532. K-diff Pairs in an Array

Problem Link

Description

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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

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Python3

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