"""
532. K-diff pairs in an Array
Medium

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
    a <= b
    b - a == k

 

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).

Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2

Example 5:

Input: nums = [-1,-2,-3], k = 1
Output: 2

 

Constraints:

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

https://leetcode.com/problems/k-diff-pairs-in-an-array/
"""


class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        nums.sort()
        pairs = 0
        for i in range(0, len(nums) - 1):
            if i != 0 and nums[i] == nums[i - 1]:
                continue
            for j in range(i + 1, len(nums)):
                if j - 1 != i and nums[j] == nums[j - 1]:
                    continue
                if nums[j] - nums[i] == k:
                    pairs += 1
                if nums[j] - nums[i] > k:
                    break
        return pairs
