LeetCode #238: Product of Array Except Self | Prefix Sum

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0:00 Problem overview
0:27 Brute force solution
1:22 Optimized O(n) solution
4:22 Code walkthrough

#programming #coding #algorithms

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I really appreciate the explanation and how you visualize the problem! Great stuff!

karlagabriel

Very clear concise explanation and visualisation! Thank you very much and please keep posting! You are highly appreciated!

dnm

the explanation of the logic is just fantastic!!

Sanai

Once again, best visuals and explanation out there!

jpkeys

Once again, best explanation and visuals out there, thank you!

jpkeys

the grid visualization really helped. finally understood this problem thank you

yilinliu

perfect . simplified .really easy to understand .kudos man . cant wait for more

bullymaguire

Though I could understand what the solutions were doing there was not much for the intuition behind it. This puts that crucial piece of the puzzle together. Thank you

ChamplooMusashi

Great explanation, i ha to watch it 2 times, but i understood it

voicubogdan

My only question is how do you even start coming up with an algorithm to solve it, i understand the solution but coming up with the solution in the first place is difficult for me, how do you think to start doing left and right products and multiply them?

azka

how on earth you came up with this solution omg :D

ruslan

thanks for this, and thanks for the captions.

skinnycucumber

Your explanation is really good but when u fill up the left and right, ur red line of box is not easy to understand. Instead of that, just say multiply every left side or multiply every right side elements. Not like one by one.
For example, when we calculate every left side of 5, say 2*3*4.
And I think until 4:12 mins are enough to understand this question.

Thank you for your explanation.

cocotv

class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
i = 0
length = len(nums)
j = length - 1
left = list([1]);right = list([1])

for i in range(length-1):
left.append( left[i] * nums[i] )
right.append( right[i] * nums[j] )
# backwards
j -= 1

right.reverse()
out = [ i * j for i, j in zip(left, right) ]
return out

guruprasaad

Could you also find the product of all elements, and divide by each num[i] to create the product array

runesbroken

Could you use C++, Java or C#, please? Python is hard to read.

tomdriver

But it's gonna be O(2n-2), right?

Khanhan

My Solution
nums = [3, 2, 4, 0]

res=[]
nums1=copy.copy(nums)
print(nums1)
for i in range(0, len(nums1)):
nums1.remove(nums[i])
prod=reduce((lambda x, y:x*y), nums1)
res.append(prod)
nums1.insert(i, nums[i])
return(res)

DevanshClasses

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