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;
}