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/find-n-unique-integers-sum-up-to-zero/discuss/2287459/Python-fastest-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
if n==1:
return [0]
if n==2:
return[-1,1]
x=[]
for i in range(n-1):
x.append(i)
x.append(-sum(x))
return x | find-n-unique-integers-sum-up-to-zero | Python fastest solution | yagnic40 | 1 | 57 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,400 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2005000/easy-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
a=[]
if(n%2==0):
pass
else:
a.append(0)
for i in range(1,n//2+1):
a.append(i)
a.append(-i)
return a | find-n-unique-integers-sum-up-to-zero | easy solution | Durgavamsi | 1 | 59 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,401 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1150132/Python3-Simple-Solution-in-O(n)-runtime-Runtime-beats-93-of-Python3-submissions | class Solution:
def sumZero(self, n: int) -> List[int]:
if(n % 2 == 1): arr = [0]
else: arr = []
for i in range(1 , n // 2 + 1):
arr.extend([i , -i])
return arr | find-n-unique-integers-sum-up-to-zero | [Python3] Simple Solution in O(n) runtime , Runtime beats 93% of Python3 submissions | Lolopola | 1 | 69 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,402 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1096029/Easy-Python3-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
num = 1
if n == 1:
ans.append(0)
return ans
if n % 2 != 0:
ans.append(0)
length = n // 2
for i in range(0, length):
ans.append(num)
ans.append(num * (-1))
num += 1
return ans | find-n-unique-integers-sum-up-to-zero | Easy Python3 Solution | nematov_olimjon | 1 | 116 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,403 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/463935/Python3-one-liner | class Solution:
def sumZero(self, n: int) -> List[int]:
return list(range(n-1)) + [-(n-1)*(n-2)//2] | find-n-unique-integers-sum-up-to-zero | [Python3] one-liner | ye15 | 1 | 77 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,404 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2844203/Simple-Python-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
answer = []
if n % 2 != 0: # Number is odd, append 0 to answer
answer.append(0)
for num in range(1, (n // 2) + 1):
answer.append(num)
answer.append(-num)
return answer | find-n-unique-integers-sum-up-to-zero | Simple Python Solution | corylynn | 0 | 2 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,405 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2800532/Python-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
l=[]
if(n%2==0):
for i in range(1,(n//2)+1):
l.append(i)
l.append(-i)
else:
l.append(0)
for i in range(1,(n//2)+1):
l.append(i)
l.append(-i)
return l | find-n-unique-integers-sum-up-to-zero | Python Solution | CEOSRICHARAN | 0 | 5 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,406 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2796581/python-super-easy | class Solution:
def sumZero(self, n: int) -> List[int]:
i = 1
arr = []
while n > 1:
arr.append(-i)
arr.append(i)
i+=1
n-=2
if n == 1:
arr.append(0)
return arr | find-n-unique-integers-sum-up-to-zero | python super easy | harrychen1995 | 0 | 5 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,407 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2755960/Python3-Solution-oror-Symmetric-Filling | class Solution:
def sumZero(self, n: int) -> List[int]:
rang, remind = n//2, n%2
if remind == 0:
new = []
else:
new = [0]
for i in range(1, rang+1):
new.append(-i)
new.append(i)
return new | find-n-unique-integers-sum-up-to-zero | Python3 Solution || Symmetric Filling | shashank_shashi | 0 | 3 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,408 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2750430/Python3-One-Line-Fast-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
return [i for i in range (1,n//2+1)]+[-i for i in range (1,n//2+1)]+([0] if n%2!=0 else []) | find-n-unique-integers-sum-up-to-zero | [Python3] One-Line Fast Solution | keioon | 0 | 6 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,409 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2708793/easy-Python | class Solution:
def sumZero(self, n: int) -> List[int]:
return list(range(1,n//2+1))+[0]*(n&1)+list(range(-(n//2),0)) | find-n-unique-integers-sum-up-to-zero | easy Python | MaryLuz | 0 | 3 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,410 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2705100/Python-simple-approach | class Solution:
def sumZero(self, n: int) -> List[int]:
l1 = []
if n%2 == 1:
l1.append(0)
for i in range(n//2):
l1.extend([i+1, -(i+1)])
return l1 | find-n-unique-integers-sum-up-to-zero | Python simple approach | swiftv99 | 0 | 5 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,411 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2703988/Python-naive-brute-forse | class Solution:
def sumZero(self, n: int) -> List[int]:
if n==1: return [0]
if n%2==0:
out=[]
else:
out=[0]
for i in range(1,n//2+1):
out.extend([i,-i])
return out | find-n-unique-integers-sum-up-to-zero | Python, naive brute forse | Leox2022 | 0 | 1 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,412 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2692924/Easy-and-simple-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
res=[]
if(n%2==1):
res.append(0)
for i in range(1,n//2+1):
res.append(i)
res.append(-i)
return res | find-n-unique-integers-sum-up-to-zero | Easy and simple solution | Raghunath_Reddy | 0 | 7 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,413 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2660724/Python-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
res = [0] if n % 2 == 1 else []
cur = 1
while len(res) < n:
res.append(cur)
res.append(-1*cur)
cur += 1
return res | find-n-unique-integers-sum-up-to-zero | Python solution | Mark5013 | 0 | 31 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,414 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2622651/SIMPLE-PYTHON3-SOLUTION-easy-approach-explained | class Solution:
def sumZero(self, n: int) -> List[int]:
if n == 1:
return [0]
res = [0 for x in range(n)] # fill array with 0
c = 1
# Fill array with disctinc numbers in the following pattern . [-1,-2,0, 2,1]
left, right = 0, len(res)-1
while left < right:
res[left] = c * -1
res[right] = c
c +=1
left +=1
right -=1
if len(res) % 2 != 0:
mid = len(res) // 2
res[mid] = 0
return res | find-n-unique-integers-sum-up-to-zero | ✅✔ SIMPLE PYTHON3 SOLUTION ✅✔ easy approach explained | rajukommula | 0 | 44 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,415 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2601420/Unique-Integer-sum-to-0-oror-Python-oror-Easy-and-simple-approach | class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
limit = int(n/2)
for i in range(1, limit+1):
ans.append(-i)
ans.append(i)
if(n%2 == 1):
ans.append(0)
