532. K-diff Pairs in an Array

Description

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

k-diff-pairs-in-an-array.py

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        s = Counter(nums)
        res = 0

        for x in s:
            if k == 0 and s[x] >= 2 or k != 0 and k + x in s:
                res += 1

        return res