leetcode/apps/docs/content/array/1426-counting-elements.mdx

---
title: '1426. Counting Elements'
description: Given an integer array arr, count how many elements x there are, such that x + 1 is also in arr. If there are duplicates in arr, count them separately
sidebar:
  label: 'Counting Elements'
  badge: 'Easy'
---

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

### Example 1:
- Input: `arr = [1,2,3]`
- Output: `2`
- Explanation: `1` and `2` are counted cause `2` and `3` are in `arr`.

### Example 2:
- Input: `arr = [1,1,3,3,5,5,7,7]`
- Output: `0`
- Explanation: No numbers are counted, cause there is no `2`, `4`, `6`, or `8` in `arr`.

### Constraints:

- `1 <= arr.length <= 1000`
- `0 <= arr[i] <= 1000`

## Approach

```mermaid
flowchart TD
  S(["countElements(arr)"]) --> I["arrSet = Set(arr), count = 0"]
  I --> L{"more n in arr?"}
  L -- no --> E(["return count"])
  L -- yes --> H{"arrSet has n + 1?"}
  H -- yes --> C["count++"]
  C --> L
  H -- no --> L
```

## Solution

```js
/**
 * @param {number[]} arr
 * @return {number}
 */
var countElements = function(arr) {
  let arrSet = new Set(arr);
  let count = 0; 

  for (let n of arr) {
    let sum = n + 1;
    if (arrSet.has(sum)) {
      count++;
    }
  }
  return count;
};
```