return ans | find-n-unique-integers-sum-up-to-zero | Unique Integer sum to 0 || Python || Easy and simple approach | vanshika_2507 | 0 | 13 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,416 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2550244/Python3-or-Two-Pointers | class Solution:
def sumZero(self, n: int) -> List[int]:
l,r=1,-1
ans=[]
while l<=n//2:
ans.append(l)
ans.append(r)
l+=1
r-=1
if n%2!=0:
ans.append(0)
return ans | find-n-unique-integers-sum-up-to-zero | [Python3] | Two Pointers | swapnilsingh421 | 0 | 18 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,417 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2426830/Python-easy-O(n)-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
result = []
for num in range(n//2):
result.append(num + 1)
result.append(-num - 1)
if n % 2 != 0:
result.append(0)
return result | find-n-unique-integers-sum-up-to-zero | Python easy O(n) solution | samanehghafouri | 0 | 33 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,418 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2092131/PYTHON-or-Super-simple-python-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
sum = 0
res = []
for i in range(1, n):
res.append(i)
sum += i
res.append(-sum)
return res | find-n-unique-integers-sum-up-to-zero | PYTHON | Super simple python solution | shreeruparel | 0 | 129 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,419 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/2080772/O(N)-simple-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
res = []
for i in range(1, n // 2 + 1):
res.extend([i, -i])
if n % 2:
res.append(0)
return res | find-n-unique-integers-sum-up-to-zero | O(N) simple solution | andrewnerdimo | 0 | 76 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,420 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1954197/Python-Easy-And-Fast-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
result = []
for i in range(1, int(abs(n/2)) + 1):
result.extend([i, -i])
if len(result) != n:
result.append(0)
return result | find-n-unique-integers-sum-up-to-zero | Python Easy And Fast Solution | hardik097 | 0 | 52 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,421 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1927896/easy-python-code | class Solution:
def sumZero(self, n: int) -> List[int]:
output = []
if n%2 == 0:
for i in range(1,(n//2)+1):
output.append(i)
output.append(i-2*i)
else:
output.append(0)
for i in range(1,(n//2)+1):
output.append(i)
output.append(i-2*i)
return output | find-n-unique-integers-sum-up-to-zero | easy python code | dakash682 | 0 | 61 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,422 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1887183/Python-solution-using-consecutive-integers | class Solution:
def sumZero(self, n: int) -> List[int]:
res = []
if n % 2 == 0:
res.extend([x for x in range(1, (n // 2) + 1)])
res.extend([y for y in range(-(n // 2), 0, 1)])
return sorted(res)
res.append(0)
res.extend([x for x in range(1, (n // 2) + 1)])
res.extend([y for y in range(-(n // 2), 0, 1)])
return sorted(res) | find-n-unique-integers-sum-up-to-zero | Python solution using consecutive integers | alishak1999 | 0 | 91 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,423 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1873738/python-simpe-short-onliner-or-rangeorwell-explainedorbruteforce-and-best-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
k=[i+1 for i in range(n//2)]
k+=[-(i+1) for i in range(n//2)]
return k if n%2==0 else k+[0] | find-n-unique-integers-sum-up-to-zero | python simpe short onliner | range|well explained|bruteforce and best solution | YaBhiThikHai | 0 | 63 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,424 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1873738/python-simpe-short-onliner-or-rangeorwell-explainedorbruteforce-and-best-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
return range(1-n,n,2)
# intution
# n=0 [0]
# n=1 [-1,1] obeservation: n[-1]-n[0]=2=step
# n=3 [-2,0,2]
# n=4 [-3,-1,1,3]
# n=5 [-4,-2,0,2,4] note: start=1-n and end=n-1 | find-n-unique-integers-sum-up-to-zero | python simpe short onliner | range|well explained|bruteforce and best solution | YaBhiThikHai | 0 | 63 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,425 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1856397/Python-Solution | class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
counter = 1
if n % 2 == 1:
ans.append(0)
n -= 1
times = n // 2
for i in range(times):
ans.append(-1 * counter)
ans.append(counter)
counter += 1
return ans | find-n-unique-integers-sum-up-to-zero | Python Solution | DietCoke777 | 0 | 70 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,426 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1849902/Python-One-Line!-Simple-and-Elegant | class Solution(object):
def sumZero(self, n):
return range(-n//2+n%2,0) + range(0,n%2) + range(1,n//2+1) | find-n-unique-integers-sum-up-to-zero | Python - One Line! Simple and Elegant | domthedeveloper | 0 | 61 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,427 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1809367/3-Lines-Python-Solution-oror-94-Faster(32ms)-oror-Memory-less-than-80 | class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
for i in range(1,n//2+1): ans.extend({i,-i})
return ans + [0]*(n%2) | find-n-unique-integers-sum-up-to-zero | 3-Lines Python Solution || 94% Faster(32ms) || Memory less than 80% | Taha-C | 0 | 68 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,428 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1754946/Faster-than-90-Python | class Solution:
def sumZero(self, n: int) -> List[int]:
ans = []
if n % 2 == 0:
b = int(-n/2)
for i in range(0,n):
if i == n/2:
b = 1
ans.append(b)
b += 1
return ans
b = int((n//2)*-1)
for i in range(0,n):
ans.append(b)
b += 1
return ans | find-n-unique-integers-sum-up-to-zero | Faster than 90% Python | ra6115 | 0 | 97 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,429 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1730008/Python-dollarolution | class Solution:
def sumZero(self, n: int) -> List[int]:
l, count = [], 0
if n % 2 == 0:
for i in range(1,n//2+1):
l += [i,-i]
else:
for i in range(n-1):
l.append(i)
count += i
l.append(-count)
return l | find-n-unique-integers-sum-up-to-zero | Python $olution | AakRay | 0 | 72 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,430 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1687269/Trivial-O(n)-python | class Solution:
def sumZero(self, n: int) -> List[int]:
arr = []
for i in range(n//2):
arr.append(i + 1)
arr.append(-1 * (i + 1))
if len(arr) < n:
arr.append(0)
return arr | find-n-unique-integers-sum-up-to-zero | Trivial O(n) python | snagsbybalin | 0 | 35 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,431 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1290501/Python-3-%3A-Simple-and-easy-to-understand | class Solution:
def sumZero(self, n: int) -> List[int]:
if n % 2 == 0 :
return [ i for i in range(-(n//2),(n//2)+1) if i != 0 ]
else :
