LeetCode-in-Java

532. K-diff Pairs in an Array

Medium

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

import java.util.HashSet;

public class Solution {
    public int findPairs(int[] nums, int k) {
        int res = 0;
        HashSet<Integer> set = new HashSet<>();
        HashSet<Integer> twice = new HashSet<>();
        for (int n : nums) {
            if (set.contains(n)) {
                if (k == 0 && !twice.contains(n)) {
                    res++;
                    twice.add(n);
                } else {
                    continue;
                }
            } else {
                if (set.contains(n - k)) {
                    res++;
                }
                if (set.contains(n + k)) {
                    res++;
                }
            }
            set.add(n);
        }
        return res;
    }
}