k_diff_pairs_in_an_array.py

"""
Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

0 <= i, j < nums.length
i != j
|nums[i] - nums[j]| == k
Notice that |val| denotes the absolute value of val.

 

Example 1:

Input: nums = [3,1,4,1,5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:

Input: nums = [1,2,3,4,5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:

Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
 

Constraints:

1 <= nums.length <= 104
-107 <= nums[i] <= 107
0 <= k <= 107

"""

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        if k < 0:
            return 0

        freq_dict = collections.Counter(nums)
        pairs = 0

        for num, freq in freq_dict.items():
            if (k == 0 and freq > 1) or (k != 0 and num + k in freq_dict):
                pairs = pairs + 1
        return pairs