# K-Diff Pairs In An Array Problem

## Description

LeetCode Problem 532.

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:
``````1
2
3
4
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:

``````1
2
3
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:

``````1
2
3
Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
``````

Example 4:

``````1
2
Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2
``````

Example 5:

``````1
2
Input: nums = [-1,-2,-3], k = 1
Output: 2
``````

Constraints:

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

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
class Solution {
public:
int findPairs(vector<int>& A, int k) {
unordered_map<int, int> m;
for (int n : A) m[n]++;
int ans = 0;
for (auto &[n, cnt] : m) {
ans += k ? m.count(n - k) : cnt > 1;
}
return ans;
}
};
``````