bounds = new
return -1 | nearest-exit-from-entrance-in-maze | Python BFS solution | markkorvin | 0 | 48 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,198 |
https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/discuss/1823249/Python-or-BFS-Solution | class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows = len(maze)
cols = len(maze[0])
sr, sc = entrance #starting row, starting column
visited = {(sr,sc)}
q = deque([(sr,sc,0)]) # 0 = initial distance
while q:
x, y, d = q.popleft() # d = distance
for dx, dy in [(x+1,y),(x-1,y),(x,y+1),(x,y-1)]:
if 0 <= dx < rows and 0 <= dy < cols and (dx,dy) not in visited and maze[dx][dy] == ".":
visited.add((dx,dy))
q.append((dx,dy,d+1))
if dx == 0 or dx == rows - 1 or dy == 0 or dy == cols - 1: # check if reached the border
return d + 1
return -1 | nearest-exit-from-entrance-in-maze | Python | BFS Solution | jgroszew | 0 | 42 | nearest exit from entrance in maze | 1,926 | 0.49 | Medium | 27,199 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.