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

**Solution:**

```
class Solution {
public int findPairs(int[] nums, int k) {
Map<Integer,Integer> map = new HashMap<>();
for(int n:nums)
{
map.put(n,map.getOrDefault(n,0)+1);
}
int result=0;
for(int x:map.keySet())
{
if((k>0 && map.containsKey(x+k)) || (k==0 && map.get(x)>1))
result++;
}
return result;
}
}
```

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