2026-09-18 — Product of Array Except Self

The division approach fails on zeroes: with one zero, every other answer is 0 and you would be dividing by zero to get the one non-zero entry. With two zeroes, everything is 0. Handling that with special cases is possible but ugly — and the interviewer's real question is whether you see the alternative.

The product of everything except i is:

(everything to the LEFT of i)  ×  (everything to the RIGHT of i)

Both of those can be built in a single pass each.

def product_except_self(nums: list[int]) -> list[int]:
    n = len(nums)
    answer = [1] * n

    # Pass 1: answer[i] = product of everything LEFT of i.
    running = 1
    for i in range(n):
        answer[i] = running
        running *= nums[i]

    # Pass 2: multiply in the product of everything RIGHT of i.
    running = 1
    for i in range(n - 1, -1, -1):
        answer[i] *= running
        running *= nums[i]

    return answer

Tracing [1, 2, 3, 4]:

How it works. The first pass writes the left product into answer[i] before folding nums[i] into the running product, so nums[i] itself is never included. The second pass does the same from the right, multiplying into what is already there. No division, and zeroes need no special handling at all — a zero simply propagates into the running product like any other value.

Complexity. Time: O(n) — two passes. Space: O(1) extra — the running variables; the output array is required by the problem and does not count.

Edge cases that matter. - One zero — every position except that one gets 0; the zero's own position gets the product of the rest. Falls out automatically. - Two or more zeroes — everything is 0. Also automatic. - Two elements, the minimum interesting size. - Negative numbers — signs take care of themselves.

Common wrong answer. total = math.prod(nums) then total // nums[i]. It is O(n) and looks clever, but it is explicitly disallowed, it crashes on a zero, and with two zeroes there is no total to divide anyway.