Counting Element

Given an integer array arr, count element x such that x + 1 is also in arr.

If there're duplicates in arr, count them seperately.

**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
```

**Solutions**

Use hashtable to search existence of

`element + 1`

, take`O(2n)`

time.

```
class Solution {
func countElements(_ arr: [Int]) -> Int {
var hash = Set(arr)
var res = 0
for ele in arr {
if hash.contains(ele + 1) {
res += 1
}
}
return res
}
}
```