LeetCode-in-Java

2215. Find the Difference of Two Arrays

Easy

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]

Output: [[1,3],[4,6]]

Explanation:

For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].

For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]

Output: [[3],[]]

Explanation:

For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].

Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• -1000 <= nums1[i], nums2[i] <= 1000

Solution

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class Solution {
    public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
        Set<Integer> set1 = createSet(nums1);
        Set<Integer> set2 = createSet(nums2);
        return Arrays.asList(getMissing(set1, set2), getMissing(set2, set1));
    }

    private Set<Integer> createSet(int[] array) {
        Set<Integer> set = new HashSet<>();
        for (int x : array) {
            set.add(x);
        }
        return set;
    }

    private List<Integer> getMissing(Set<Integer> first, Set<Integer> second) {
        List<Integer> list = new ArrayList<>();
        for (int x : first) {
            if (!second.contains(x)) {
                list.add(x);
            }
        }
        return list;
    }
}

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