2215

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

Companies:
Google

Related Topics:
Hash Table

Similar Questions:

• Intersection of Two Arrays (Easy)
• Intersection of Two Arrays II (Easy)

Solution 1. Hash Set

// OJ: https://leetcode.com/problems/find-the-difference-of-two-arrays/
// Author: github.com/lzl124631x
// Time: O(A + B)
// Space: O(A + B)
class Solution {
public:
    vector<vector<int>> findDifference(vector<int>& A, vector<int>& B) {
        unordered_set<int> sa(begin(A), end(A)), sb(begin(B), end(B));
        vector<vector<int>> ans(2);
        for (int n : sa) {
            if (sb.count(n) == 0) ans[0].push_back(n);
        }
        for (int n : sb) {
            if (sa.count(n) == 0) ans[1].push_back(n);
        }
        return ans;
    }
};

Discuss

https://leetcode.com/problems/find-the-difference-of-two-arrays/discuss/1886960/