# 532. K-diff Pairs in an Array

#### Medium

***

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`.

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**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).
```

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**Constraints:**

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

```python
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
```