Count Number of Pairs With Absolute Difference K

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

• x if x >= 0.
• -x if x < 0.

Example 1:

Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:

• [1,2,2,1]
• [1,2,2,1]
• [1,2,2,1]
• [1,2,2,1]

Example 2:

Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.

Example 3:

Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:

• [3,2,1,5,4]
• [3,2,1,5,4]
• [3,2,1,5,4]

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100
• 1 <= k <= 99

SOLUTION:

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        exists = {}
        ctr = 0
        for i, num in enumerate(nums):
            exists[num] = exists.get(num, []) + [i]
        for i, num in enumerate(nums):
            for val in [num - k, num + k]:
                if val in exists:
                    for j in exists[val]:
                        if j > i:
                            ctr += 1
        return ctr