---
title: '238. 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]
sidebar:
  label: 'Product of Array Except Self'
  badge: 'Medium'
---

Array

::::warning
You must write an algorithm that runs in O(n) time and without using the division operation.
Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
::::

### 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 <= 10^5`
- `-30 <= nums[i] <= 30`
- The input is generated such that `answer[i]` is guaranteed to fit in a 32-bit integer.

## Approach

```mermaid
flowchart TD
  S(["productExceptSelf(nums)"]) --> I["answer = ones(n), rightArr = ones(n)"]
  I --> P1["Pass 1 — left to right: answer[i] = nums[i-1] * answer[i-1]"]
  P1 --> P2["Pass 2 — right to left: rightArr[i] = nums[i+1] * rightArr[i+1]"]
  P2 --> P3["Pass 3: answer[i] *= rightArr[i]"]
  P3 --> E(["return answer — no division needed"])
```

## Solution

```js
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function(nums) {
  const n = nums.length;
  const answer = new Array(n).fill(1);
  const rightArr = new Array(n).fill(1);

  for (let i = 1; i < n; i++) {
    answer[i] = nums[i - 1] * answer[i - 1];
  }
  for (let i = n - 2; i >= 0; i--) {
    rightArr[i] = nums[i + 1] * rightArr[i + 1];
  }
  for (let i = 0; i < n; i++) {
    answer[i] *= rightArr[i];
  }
  return answer;
};
```

## Explanation

[Product of Array Except Self - Leetcode 238 - Python](https://www.youtube.com/watch?v=bNvIQI2wAjk)