return [ i for i in range(-(n//2),(n//2)+1) ] | find-n-unique-integers-sum-up-to-zero | Python 3 : Simple and easy to understand | rohitkhairnar | 0 | 248 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,432 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1230078/Python | class Solution:
def sumZero(self, n: int) -> List[int]:
result = []
for i in range(1, (n // 2)+1):
result += [i, -i]
if len(result) < n: result += [0]
return result | find-n-unique-integers-sum-up-to-zero | Python | dev-josh | 0 | 208 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,433 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1167891/Python-pythonic | class Solution:
def sumZero(self, n: int) -> List[int]:
result = [x for i in range(1, n//2+1) for x in {i, -i}]
if n % 2:
result.append(0)
return result | find-n-unique-integers-sum-up-to-zero | [Python] pythonic | cruim | 0 | 98 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,434 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1159253/simple-to-understand | class Solution:
def sumZero(self, n: int) -> List[int]:
res=[]
i,j=0,0
if n%2!=0:
res.append(0)
while n-1 :
res.append(i+1)
res.append(j-1)
n-=2
i+=1
j-=1
else:
while n:
res.append(i+1)
res.append(j-1)
n-=2
i+=1
j-=1
res.sort()
return(res) | find-n-unique-integers-sum-up-to-zero | simple to understand | janhaviborde23 | 0 | 78 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,435 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/1027862/Python3-simple-solution | class Solution:
def sumZero(self, n: int) -> List[int]:
l = []
if n % 2 == 0:
for i in range(-(n//2),n//2+1,1):
l.append(i)
l.remove(0)
return l
else:
for i in range(-(n//2),n//2+1,1):
l.append(i)
return l | find-n-unique-integers-sum-up-to-zero | Python3 simple solution | EklavyaJoshi | 0 | 113 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,436 |
https://leetcode.com/problems/find-n-unique-integers-sum-up-to-zero/discuss/909368/Easiest-solution-Python-O(N)-100-space | class Solution:
def sumZero(self, n: int) -> List[int]:
l = []
if n % 2 == 0:
for i in range(1, n//2 + 1):
l.append(i * 2)
l.append(-i * 2)
else:
for i in range(-n//2 + 1, n//2 + 1):
l.append(i)
return l | find-n-unique-integers-sum-up-to-zero | Easiest solution Python O(N), 100% space | vanigupta20024 | 0 | 179 | find n unique integers sum up to zero | 1,304 | 0.771 | Easy | 19,437 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/523589/python-only-2-lines-easy-to-read-with-explanation.-Can-it-be-any-shorter | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
h = lambda r: h(r.left) + [r.val] + h(r.right) if r else []
return sorted( h(root1) + h(root2) ) | all-elements-in-two-binary-search-trees | python, only 2 lines, easy to read, with explanation. Can it be any shorter? | rmoskalenko | 2 | 155 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,438 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/708517/Easy-Python-O(n)-Merge-Sorted-Lists | class Solution:
# Time: O(n)
# Space: O(n)
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
l1 = self.dfs(root1, [])
l2 = self.dfs(root2, [])
res = []
while l1 or l2:
if not l1:
res.append(l2.pop(0))
elif not l2:
res.append(l1.pop(0))
else:
res.append(l1.pop(0) if l1[0] < l2[0] else l2.pop(0))
return res
def dfs(self, node, vals):
if not node:
return
self.dfs(node.left, vals)
vals.append(node.val)
self.dfs(node.right, vals)
return vals | all-elements-in-two-binary-search-trees | Easy Python O(n) Merge Sorted Lists | whissely | 1 | 162 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,439 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/464010/Python-3-(DFS)-(beats-100)-(eight-lines) | class Solution:
def getAllElements(self, R1: TreeNode, R2: TreeNode) -> List[int]:
A = []
def dfs(T):
A.append(T.val)
if T.left != None: dfs(T.left)
if T.right != None: dfs(T.right)
if R1 != None: dfs(R1)
if R2 != None: dfs(R2)
return sorted(A)
- Junaid Mansuri
- Chicago, IL | all-elements-in-two-binary-search-trees | Python 3 (DFS) (beats 100%) (eight lines) | junaidmansuri | 1 | 323 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,440 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/2832705/Python3-Solution | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def bstToArrayInorder(root, arr):
if root is None:
return
bstToArrayInorder(root.left, arr)
arr.append(root.val)
bstToArrayInorder(root.right, arr)
return
arr = []
bstToArrayInorder(root1, arr)
bstToArrayInorder(root2, arr)
return sorted(arr) | all-elements-in-two-binary-search-trees | Python3 Solution | sipi09 | 0 | 2 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,441 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/2701853/This-solution-is-much-faster | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
result = []
def inorder(root: TreeNode):
if root:
inorder(root.left)
result.append(root.val)
inorder(root.right)
inorder(root1)
inorder(root2)
return sorted(result) | all-elements-in-two-binary-search-trees | This solution is much faster | namashin | 0 | 3 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,442 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/2701814/Python-and-Golang-step-by-step-Solution | class Solution:
def inorder_generator(self, root: TreeNode) -> Generator:
if root:
yield from self.inorder_generator(root.left)
yield root.val
yield from self.inorder_generator(root.right)
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
# Get root1 and root2 all elements
# by generator in order to save memory.
root1_list = [val for val in self.inorder_generator(root1)]
root2_list = [val for val in self.inorder_generator(root2)]
# Merge, Sort it and Return
return sorted(root1_list + root2_list) | all-elements-in-two-binary-search-trees | Python and Golang step by step Solution | namashin | 0 | 1 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,443 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1782710/Python3-solution-using-sort | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def dfs(root):
if not root:
return
dfs(root.left)
s.append(root.val)
dfs(root.right)
s=[]
dfs(root1)
dfs(root2)
return sorted(s) | all-elements-in-two-binary-search-trees | Python3 solution using sort | Karna61814 | 0 | 31 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,444 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1722548/python-easy-O(n%2Bm)-time-O(n%2Bm)-space-solution-using-inorder-traversal | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
res = []
def inorder(node):
if node:
inorder(node.left)
res.append(node.val)
inorder(node.right)
inorder(root1)
res.append(float('-inf'))
inorder(root2)
ans = []
n = len(res)
i, j = 0, res.index(float('-inf'))+1
k = res.index(float('-inf'))
while i < k and j < n:
if i < k and j < n and res[i] == res[j]:
ans.append(res[i])
ans.append(res[i])
i += 1
j += 1
elif i < k and j< n and res[i] < res[j]:
ans.append(res[i])
i += 1
elif i< k and j< n and res[i] > res[j]: # res[i] > res[j]
ans.append(res[j])
j += 1
if i < k:
while i < k:
ans.append(res[i])
i +=1
elif j < n:
while j < n:
ans.append(res[j])
j += 1
