190. Product of Array Except Self

answer[i] is everything to the left of i multiplied by everything to the right of i. Build the two halves separately, then combine.

Approach

1. Create prefix of length n with prefix[0] = 1 and prefix[i] = prefix[i-1] * nums[i-1].
2. Create suffix similarly from the right: suffix[n-1] = 1 and suffix[i] = suffix[i+1] * nums[i+1].
3. answer[i] = prefix[i] * suffix[i].

Walk-through

nums = [1, 2, 3, 4]

Complexity

• Time: O(n).
• Space: O(n) for the two helper arrays.

import java.util.*;

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] prefix = new int[n];
        prefix[0] = 1;
        for (int i = 1; i < n; i++) prefix[i] = prefix[i - 1] * nums[i - 1];
        int[] suffix = new int[n];
        suffix[n - 1] = 1;
        for (int i = n - 2; i >= 0; i--) suffix[i] = suffix[i + 1] * nums[i + 1];
        int[] result = new int[n];
        for (int i = 0; i < n; i++) result[i] = prefix[i] * suffix[i];
        return result;
    }
}

We can avoid the two helper arrays by writing prefix products directly into the output, then folding in the suffix product with a single backward pass.

Approach

1. Pass 1: answer[i] = product of nums[0..i-1].
2. Pass 2: maintain a running right = 1, sweep from the end and multiply answer[i] *= right, then right *= nums[i].

Complexity

• Time: O(n).
• Space: O(1) extra (output not counted).