day-186.cpp

/*
  K-diff Pairs in an Array


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
a <= b
b - a == k


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).
Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2
Example 5:

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


Constraints:

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


*/

// O(N) solution using hash map
class Solution {
   public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int, int> hashMap;
        int N = nums.size();
        for (int idx = 0; idx < N; idx++) {
            hashMap[nums[idx]] += 1;
        }

        int answer = 0;
        for (auto [key, _] : hashMap) {
            if ((k > 0 && hashMap.count(key - k)) ||
                (k == 0 && hashMap[key] >= 2)) {
                answer += 1;
            }
        }

        return answer;
    }
};