K-diff Pairs in an Array

k_diff_pairs_in_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
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
"""
from typing import List


class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        if not nums:
            return 0
        nums.sort()
        left, right, n, res = 0, 1, len(nums), 0
        while left < n and right < n:
            if left == right or nums[right] - nums[left] < k:
                right += 1
            elif nums[right] - nums[left] > k:
                left += 1
            else:
                left += 1
                res += 1
                while left < n and nums[left] == nums[left - 1]:
                    left += 1
        return res