return ans | all-elements-in-two-binary-search-trees | python easy O(n+m) time, O(n+m) space solution using inorder traversal | byuns9334 | 0 | 23 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,445 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1722061/Easy-Python-using-Inorder-Traversal-or-Faster-than-100 | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
result = []
self.helper(root1, result)
self.helper(root2, result)
return sorted(result)
def helper(self, root, result):
if not root:
return
if root.left:
self.helper(root.left, result)
result.append(root.val)
if root.right:
self.helper(root.right, result) | all-elements-in-two-binary-search-trees | Easy Python using Inorder Traversal | Faster than 100% | Fyzzys | 0 | 14 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,446 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1721958/Python-orRecursion-or-Simple-Solution-or-TreeToList-O(n)or-Beginner | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def treeToList(root):
if not root:
return []
return treeToList(root.left)+[root.val]+treeToList(root.right)
def mergeLists(l1, l2):
p1 = 0
p2 = 0
res = list()
while(p1 < len(l1) and p2 < len(l2)):
if l1[p1] > l2[p2]:
res.append(l2[p2])
p2 += 1
else:
res.append(l1[p1])
p1 +=1
while(p1 < len(l1)):
res.append(l1[p1])
p1 +=1
while(p2 < len(l2)):
res.append(l2[p2])
p2 +=1
return res
t1 = treeToList(root1)
t2 = treeToList(root2)
return mergeLists(t1,t2) | all-elements-in-two-binary-search-trees | Python |Recursion | Simple Solution | TreeToList O(n)| Beginner | letyrodri | 0 | 17 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,447 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1721869/python-with-preorder-traverse-method-(faster-than-68.01) | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def preorder_(root, node_list):
if not root:
return
node_list.append(root.val)
preorder_(root.left, node_list)
preorder_(root.right, node_list)
ans = []
preorder_(root1, ans)
preorder_(root2, ans)
return sorted(ans)
``` | all-elements-in-two-binary-search-trees | python with preorder traverse method (faster than 68.01%) | ctw01 | 0 | 10 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,448 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1721466/Python-3-Just-flatten-the-trees-into-lists-and-sort-them-(312ms-18MB). | class Solution:
def flatten_tree(self, root: TreeNode) -> List[int]:
result = []
trees = [root]
while trees:
tree_copies = trees[:]
trees = []
for subtree in tree_copies:
if subtree:
result.append(subtree.val)
trees.append(subtree.left)
trees.append(subtree.right)
if not any(trees):
break
return result
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
list1 = self.flatten_tree(root1)
list2 = self.flatten_tree(root2)
return sorted(list1 + list2) | all-elements-in-two-binary-search-trees | [Python 3] Just flatten the trees into lists and sort them (312ms, 18MB). | seankala | 0 | 15 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,449 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720820/Python3-oror-Recursive-Solution | class Solution:
def getList(self, root,lst):
if root:
lst.append(root.val)
self.getList(root.left,lst)
self.getList(root.right,lst)
return lst
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
lst=self.getList(root1,[])
lst=self.getList(root2,lst)
return sorted(lst) | all-elements-in-two-binary-search-trees | [ ✅ ✅ ✅ Python3 || Recursive Solution ] 🆗 🆗 🆗 | alchimie54 | 0 | 16 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,450 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720684/Python-Simple-Python-Solution-Using-Preorder-Traversal-and-Sorting | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
ans=[]
def PrintTree(node):
if node != None:
ans.append(node.val)
PrintTree(node.left)
PrintTree(node.right)
PrintTree(root1)
PrintTree(root2)
ans=sorted(ans)
return ans | all-elements-in-two-binary-search-trees | [ Python ] ✔✔ Simple Python Solution Using Preorder Traversal and Sorting | ASHOK_KUMAR_MEGHVANSHI | 0 | 19 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,451 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720407/Python-3-Solution-Linear-Time-Recursive-and-Iterative-Inorder-Traversal | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
list1, list2 = [], []
self.inOrder(root1, list1)
self.inOrder(root2, list2)
return self.mergeSortedList(list1, list2)
def inOrder(self, node, res):
if node is None: return
self.preOrder(node.left, res)
res.append(node.val)
self.preOrder(node.right, res)
def mergeSortedList(self, list1, list2):
i, j = 0, 0
res = []
while i < len(list1) and j < len(list2):
if list1[i] <= list2[j]:
res.append(list1[i])
i += 1
else:
res.append(list2[j])
j += 1
while i < len(list1):
res.append(list1[i])
i += 1
while j < len(list2):
res.append(list2[j])
j += 1
return res | all-elements-in-two-binary-search-trees | [Python 3] Solution Linear Time Recursive & Iterative Inorder Traversal | gcobs0834 | 0 | 15 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,452 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720407/Python-3-Solution-Linear-Time-Recursive-and-Iterative-Inorder-Traversal | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
stack1, stack2, res = [], [], []
tree1Node = self.getNextNode(stack1, root1)
tree2Node = self.getNextNode(stack2, root2)
while tree1Node and tree2Node and len(stack1) and len(stack2):
if tree1Node.val <= tree2Node.val:
stack1.pop()
res.append(tree1Node.val)
tree1Node = self.getNextNode(stack1, tree1Node.right)
else:
stack2.pop()
res.append(tree2Node.val)
tree2Node = self.getNextNode(stack2, tree2Node.right)
while tree1Node and len(stack1):
stack1.pop()
res.append(tree1Node.val)
tree1Node = self.getNextNode(stack1, tree1Node.right)
while tree2Node and len(stack2):
stack2.pop()
res.append(tree2Node.val)
tree2Node = self.getNextNode(stack2, tree2Node.right)
return res
def getNextNode(self, stack, node):
while node is not None:
stack.append(node)
node = node.left
return stack[-1] if stack else None | all-elements-in-two-binary-search-trees | [Python 3] Solution Linear Time Recursive & Iterative Inorder Traversal | gcobs0834 | 0 | 15 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,453 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720319/Python-3-EASY-Intuitive-Solution | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
res = []
def BST_to_list(root: TreeNode) -> None:
nonlocal res
if root:
res.append(root.val)
BST_to_list(root.left)
BST_to_list(root.right)
BST_to_list(root1)
BST_to_list(root2)
res.sort()
return res | all-elements-in-two-binary-search-trees | ✅ [Python 3] EASY Intuitive Solution | JawadNoor | 0 | 10 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,454 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720259/Python-3-simple-dfs | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def dfs(root):
if root is None:
return []
return dfs(root.left) + [root.val] + dfs(root.right)
list1, list2 = dfs(root1), dfs(root2)
n1, n2 = len(list1), len(list2)
i = j = 0
res = []
