0532 - K-diff Pairs in an Array (Medium)

Problem Link

https://leetcode.com/problems/k-diff-pairs-in-an-array/

Problem Statement

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 <= 10^4
• -10^7 <= nums[i] <= 10^7
• 0 <= k <= 10^7

Approach 1: Hash Map

Let's build a frequency hash map $m$. The key $x$ is each unique number and the value $y$ is the occurrence of that number. For example, for input $[1,3,1,5,4]$, we will have the following $(x, y)$:

key	value
1	2
3	1
4	1
5	1

We iterate $m$ to check each $(x, y)$. There are two cases:

1. If $k$ is $0$, that means $nums[i] == nums[j]$, so we need at least two occurrences of that number.
2. If $k$is greater than $0$, then we need to check if the complement $x + k$ exists in hash map or not.

Written by @wingkwong

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int, int> m;
        int ans = 0;
        for (auto x : nums) m[x]++;
        for (auto x : m) {
            // if k = 0, the element in the pair must be equal
            // if k > 0, check if b = a + k exists
            ans += (!k && x.second > 1) || (k && m.count(x.first + k));
        }
        return ans;
    }
};