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`

Solution:

```
func countElements(arr []int) int {
lenA := len(arr)
dict := map[int]int{}
for i:= 0; i < lenA; i++ {
dict[arr[i]]++;
}
result := 0;
for i:= 0; i < lenA; i++ {
if (dict[arr[i]+1] > 0) {
result++;
}
}
return result;
}
```