while i < n1 and j < n2:
if list1[i] <= list2[j]:
res.append(list1[i])
i += 1
else:
res.append(list2[j])
j += 1
for k in range(i, n1):
res.append(list1[k])
for k in range(j, n2):
res.append(list2[k])
return res | all-elements-in-two-binary-search-trees | Python 3 simple dfs | dereky4 | 0 | 35 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,455 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720152/Python-O(n)-Solution | class Solution:
def inorder(self, node, arr):
if node:
self.inorder(node.left, arr)
arr.append(node.val)
self.inorder(node.right, arr)
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
l = []
r = []
self.inorder(root1, l)
self.inorder(root2, r)
n = len(l)
m = len(r)
res = [0] * (m + n)
i, j, k = n - 1, m - 1, m + n - 1
while i >= 0 and j >= 0:
if l[i] > r[j]:
res[k] = l[i]
i -= 1
else:
res[k] = r[j]
j -= 1
k -= 1
while i >= 0:
res[k] = l[i]
i -= 1
k -= 1
while j >= 0:
res[k] = r[j]
j -= 1
k -= 1
return res | all-elements-in-two-binary-search-trees | Python O(n) Solution | imrajeshberwal | 0 | 50 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,456 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1720133/Python-Solution-DFS%2BSort | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
items = []
def dfs(root):
if root is None:
return
items.append(root.val)
dfs(root.left)
dfs(root.right)
dfs(root1)
dfs(root2)
items.sort()
return items | all-elements-in-two-binary-search-trees | Python Solution, DFS+Sort | pradeep288 | 0 | 30 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,457 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/830861/All-elements-in-2-BST%3A-naive-99-python3-brute-force-solution-(324ms) | class Solution:
def push(self, root, result):
if root.left: self.push(root.left, result)
result.append(root.val)
if root.right: self.push(root.right, result)
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
result = []
if root1: self.push(root1, result)
if root2: self.push(root2, result)
result.sort()
return result | all-elements-in-two-binary-search-trees | All elements in 2 BST: naive 99% python3 brute force solution (324ms) | leetavenger | 0 | 44 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,458 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/722507/python3Java-BFS | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
ans=[]
stack=[root1,root2]
while stack:
temp=stack.pop(0)
if not temp:continue
ans.append(temp.val)
if temp.left:stack.append(temp.left)
if temp.right:stack.append(temp.right)
return sorted(ans)
··· | all-elements-in-two-binary-search-trees | python3/Java BFS | 752937603 | 0 | 96 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,459 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/668088/Python-3.-All-Elements-in-Two-Binary-Trees.-Easy-DFS-Solution.-Beats-93. | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def traverse(node):
if not node:
return None
traverse(node.left)
listBoth.append(node.val)
traverse(node.right)
listBoth=[]
traverse(root1)
traverse(root2)
listBoth.sort()
return listBoth | all-elements-in-two-binary-search-trees | [Python 3]. All Elements in Two Binary Trees. Easy DFS Solution. Beats 93%. | tilak_ | 0 | 93 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,460 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/463908/Python3-inorder-traversal-and-merge | class Solution:
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def fn(node):
ans, stack = [], []
while node or stack:
if node:
stack.append(node)
node = node.left
else:
node = stack.pop()
ans.append(node.val)
node = node.right
return ans
vals1 = fn(root1)
vals2 = fn(root2)
ans = []
i = j = 0
while i < len(vals1) or j < len(vals2):
if j == len(vals2) or i < len(vals1) and vals1[i] < vals2[j]:
ans.append(vals1[i])
i += 1
else:
ans.append(vals2[j])
j += 1
return ans | all-elements-in-two-binary-search-trees | [Python3] inorder traversal & merge | ye15 | 0 | 80 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,461 |
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/discuss/1284721/Python-Faster-Than-92-Recursive-In-Order | class Solution:
def __init__(self):
self.l1 = []
self.l2 = []
def inOrder(self,root, s):
if root == None:
return
self.inOrder(root.left, s)
s.append(root.val)
self.inOrder(root.right, s)
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
self.inOrder(root1, self.l1)
self.inOrder(root2, self.l2)
l = self.l1 + self.l2
l.sort()
return l | all-elements-in-two-binary-search-trees | Python Faster Than 92% Recursive In Order | paramvs8 | -1 | 96 | all elements in two binary search trees | 1,305 | 0.798 | Medium | 19,462 |
https://leetcode.com/problems/jump-game-iii/discuss/571683/Python3-3-Lines-DFS.-O(N)-time-and-space.-Recursion | class Solution:
def canReach(self, arr: List[int], i: int) -> bool:
if i < 0 or i >= len(arr) or arr[i] < 0: return False
arr[i] *= -1 # Mark visited
return arr[i] == 0 or self.canReach(arr, i - arr[i]) or self.canReach(arr, i + arr[i]) | jump-game-iii | [Python3] 3 Lines DFS. O(N) time and space. Recursion | jimmyyentran | 5 | 245 | jump game iii | 1,306 | 0.631 | Medium | 19,463 |
https://leetcode.com/problems/jump-game-iii/discuss/2732996/Easy-python-solution-using-BFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
n=len(arr)
visited=[0]*n
lst=[start]
visited[start]=1
while lst:
x=lst.pop(0)
if arr[x]==0:
return True
if x+arr[x]<n and visited[x+arr[x]]==0:
lst.append(x+arr[x])
visited[x+arr[x]]=1
if x-arr[x]>=0 and visited[x-arr[x]]==0:
lst.append(x-arr[x])
visited[x-arr[x]]=1
return False | jump-game-iii | Easy python solution using BFS | beneath_ocean | 2 | 99 | jump game iii | 1,306 | 0.631 | Medium | 19,464 |
https://leetcode.com/problems/jump-game-iii/discuss/1968978/Python3-oror-BFS-oror-Faster-than-89.62 | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
vis = [0]*len(arr)
pos = [start]
while pos:
nextPos = []
while pos:
x = pos.pop()
if arr[x] == 0: return True
vis[x] = 1
for y in (x-arr[x], x+arr[x]):
if 0 <= y < len(arr) and not vis[y]: nextPos.append(y)
pos = nextPos
return False | jump-game-iii | Python3 || BFS || Faster than 89.62% | bvian | 2 | 58 | jump game iii | 1,306 | 0.631 | Medium | 19,465 |
https://leetcode.com/problems/jump-game-iii/discuss/1389143/Python-BFS-Approach | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
queue = [start]
visited = set()
while queue:
u = queue.pop(0)
if arr[u] == 0:
return True
visited.add(u)
nextjump = u + arr[u]
if nextjump < len(arr) and nextjump not in visited:
if arr[nextjump] == 0:
return True
visited.add(nextjump)
queue.append(nextjump)
nextjump = u - arr[u]
if nextjump >= 0 and nextjump not in visited:
if arr[nextjump] == 0:
return True
visited.add(nextjump)
queue.append(nextjump)
return False | jump-game-iii | [Python] BFS Approach | mizan-ali | 2 | 120 | jump game iii | 1,306 | 0.631 | Medium | 19,466 |
https://leetcode.com/problems/jump-game-iii/discuss/463924/Python-traversal-(DFS) | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
stack = [start]
arr[start] *= -1 # mark "visited"
while stack:
