Given an array of integersnumsand an integerk, returnthe number ofuniquek-diff pairs in the array.

Ak-diffpair isan 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

two pointers,time = O(n), space = O(n)

class Solution {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k < 0) {
            
 
return 0;
        }
        Map<Integer, Integer> map = new HashMap<>();
        int count = 0;
        for (int i : nums) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (k == 0) {
                if (entry.getValue() >= 2) {
                    count++;
                } 
            } else {
                if (map.containsKey(entry.getKey() + k)) {
                    count++;
                }
            }
        }
        return count;
    }
}