Product Except Self With Zeros

You are given an integer array nums. Return an array result where result[i] is the product of all elements in nums except nums[i].
The source solution explicitly handled three zero cases:
If there are no zeroes, each position receives the product of all numbers divided by the current value.
If there is exactly one zero, only the zero position receives the product of all non-zero values.
If there is more than one zero, every output value is 0.

Function
productExceptSelf(nums: int[]) → int[]

Examples
Example 1
nums = [1,2,0,4]
return = [0,0,8,0]
The source shared the rule but did not include this exact sample. FastPrep added this small example so the behavior can be checked directly.

Constraints
The source expects integer input.
The source solution uses integer division in the no-zero case.

The trick is counting zeros before you multiply anything. Walk the array once, multiply all non-zero values, and count the zeros. Then branch. No zeros: each output is total divided by nums[i], using integer division. Exactly one zero: the zero's position gets the product of the non-zero values, everything else is 0. Two or more zeros: the whole array is 0. That's O(n) time and O(1) extra space beyond the output. The common pitfall is multiplying zeros into the total, which turns every answer into 0 and breaks the single-zero case. Another is dividing by zero in the no-zero branch by mistake. Check your code against [1,2,0,4], which should return [0,0,8,0]. If you'd rather skip division, prefix and suffix products also work, and they handle zeros for free. StealthCoder is the hedge if the live OA rattles you and the case split slips your mind.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.