i = stack.pop()
if arr[i] == 0: return True
for ii in i - arr[i], i + arr[i]:
if 0 <= ii < len(arr) and arr[ii] >= 0:
stack.append(ii)
arr[ii] *= -1
return False | jump-game-iii | [Python] traversal (DFS) | ye15 | 2 | 86 | jump game iii | 1,306 | 0.631 | Medium | 19,467 |
https://leetcode.com/problems/jump-game-iii/discuss/2449212/python-perfect-answer-BFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
qu=deque([start])
vis=set()
while qu:
r=len(qu)
for i in range(r):
temp=qu.popleft()
vis.add(temp)
if arr[temp]==0:
return True
if temp+arr[temp] in range(len(arr)) and temp+arr[temp] not in vis:
qu.append(temp+arr[temp])
if temp-arr[temp] in range(len(arr)) and temp-arr[temp] not in vis:
qu.append(temp-arr[temp])
return False | jump-game-iii | python perfect answer BFS | benon | 1 | 63 | jump game iii | 1,306 | 0.631 | Medium | 19,468 |
https://leetcode.com/problems/jump-game-iii/discuss/2025140/3-Lines-Python-Solution-oror-85-Faster-oror-Memory-less-than-90 | class Solution:
def canReach(self, A: List[int], start: int) -> bool:
seen=set() ; queue=deque([start])
while queue:
i=queue.popleft()
if A[i]==0: return True
seen.add(i)
for x in {i-A[i],i+A[i]}:
if x not in seen and 0<=x<len(A): queue.append(x) | jump-game-iii | 3-Lines Python Solution || 85% Faster || Memory less than 90% | Taha-C | 1 | 60 | jump game iii | 1,306 | 0.631 | Medium | 19,469 |
https://leetcode.com/problems/jump-game-iii/discuss/2025140/3-Lines-Python-Solution-oror-85-Faster-oror-Memory-less-than-90 | class Solution:
def canReach(self, A: List[int], i: int) -> bool:
if i<0 or i>=len(A) or A[i]<0: return False
A[i]*=-1
return A[i]==0 or self.canReach(A,i+A[i]) or self.canReach(A,i-A[i]) | jump-game-iii | 3-Lines Python Solution || 85% Faster || Memory less than 90% | Taha-C | 1 | 60 | jump game iii | 1,306 | 0.631 | Medium | 19,470 |
https://leetcode.com/problems/jump-game-iii/discuss/1619642/Python3-5-lines-or-Short-and-Clean-or-DFS-or-O(n)-Time-or-O(n)-Space | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
if not 0 <= start < len(arr) or arr[start] < 0: return False
arr[start] *= -1
return not arr[start] or self.canReach(arr, start - arr[start]) or self.canReach(arr, start + arr[start]) | jump-game-iii | [Python3] 5 lines | Short and Clean | DFS | O(n) Time | O(n) Space | PatrickOweijane | 1 | 102 | jump game iii | 1,306 | 0.631 | Medium | 19,471 |
https://leetcode.com/problems/jump-game-iii/discuss/1619531/Simple-Python-DFS-solution-with-trace-image-and-comments | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
def isValidMove(idx):
# to check if a particular move is allowed or not
if idx >= 0 and idx < len(arr):
return True
return False
visited = set()
def canReachHelper(start, end=0):
# visited set to solve cycles
if start in visited:
return False
# found valid solution
if arr[start] == end:
return True
steps = arr[start]
visited.add(start)
# jump right side of array
new_positive_move = start + steps
# jump left side of array
new_negative_move = start - steps
if isValidMove(new_positive_move):
if canReachHelper(new_positive_move):
# early return incase a path is found
return True
if isValidMove(new_negative_move):
return canReachHelper(new_negative_move)
return canReachHelper(start) | jump-game-iii | Simple Python DFS solution with trace image and comments | vineeth_moturu | 1 | 102 | jump game iii | 1,306 | 0.631 | Medium | 19,472 |
https://leetcode.com/problems/jump-game-iii/discuss/1619219/Python3-Solution-DFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
if start >= 0 and start < len(arr) and arr[start] >= 0:
if arr[start] == 0:
return True
arr[start] = -arr[start]
return self.canReach(arr, start + arr[start]) or self.canReach(arr, start - arr[start])
return False | jump-game-iii | Python3 Solution - DFS | Cheems_Coder | 1 | 79 | jump game iii | 1,306 | 0.631 | Medium | 19,473 |
https://leetcode.com/problems/jump-game-iii/discuss/529895/Python3-simple-solution-using-a-while()-loop | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
seen, temp = set(), [start]
while temp:
i = temp.pop()
if arr[i] == 0: return True
else: seen.add(i)
if 0 <= i - arr[i] < len(arr) and i - arr[i] not in seen:
temp.append(i - arr[i])
if 0 <= i + arr[i] < len(arr) and i + arr[i] not in seen:
temp.append(i + arr[i])
return False | jump-game-iii | Python3 simple solution using a while() loop | jb07 | 1 | 70 | jump game iii | 1,306 | 0.631 | Medium | 19,474 |
https://leetcode.com/problems/jump-game-iii/discuss/2837860/Python-(Simple-DFS) | class Solution:
def canReach(self, arr, start):
n, stack, visited = len(arr), [start], set()
visited.add(start)
while stack:
idx = stack.pop(0)
if arr[idx] == 0:
return True
if 0 <= idx + arr[idx] < n and idx + arr[idx] not in visited:
stack.append(idx + arr[idx])
visited.add(idx + arr[idx])
if 0 <= idx - arr[idx] < n and idx - arr[idx] not in visited:
stack.append(idx - arr[idx])
visited.add(idx - arr[idx])
return False | jump-game-iii | Python (Simple DFS) | rnotappl | 0 | 4 | jump game iii | 1,306 | 0.631 | Medium | 19,475 |
https://leetcode.com/problems/jump-game-iii/discuss/2825476/python-super-easy-BFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
queue = [start]
visit = set()
visit.add(start)
while queue:
size = queue
for i in range(len(size)):
curr = queue.pop(0)
if arr[curr] == 0:
return True
if curr + arr[curr] < len(arr) and curr + arr[curr] not in visit:
queue.append(curr+arr[curr])
visit.add(curr+arr[curr])
if curr - arr[curr] >= 0 and curr - arr[curr] not in visit:
queue.append(curr-arr[curr])
visit.add(curr-arr[curr])
return False | jump-game-iii | python super easy BFS | harrychen1995 | 0 | 3 | jump game iii | 1,306 | 0.631 | Medium | 19,476 |
https://leetcode.com/problems/jump-game-iii/discuss/2713608/Recursion | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
visited = set()
def solve(idx):
if idx < 0 or idx > len(arr)-1 or idx in visited or len(visited) == len(arr):
return False
if arr[idx] == 0:
return True
visited.add(idx)
if solve(idx+arr[idx]) or solve(idx-arr[idx]):
return True
return False
return solve(start) | jump-game-iii | Recursion | mukundjha | 0 | 3 | jump game iii | 1,306 | 0.631 | Medium | 19,477 |
https://leetcode.com/problems/jump-game-iii/discuss/2618850/Easy-DFS-or-BFS-explanation-90%2B | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
visited = set()
q = deque()
q.append(start)
while(q):
idx = q.popleft()
visited.add(idx)
if arr[idx] == 0: # win condition
return True
nxt = [(idx+arr[idx]), (idx-arr[idx])]
for x in nxt:
if x >= 0 and x < len(arr) and x not in visited:
q.append(x)
return False | jump-game-iii | Easy DFS | BFS explanation 90%+ | gabrielpetrov99 | 0 | 21 | jump game iii | 1,306 | 0.631 | Medium | 19,478 |
https://leetcode.com/problems/jump-game-iii/discuss/2618850/Easy-DFS-or-BFS-explanation-90%2B | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
visited = set()
def dfs(i):
visited.add(i)
if arr[i] == 0: # win condition
return True
nxt = [(i+arr[i]), (i-arr[i])]
for x in nxt:
if x >= 0 and x < len(arr) and x not in visited and dfs(x):
