238. Product of Array Except Self

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

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

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

Solution:

We multiply the elements from two directions.

In the first pass, for each element,  we multiply all elements from its left.

In the second pass, for each element, we multiply all elements from its right.

Hence, in any position, the corresponding value is the product of all elements except the element in this position.

The time complexity is O(n) and the space complexity is O(1).



Code:

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        int left = 1;
        for (int i = 0; i < n; i++) {
            result[i] = left;
            left *= nums[i];
        }
        int right = 1;
        for (int i = n - 1; i >= 0; i--) {
            result[i] *= right;
            right *= nums[i];
        }
        return result;
    }
}