## Counting Elements Problem statement:

Given an integer array `arr`

, count element `x`

such that `x + 1`

is also in `arr`

.

If there’re duplicates in `arr`

, count them separately.

**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's no 2, 4, 6, or 8 in arr.
```

**Example 3:**

```
**Input:** arr = [1,3,2,3,5,0]
**Output:** 3
**Explanation:** 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
```

**Example 4:**

```
**Input:** arr = [1,1,2,2]
**Output:** 2
**Explanation:** Two 1s are counted cause 2 is in arr.
```

**Constraints:**

`1 <= arr.length <= 1000`

`0 <= arr[i] <= 1000`

## Counting Elements ES6 solution

```
/\*\* \* @param {number[]} arr \* @return {number} \*/
var countElements = function(arr) {
let count = 0;
for(let i = 0; i < arr.length; i++){
if(arr.includes(arr[i]+1)) count++;
}
}
return count;
};
```

## Discussion (0)