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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

  • |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 <= 104

  • -107 <= nums[i] <= 107

  • 0 <= k <= 107

Solution:

class Solution {
    public int findPairs(int[] nums, int k) {
        Map<Integer,Integer> map = new HashMap<>();
        for(int n:nums)
        {
          map.put(n,map.getOrDefault(n,0)+1);   
        }
        int result=0;
        for(int x:map.keySet())
        {
            if((k>0 && map.containsKey(x+k)) || (k==0 && map.get(x)>1))
             result++;  
        }
        return result;
    }
}




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