return True
return False
return dfs(start) | jump-game-iii | Easy DFS | BFS explanation 90%+ | gabrielpetrov99 | 0 | 21 | jump game iii | 1,306 | 0.631 | Medium | 19,479 |
https://leetcode.com/problems/jump-game-iii/discuss/2601366/Jump-Game-3-oror-Easy-solution-oror-O(n)-Time-and-O(1)-Space-complexity-oror-Python | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
def traverse_path(idx, arr):
# Return if not a valid index
if(idx < 0 or idx >= len(arr)):
return False
# Return if index is already visited
if(arr[idx] < 0):
return False
# Return if we found destination
if(arr[idx] == 0):
return True
jumps = arr[idx]
# marking value as negative so that it won't be visited again
arr[idx] = -arr[idx]
# from every index we have 2 options left jump or right jump
return traverse_path(idx + jumps, arr) or traverse_path(idx - jumps, arr)
return traverse_path(start, arr) | jump-game-iii | Jump Game 3 || Easy solution || O(n) Time and O(1) Space complexity || Python | vanshika_2507 | 0 | 23 | jump game iii | 1,306 | 0.631 | Medium | 19,480 |
https://leetcode.com/problems/jump-game-iii/discuss/2564897/Recursive-solution | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
n = len(arr)
def recurse(start: int, seen: set) -> bool:
if start in seen:
return False
seen.add(start)
if start < 0 or start >= n:
return False
if arr[start] == 0:
return True
return recurse(start - arr[start], seen) or recurse(start + arr[start], seen)
return recurse(start, set()) | jump-game-iii | Recursive solution | mansoorafzal | 0 | 20 | jump game iii | 1,306 | 0.631 | Medium | 19,481 |
https://leetcode.com/problems/jump-game-iii/discuss/2550601/Simplest-Approach | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
visited = set()
dp = {}
def solve(start, visited):
if start in dp:
return dp[start]
if start in visited:
return
visited.add(start)
if start<0 or start>=len(arr):
return False
if arr[start] == 0:
return True
dp[start] = solve(start-arr[start],visited) or solve(start+arr[start],visited)
return dp[start]
return solve(start,visited) | jump-game-iii | Simplest Approach | Abhi_009 | 0 | 22 | jump game iii | 1,306 | 0.631 | Medium | 19,482 |
https://leetcode.com/problems/jump-game-iii/discuss/2465683/Straightforward-bfs-python3-solution | class Solution:
# O(n) time,
# O(1) space,
# Approach: BFS,
def canReach(self, arr: List[int], start: int) -> bool:
n = len(arr)
def getNeighbours(i:int) -> List[int]:
n = len(arr)
neighbours = []
# print(i)
if (i + arr[i]) < n:
neighbours.append(i + arr[i])
if (i - arr[i]) > -1:
neighbours.append(i - arr[i])
return neighbours
qu = deque()
qu.append(start)
vstd = set()
vstd.add(start)
while qu:
m = len(qu)
for i in range(m):
nod = qu.popleft()
if arr[nod] == 0:
return True
nbrs = getNeighbours(nod)
for nb in nbrs:
if nb in vstd: continue
vstd.add(nb)
qu.append(nb)
return False | jump-game-iii | Straightforward bfs python3 solution | destifo | 0 | 8 | jump game iii | 1,306 | 0.631 | Medium | 19,483 |
https://leetcode.com/problems/jump-game-iii/discuss/2365033/Python3-or-Solved-using-BFS-%2B-Queue-%2B-Model-Problem-as-Directed-Graph | class Solution:
#n = len(arr)
#Time-Complexity: O(n), in worst case we visit each and every index position and finds out
#there's no indices with integer 0 !
#Space-Complexity: O(n + n), by same argument as T.C! -> O(n)
def canReach(self, arr: List[int], start: int) -> bool:
#We can model this as a general directed graph problem!
#If we are at node i(at position index i), we can either
#jump to two descendants: arr[i] + i or arr[i] - i!
#We will only add to queue index positions not already visited
#and in-bounds!
visited = set()
q = collections.deque()
q.append(start)
visited.add(start)
#as long as queue is not empty, keep bfs going!
while q:
cur_index = q.popleft()
#check if at current index has value 0! If so, immediately
#break and return True
if(arr[cur_index] == 0):
return True
#otherwise, process the two descendants and only add to queue
#if it's not already visited and is in-bounds!
neighbor1 = arr[cur_index] + cur_index
neighbor2 = cur_index - arr[cur_index]
if(neighbor1 not in visited and 0<=neighbor1 < len(arr)):
q.append(neighbor1)
visited.add(neighbor1)
if(neighbor2 not in visited and 0<=neighbor2 < len(arr)):
q.append(neighbor2)
visited.add(neighbor2)
#once bfs is over, we tried every possible path from start!
#could not reach index position with value of 0!
return False | jump-game-iii | Python3 | Solved using BFS + Queue + Model Problem as Directed Graph | JOON1234 | 0 | 8 | jump game iii | 1,306 | 0.631 | Medium | 19,484 |
https://leetcode.com/problems/jump-game-iii/discuss/2118392/Breadth-First-Search-(BFS)-Algorithm-in-Python | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
max_size = len(arr) - 1
visited = set()
queue = []
queue.append(start)
visited.add(start)
while not len(queue) == 0:
curr = queue.pop(0)
print("\n")
step = arr[curr]
if step == 0:
return True
lower = curr - step
upper = curr + step
next_nodes = []
if lower >= 0:
next_nodes.append(lower)
if upper <= max_size:
next_nodes.append(upper)
for next_node in next_nodes:
if next_node not in visited:
queue.append(next_node)
visited.add(next_node)
return False | jump-game-iii | Breadth First Search (BFS) Algorithm in Python | juliancarax | 0 | 38 | jump game iii | 1,306 | 0.631 | Medium | 19,485 |
https://leetcode.com/problems/jump-game-iii/discuss/2090895/Python-Solution-in-BFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
# Time O(n)
# Space O(n)
n = len(arr)
que = deque()
visited = [False] * n
que.append(start)
while que:
tmp = que.popleft()
if arr[tmp] == 0:
return True
if visited[tmp]:
continue
if tmp + arr[tmp] < n:
que.append(tmp + arr[tmp])
if tmp - arr[tmp] >= 0:
que.append(tmp - arr[tmp])
visited[tmp] = True
return False | jump-game-iii | Python Solution in BFS | blazers08 | 0 | 26 | jump game iii | 1,306 | 0.631 | Medium | 19,486 |
https://leetcode.com/problems/jump-game-iii/discuss/1946494/PYTHON3-SOLUTION-or-Beats-100-of-python3-solutions | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
queue = deque()
visited = set()
queue.append(start)
m = len(arr)
while queue:
index = queue.popleft()
left, right = index - arr[index], index + arr[index]
if arr[index] == 0:
return True
if 0 <= left < m and left not in visited:
queue.append(left)
visited.add(left)
if 0 <= right < m and right not in visited:
queue.append(right)
visited.add(right)
return False | jump-game-iii | PYTHON3 SOLUTION | Beats 100% of python3 solutions | dericktolentino2002 | 0 | 18 | jump game iii | 1,306 | 0.631 | Medium | 19,487 |
https://leetcode.com/problems/jump-game-iii/discuss/1835216/Python-or-BFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
n = len(arr)
q = deque([start])
visited = {start}
while q:
idx = q.popleft()
if arr[idx] == 0:
return True
forward = idx + arr[idx] # jump forwards
back = idx - arr[idx] # jump backwards
if forward not in visited and forward < n:
visited.add(forward)
q.append(forward)
if back not in visited and back >= 0:
visited.add(back)
