Each shelf's product of all the other shelves

Make two passes. Left to right, write the product of everything before i; right to left, multiply in the product of everything after i, carried in one variable. Each slot ends as before × after. O(n) time and O(1) extra space besides the output, and zeros need no special case because nothing is divided.

Division fails on zeros and hides precision issues with floats, which is why the question forbids it.

The output list doubles as the 'before' array, so only one running variable is needed for 'after'.

Two zeros make every slot zero; one zero makes every slot but its own zero. The two-pass method handles both without branching.

def others_product(mult: list[int]) -> list[int]:
    out = [1] * len(mult)
    before = 1
    for i in range(len(mult)):
        out[i] = before
        before *= mult[i]
    after = 1
    for i in range(len(mult) - 1, -1, -1):
        out[i] *= after
        after *= mult[i]
    return out

Common mistakes

• Using division and then patching zero cases with special branches.
• Off-by-one: including the element itself in its own 'before' or 'after' product.

Say 'before times after' first and draw it on three cells. The code then reads as a direct translation.

Follow-up questions

Think of your answer first, then open the hint.

1. How would you handle overflow in a fixed-width language?

   Show hint

   Use 64-bit integers or a modulus if the question allows; Python ints do not overflow.

2. Can you do it in one pass?

   Show hint

   Fill from both ends at once with two running products; the total work is the same.