0238. Product of Array Except Self

https://leetcode.com/problems/product-of-array-except-self

Description

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:

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

Example 2:

**Input:** nums = [-1,1,0,-3,3]
**Output:** [0,0,9,0,0]

Constraints:

• 2 <= nums.length <= 105

• -30 <= nums[i] <= 30

• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in O(1)extra space complexity? (The output array does not count as extra space for space complexity analysis.)

ac

https://leetcode.com/problems/product-of-array-except-self/discuss/65622/Simple-Java-solution-in-O(n)-without-extra-space/67603

// most important clue: break into left * right, which hard to come up with
class Solution {
    public int[] productExceptSelf(int[] nums) {
        // left * right
        // walk -> int[] left
        // walk backwards -> right * left

        // corner cases, no here, n>1

        // int[] left
        int[] left = new int[nums.length];
        left[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            left[i] = left[i-1] * nums[i-1];
        }

        // right * left
        int right = 1;
        left[left.length - 1] *= right; 
        for (int i = left.length - 1 - 1; i >= 0; i--) {
            right *= nums[i+1];
            left[i] = left[i] * right;
        }

        // return left
        return left;
    }
}