q.append(back)
return False | jump-game-iii | Python | BFS | jgroszew | 0 | 40 | jump game iii | 1,306 | 0.631 | Medium | 19,488 |
https://leetcode.com/problems/jump-game-iii/discuss/1773944/Python-easy-to-read-and-understand-or-DFS | class Solution:
def dfs(self, arr, i, visited):
if i < 0 or i >= len(arr) or visited[i] == True:
return
if arr[i] == 0:
self.ans = True
return
visited[i] = True
self.dfs(arr, i+arr[i], visited)
self.dfs(arr, i-arr[i], visited)
visited[i] = False
def canReach(self, arr: List[int], start: int) -> bool:
self.ans = False
n = len(arr)
visited = [False for _ in range(n)]
self.dfs(arr, start, visited)
return self.ans | jump-game-iii | Python easy to read and understand | DFS | sanial2001 | 0 | 77 | jump game iii | 1,306 | 0.631 | Medium | 19,489 |
https://leetcode.com/problems/jump-game-iii/discuss/1681478/PYTHON-Recursive-solution-DFS | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
self.seen = set()
def dfs(arr, index):
if index > len(arr)-1 or index < 0 or (index in self.seen):
return False
if arr[index] == 0:
return True
self.seen.add(index)
return dfs(arr, index + arr[index]) or dfs(arr, index - arr[index])
return dfs(arr, start) | jump-game-iii | [PYTHON] Recursive solution DFS | satyu | 0 | 47 | jump game iii | 1,306 | 0.631 | Medium | 19,490 |
https://leetcode.com/problems/jump-game-iii/discuss/1619921/Python3-Dfs-recursion-solution | class Solution:
def traversal(self, arr, idx):
if idx < 0 or idx >= len(arr) or arr[idx] < 0:
return False
if arr[idx] == 0:
return True
arr[idx] *= -1
return self.traversal(arr, idx + arr[idx]) or self.traversal(arr, idx - arr[idx])
def canReach(self, arr: List[int], start: int) -> bool:
return self.traversal(arr, start) | jump-game-iii | [Python3] Dfs recursion solution | maosipov11 | 0 | 28 | jump game iii | 1,306 | 0.631 | Medium | 19,491 |
https://leetcode.com/problems/jump-game-iii/discuss/1619524/Iterative-and-recursive-DFS-in-Python | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
def helper(i: int = start):
if i >= len(arr) or i < 0 or arr[i] < 0:
return False
if arr[i] == 0:
return True
arr[i] = -arr[i]
return helper(i + arr[i]) or helper(i - arr[i])
return helper() | jump-game-iii | Iterative and recursive DFS in Python | mousun224 | 0 | 29 | jump game iii | 1,306 | 0.631 | Medium | 19,492 |
https://leetcode.com/problems/jump-game-iii/discuss/1619524/Iterative-and-recursive-DFS-in-Python | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
stack = [start]
while stack:
v = stack.pop()
if arr[v] < 0:
continue
if arr[v] == 0:
return True
arr[v] = -arr[v]
for step in (arr[v], -arr[v]):
if 0 <= (v + step) < len(arr):
stack += (v + step),
return False | jump-game-iii | Iterative and recursive DFS in Python | mousun224 | 0 | 29 | jump game iii | 1,306 | 0.631 | Medium | 19,493 |
https://leetcode.com/problems/jump-game-iii/discuss/1618956/Python-BFS-with-explanation-(96.12-faster-84.22less-memory) | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
jumps = [start]
while jumps:
current = jumps.pop(0)
if arr[current] == -1:
continue
elif arr[current] == 0:
return True
next_steps = current+arr[current], current-arr[current]
# Add next steps to the queue
if next_steps[0]<len(arr):
jumps.append(next_steps[0])
if next_steps[1]>=0:
jumps.append(next_steps[1])
# Mark the current index as visited
arr[current] = -1
return False | jump-game-iii | Python BFS with explanation (96.12% faster, 84.22%less memory) | kaushalya | 0 | 63 | jump game iii | 1,306 | 0.631 | Medium | 19,494 |
https://leetcode.com/problems/jump-game-iii/discuss/1618941/Python3-ITERATIVE-BFS-Explained-(beats100!) | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
n = len(arr)
vis = set()
q = deque([start])
while q:
i = q.popleft()
if arr[i] == 0:
return True
vis.add(i)
l, r = i - arr[i], i + arr[i]
if not l in vis and l > -1:
q.append(l)
if not r in vis and r < n:
q.append(r)
return False | jump-game-iii | ✔️ [Python3] ITERATIVE BFS, Explained (beats100%!) | artod | 0 | 133 | jump game iii | 1,306 | 0.631 | Medium | 19,495 |
https://leetcode.com/problems/jump-game-iii/discuss/1547883/BFS-268ms-99.7-runtime-93-memory | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
# let's do BFS, it's a binary tree
node_queue = [start]
for i in node_queue:
# Check if element is blind
if arr[i] is None:
continue
node = arr[i]
if node == 0:
return True
left = i - node
right = i + node
# Bound checks
if left >= 0:
node_queue.append(left)
if right <= len(arr) - 1:
node_queue.append(right)
# Blind the element from further expansion
arr[i] = None
return False | jump-game-iii | BFS, 268ms - 99.7% runtime, 93% memory | HoundThe | 0 | 51 | jump game iii | 1,306 | 0.631 | Medium | 19,496 |
https://leetcode.com/problems/jump-game-iii/discuss/1465220/PyPy3-Solution-using-DFS-w-comments | class Solution:
def canReach(self, arr: List[int], start: int) -> bool:
# init
n = len(arr)
# DFS
def dfs(idx, isVisited=set()):
# Check if idx is not yet visited and
# it is within boundaries
if idx not in isVisited and 0 <= idx < n:
# Add to is visited set
isVisited.add(idx)
# Check if destination reached, if yes, return true
if arr[idx] == 0:
return True
# else run dfs on i+arr[i] and i-arr[i] index
else:
return dfs(idx+arr[idx], isVisited) or dfs(idx-arr[idx], isVisited)
return False # By default return false
return dfs(start) # Start dfs at start index | jump-game-iii | [Py/Py3] Solution using DFS w/ comments | ssshukla26 | 0 | 60 | jump game iii | 1,306 | 0.631 | Medium | 19,497 |
https://leetcode.com/problems/jump-game-iii/discuss/1442807/Python-Recursive-DFS-Solution | class Solution:
#gets the next valid moves
def getNextInd(self, i, arr):
ret = []
r, l = i + arr[i], i - arr[i]
if r < len(arr):
ret.append(r)
if l >= 0:
ret.append(l)
return ret
def canReach(self, arr: List[int], start: int) -> bool:
vis = [False for _ in range(len(arr))]
def recur(start):
#if already visited, it doesn't need to be reexplored
if vis[start]:
return False
vis[start] = True
if arr[start] == 0:
return True
moves = self.getNextInd(start, arr)
reached = False
for i in moves:
reached = reached or recur(i)
return reached | jump-game-iii | Python Recursive DFS Solution | mguthrie45 | 0 | 109 | jump game iii | 1,306 | 0.631 | Medium | 19,498 |
https://leetcode.com/problems/jump-game-iii/discuss/1375836/Python3-solution-using-recursion | class Solution:
def canReach(self, arr, start):
def check(index, arr, explored):
if arr[index] == 0:
return True
else:
a = None
b = None
if index + arr[index] < len(arr) and explored[index + arr[index]] == 0:
explored[index + arr[index]] = 1
a = check(index + arr[index], arr, explored)
if index - arr[index] >= 0 and explored[index - arr[index]] == 0:
explored[index - arr[index]] = 1
b = check(index - arr[index], arr, explored)
return a or b
explored = [0]*len(arr)
x = check(start, arr, explored)
return x | jump-game-iii | Python3 solution using recursion | EklavyaJoshi | 0 | 40 | jump game iii | 1,306 | 0.631 | Medium | 19,499 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.