"""
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.
"""

from collections import defaultdict

class Solution:
    def countElements(self, arr: List[int]) -> int:
        found = defaultdict(int)
        count = 0
        for elem in arr:
            found[elem] += 1

        for elem, n in found.items():
            complement = elem + 1
            if complement in found:
                count += n
        return count
ADd day7