[LeetCode] 532. K-diff Pairs in an Array - 문제풀이

Description

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 ofunique 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

Solution 1. Hash Table

public int findPairs(int[] nums, int k) {
    
    int len = nums.length;
    Map<Integer,Integer> map = new HashMap<>();
    int cnt = 0;
    
    for(int num : nums){
        map.put(num,map.getOrDefault(num,0)+1);
    }
    
    for(Integer key : map.keySet()){
        if(k==0){
            if(map.get(key) > 1){
                cnt++;
            }
        }else{
            if (map.containsKey(key + k)) {
                cnt++;
            }
        }
    }        

    return cnt;
}

map을 Hash Table로 이용하여 각 요소별 cnt를 기록하고 각 요소와 fair( +k)e되는 요소가 있을 경우 cnt++해줍니다. 0일경우는 자기작신은 제외해야하므로 2개이상인 요소만 cnt에 추가합니다.

Reference

K-diff Pairs in an Array - LeetCode