Return a new array consisting of the product of all elements in the array excluding the current index

getProductArray.java

public int[] getProducts(int[] nums){

        if(nums == null) return new int[0];
        if(nums.length == 0) return new int[0];
        if(nums.length == 1) return new int[]{0};


        int[] products = new int[nums.length];

        //any number multiplied by 1 is the same number; therefore allProducts cannot be initialized to 0
        int allProducts = 1, start = 0, end = nums.length-1, countOfZeros = 0;

        while(start<=end){

            if(start==end){

                if(nums[start] == 0){
                    countOfZeros++; //ignore 0s in the array, keep a count in case all/some elements are 0;
                }
                else{
                    allProducts *= nums[start];
                }

            }
            else{

                if(nums[start] == 0){
                    countOfZeros++;
                }
                else{
                    allProducts *= nums[start];
                }

                if(nums[end] == 0){
                    countOfZeros++; //ignore 0s in the array, keep a count in case all elements are 0;
                }
                else{
                    allProducts *= nums[end];
                }
            }
            start++; end--;

        }

        if (countOfZeros >  1) return products; //product of any other number including a 0 still results in 0

        // only countOfZeros = 1 can result in the index having a product
        for(int index = 0; index < nums.length; index++){
            products[index] = (countOfZeros == 1) ?
                                    (nums[index] == 0) ? allProducts : 0
                              : allProducts/nums[index];
        }

   
        return products;
    }

Hello @deedee47, thank you for participating in Week 4 of Algorithm Fridays.

This is a decent implementation. One thing to note however is that your solution fails the test case for when one of the elements of the array is 0. For example, if I run:

getProducts([4, 3, 0]); ❌ // it should return [0, 0, 12] but yours returns [0, 0, 0]

But apart from that, this was a good attempt.

Hi @meekg33k, wow, the cases really run deep.

Based on the way I implemented the solution, I assumed that once a 0 is present, it makes the product 0 as well. I have considered a different approach based on the case you provided. if there's only one 0 present, then it can have a product otherwise, everything else is 0.

Thanks for the insight.
It is really interesting to participate