K-diff Pairs in an Array

Problem

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:
Input: [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:[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: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

• The pairs (i, j) and (j, i) count as the same pair.
• The length of the array won't exceed 10,000.
• All the integers in the given input belong to the range: [-1e7, 1e7].

Solution

HashMap Solution: O(n) time O(n) space

public class Solution {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length < 2) return 0; 
        if (k < 0) return 0; //k is absolute difference 
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) map.put(num, map.getOrDefault(num, 0) + 1);
        int res = 0;
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (k == 0 && entry.getValue() >= 2) res++; //We only count the unique pairs
            if (k != 0 && map.containsKey(entry.getKey() + k)) res++;
        }
        return res;
    }
}

Analysis

A typical solution using HashMap, similar to the problem 2Sum
We first inflate the map with each element as key, and its frequency as value
Then we loop through the Map.Entry because we only count the unique pairs
We divide into two situations
When k == 0 it's just to find same number appearing more than twice
Notice that we just update res++ because of unique pairs
When k != 0 we update res as long as we find map.containsKey(entry.getKey() + key)
Because we do + key we guarantee the pairs counted are all unique