LeetCode 30 day Challenge | Day 15 | Product of Array except self (C++ & Java)

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LeetCode 30 day Challenge | Problem 15 | Product of Array except Self | 15 April
Facebook Coding Interview question,
google coding interview question,
leetcode,
Product of Array,
Product of Array except self,
#Amazon #Facebook #CodingInterview #LeetCode #30DayChallenge #Google #StringShifts

[@KnowledgeCenter]

Please share your solutions - no matter whether it's O(1) space or O(n) space in the comments. I have added C++ and Java codes for both the approaches in the video.

[@adarshmv261]

Appreciate your effort sir.... can you please include python code as well instead of Java or C++

[@KnowledgeCenter]

Haha. Will definitely try from next video. I hope nobody asks me for 4th language.

[@lilacu]

Lol

[@electricalcoder3025]

Great content dude....keep moving...

[@KnowledgeCenter]

Thanks, will do!

[@jaggujaggu2922]

import itertools
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
l=[]
k=list(itertools.combinations(nums,len(nums)-1))
def multiplyList(myList) :
result = 1
for x in myList:
result = result * x
return(result)
for i in k:
l.append(multiplyList(i))
l=l[::-1]
return(l)
i am a begginer, bro can u help me with this code i am not able to pass all testcases

[@KnowledgeCenter]

Your solution doesn't look O(n) time. In your code k is a list of all subsequences of n-1 length. Then you are iterating each list and multiplying.
So, multiplying all elements of 1 list is O(n). So, do this for n lists. So, overall time complexity is O(nxn) = O(n2).
All test cases will pass if you implement O(n) solution.
Try avoiding many repeated multiplications, with the approach explained in video.

[@jaggujaggu2922]

@@KnowledgeCenter tq bro