# leetcode\_532

## Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

```
Example 1:

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

## Note:

* The pairs (i, j) and (j, i) count as the same pair.
* The length of the array won't exceed 10,000.
* All the integers in the given input belong to the range: \[-1e7, 1e7].

## Solutions

1. **hash map**
2. Since every pair will be encountered twice, we only count one of them.

```cpp
class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if (k < 0) return 0;
        unordered_map<int, int> m;
        for (auto n : nums)
            m[n]++;

        int count = 0;
        for (auto & [n, c] : m)
            if (k == 0)
                count += c > 1;
            // for each smaller one, only count once
            else if (m.count(n + k))
                count++;
        return count;
    }
};
```

1. **hashset**
2. Borrowed from others
3. Store the lower one of the valid pair to permit duplicate pairs.

```cpp
class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if (k < 0) return 0;
        int count = 0;
        unordered_set<int> seen;
        unordered_set<int> pair;

        for (auto & num : nums) {
            // only store the lower one
            if (seen.count(num + k))
                pair.insert(num);
            if (seen.count(num - k))
                pair.insert(num - k);
            seen.insert(num);
        }

        return pair.size();
    }
};
```

1. **sort**