532. K-diff Pairs in an Array

Medium

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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

• |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

from collections import defaultdict
class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        result = set()
        d = self.get_hashmap(nums)
        for num in nums:
            diff = num-k
            if diff in d:
                if (diff == num and d[diff] > 1) or diff != num:
                    result.add((diff,num) if diff < num else (num, diff))
        return len(result)
        
    def get_hashmap(self, nums):
        d = defaultdict(int)
        for num in nums:
            d[num] += 1
        return d