"""
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

Hint #1  
Use hashset to store all elements.

Hint #2  
Loop again to count all valid elements.

"""

def count_elements(nums):
    nums_dict = {}
    counter = 0

    for num in nums:
        nums_dict[num] = nums_dict.get(num, 0) + 1

    for num in nums:
        if nums_dict.get(num+1):
            counter += 1
    
    return counter
