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# 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](https://leetcode.com/problems/product-of-array-except-self/discuss/65622/Simple-Java-solution-in-O%28n%29-without-extra-space/67603)

```java
// 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;
    }
}
```