238.md

238. Product of Array Except Self

Description

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.)

Thinking Process

1. Create the result array res, initialize everything to 1
2. Scan from the front of nums, assign the product of the first (i-1) elements to res[i]
3. Scan from the back of nums, update res[i] to be res[i] times the product of the last (i-1) elements

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

Complexity

1. No additional space is needed apart for the output array
2. Time complexity is O(n) as the array is scanned twice