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 &lt;= i, j &lt; nums.length
	i != j
	nums[i] - nums[j] == k


Notice that |val| denotes the absolute value of val.

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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).


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Constraints:


	1 &lt;= nums.length &lt;= 104
	-107 &lt;= nums[i] &lt;= 107
	0 &lt;= k &lt;= 107


class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        nums.sort()
        dict1={}
        s=0
        res=[]
        for n in nums:
            if dict1.get(n,None) is not None:
                res.append(n)
            dict1[n+k]=n
        res=list(set(res))
        return len(res)
 // O(n) Time Solution

 class Solution {
 		public int findPairs(int[] nums, int k) {
 			Map<Integer, Integer> map = new HashMap();
 			for (int num : nums)
 				map.put(num, map.getOrDefault(num, 0) + 1);

 			int result = 0;
 			for (int i : map.keySet())
 				if (k > 0 && map.containsKey(i + k) || k == 0 && map.get(i) > 1)
 					result++;
 			return result;
 		}
 	}
class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int,int> a;
        for(int i:nums)
            a[i]++;
        int ans=0;
        for(auto x:a)
        {
            if(k==0)
            {    
                if(x.second>1)
                ans++;
            }
             else if(a.find(x.first+k)!=a.end())
                ans++;
        }
        
        return ans;
    }
};
var findPairs = function(nums, k) {
	nums.sort((a, b) => b - a);
	const { length } = nums;
	const hash = new Set();
	let left = 0;
	let right = 1;

	while (left < length - 1) {
		while (right < length) {
			const diff = nums[left] - nums[right];

			diff === k && hash.add(`${nums[left]},${nums[right]}`);
			diff > k
				? right = length
				: right += 1;
		}
		left += 1;
		right = left + 1;
	}
	return hash.size;
};
K-diff Pairs in an Array