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## Problem

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 <= 10^4
• -10^7 <= nums[i] <= 10^7
• 0 <= k <= 10^7

## Code

class Solution {
public int findPairs(int[] nums, int k) {
HashMap<Integer, Integer> map = new HashMap<>();

for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}

int res = 0;
for (int key : map.keySet()) {
int num1 = key - k;
int num2 = key + k;

if (k == 0) {
if (map.get(key) >= 2) {
res += 2;
}
} else {
if (map.containsKey(num1)) res++;
if (map.containsKey(num2)) res++;
}
}

return res / 2;
}
}