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K-diff Pairs in an Array

Problem Statement​

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).

Contraints:​

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

Leetcode Link

Code​

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

        if k==0:
            for key,v in c.items():
                if v>1:
                    cnt+=1
        else:
            for key,v in c.items():
                if key+k in c:
                    cnt+=1
        return cnt