Product of Array Except Self

You're given an integer array nums. Return an array answer of the same length where answer[i] is the product of every element of nums except nums[i].

You may not use division, and the solution should run in O(n) time. Every product fits in a 32-bit signed integer.

"Everything except i" splits cleanly into two parts: everything to the left of i, and everything to the right of i. So answer[i] = (product of nums[0..i-1]) × (product of nums[i+1..n-1]).

Both parts are running products. One pass left to right computes every prefix product; one pass right to left computes every suffix product. Multiply them position by position and you're done. You can even skip storing the suffix array: write prefixes into answer first, then sweep from the right with a single running variable and multiply it in.

This also explains why division is banned: total ÷ nums[i] is the obvious shortcut, but it breaks as soon as the array contains a zero.

def product_except_self(nums: list[int]) -> list[int]:
    n = len(nums)
    answer = [1] * n
    for i in range(1, n):
        answer[i] = answer[i - 1] * nums[i - 1]
    suffix = 1
    for i in range(n - 1, -1, -1):
        answer[i] *= suffix
        suffix *= nums[i]
    return answer

function productExceptSelf(nums) {
  const n = nums.length;
  const answer = new Array(n).fill(1);
  for (let i = 1; i < n; i++) answer[i] = answer[i - 1] * nums[i - 1];
  let suffix = 1;
  for (let i = n - 1; i >= 0; i--) {
    answer[i] *= suffix;
    suffix *= nums[i];
  }
  return answer;
}

Say the division approach out loud and immediately explain why it fails on zeros. That shows you understand the constraint rather than just following it.

Then state the decomposition: "left product times right product." Once you've said that, the two-pass code is short. Interviewers often ask for O(1) extra space as a follow-up, so mention up front that the output array doesn't count and you can reuse it for the prefixes.