Leetcode 238 - Product of Array Except Self

01-Jul-2026

I ended up watching youtube for a bit, a while, then an hour flew by. Man, I’m making it harder for my future self, incredibly harder I wan’t to cry. Anyhow, this question took me a minute to think about but I think I got a satisfactory answer.

Two-pass solution FTW

With a quick look at the question, I was going to run a single-pass with a variable to hold the total product of the array. That way, I could simply divide for each element of the array and return the values. An easy solution! But the question strictly states to not use division operations.

With that, the next thought was to run, if possible, a minimal amount of passes through the array. If I went through each element, one-by-one, to find each product, it’ll take $O(n^2)$ times to a final array. That’s not optimal! So taking the $O(1)$ extra space as a clue, I on a two-pass solution, holding onto the current product during each pass, and multiplying to the next element.

This means for an array [1,2,3,4], at 0-idx, we’ll hold onto $1$. Then at 1-idx, we’ll multiply the current element with $1$ in our answer array [1,1,1,1], and then multiple $1$ with $2$ to hold onto $2$. Then for the next, we do the same and hold onto $6$, and so on. Then after our first pass, we’ll have answer = [1,1,2,6]. Then we go from the right and repeat our process until 0-idx, and we’ll have [24,12,8,6].

Performance evaluation

With our two-pass solution, we have a time complexity of $O(n)$ (very beautiful) and a space complexity of $O(n)$ (coming from the answer array). Overall very quick and satisfying solution.