238. Product of Array Except Self

Given an array nums of n integers where n > 1,  return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        res[0] = 1;
        for (int i = 1; i < n; i++) {
            res[i] = res[i - 1] * nums[i - 1];
        }
        int right = 1;
        for (int i = n - 1; i >= 0; i--) {
            res[i] *= right;
            right *= nums[i];
        }
        return res;
    }
    
    public int[] productExceptSelf(int[] nums) {
        // Left is an array containing the left products
        // i.e: left[i] = nums[0] * .... * nums[i-1]
        int[] left = new int[nums.length];
        
        // Right is an array containing the array products
        //i.e: right[i] = nums[i+1] * nums[i+2]  * .... * nums[len(nums) - 1]
        int[] right = new int[nums.length];
        
        left[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            left[i] = left[i-1] * nums[i-1];
        }
        
        right[nums.length - 1] = 1;
        for (int i = nums.length - 2; i >= 0; i--) {
            right[i] = right[i+1] * nums[i+1];
        }
        
        int[] product = new int[nums.length];
        for (int i = 0; i < product.length; i++) {
            product[i] = left[i] * right[i];
        }
        
        return product;
    }
}