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.)

Solution

Prefix & Suffix, time O(n), space O(n)

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] productOfFront = new int[n];
        int[] productOfBehind = new int[n];
        
        for (int i = 0; i < n; ++i) {
            if (i == 0) {
                productOfFront[i] = 1;
                productOfBehind[n - i - 1] = 1;
                continue;
            }
            
            productOfFront[i] = nums[i - 1] * productOfFront[i - 1];
            productOfBehind[n - i - 1] = nums[n - i] * productOfBehind[n - i];
        }
        
        for (int i = 0; i < n; ++i) {
            productOfFront[i] *= productOfBehind[i];
        }
        
        return productOfFront;
    }
}

Optimised, time O(n), space O(1)

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