K-diff Pairs in an Array

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

SOLUTION:

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        exists = {}
        ctr = set()
        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.add(tuple(sorted([nums[i], nums[j]])))
        return len(ctr)