Source: LeetCode
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.
Input: arr = [1,2,3] ==> Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
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.
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.
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