Product of Array Except Self

 Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

 

Example 1:

Input: nums = [1,2,3,4]

Output: [24,12,8,6]

Example 2:

Input: nums = [-1,1,0,-3,3]

Output: [0,0,9,0,0]

class Solution {

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

        int n = nums.length;

        int[] producttforward = new int[n]; // stores product of elements after index i

        int[] productbackward = new int[n]; // stores product of elements before index i

        int[] res = new int[n];

        // Forward: product of elements to the right

        producttforward[n - 1] = 1; // last element has no right side

        for (int i = n - 2; i >= 0; i--) {

            producttforward[i] = producttforward[i + 1] * nums[i + 1];

        }

        // Backward: product of elements to the left

        productbackward[0] = 1; // first element has no left side

        for (int j = 1; j < n; j++) {

            productbackward[j] = productbackward[j - 1] * nums[j - 1];

        }

        // Combine forward and backward

        for (int k = 0; k < n; k++) {

            res[k] = producttforward[k] * productbackward[k];

        }

        return res;

    }

}