leetcode/apps/docs/content/array/268-missing-number.mdx

---
title: '268. Missing Number'
description: Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array
sidebar:
  label: 'Missing Number'
  badge: 'Easy'
---

<Badge variant="accent">Array</Badge>

::::warning
Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
::::

### Example 1:
- Input: `nums = [3,0,1]`
- Output: `2`
- Explanation: `n = 3` since there are `3` numbers, so all numbers are in the range `[0,3]`. `2` is the missing number in the range since it does not appear in `nums`.

### Example 2:
- Input: `nums = [0,1]`
- Output: `2`
- Explanation: `n = 2` since there are `2` numbers, so all numbers are in the range `[0,2]`. `2` is the missing number in the range since it does not appear in `nums`.

### Example 3:
- Input: `nums = [9,6,4,2,3,5,7,0,1]`
- Output: `8`
- Explanation: `n = 9` since there are `9` numbers, so all numbers are in the range `[0,9]`. `8` is the missing number in the range since it does not appear in `nums`.

### Constraints:

- `n == nums.length`
- `1 <= n <= 10^4`
- `0 <= nums[i] <= n`
- All the numbers of `nums` are unique.

## Approach

```mermaid
flowchart TD
  S(["missingNumber(nums)"]) --> I["numSet = Set(nums), expectedCount = n + 1"]
  I --> L{"more i in 0..expectedCount-1?"}
  L -- yes --> Q{"numSet has i?"}
  Q -- no --> R(["return i"])
  Q -- yes --> L
  L -- no --> E(["return -1"])
```

## Solution

```js
/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function(nums) {
  const numSet = new Set(nums);
  const expectedCount = nums.length + 1;

  for (let i = 0; i < expectedCount; i++){
    if(!numSet.has(i)){
      return i;
    }
  }
  return -1;
};
```

## Explanation

<YouTube id="WnPLSRLSANE" title="Missing Number - Blind 75 - Leetcode 268 - Python" />