Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 3

Description

Find the Difference of Two Arrays

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

class Solution {
public:
  vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) {
    set<int> st1(nums1.begin(), nums1.end());
    set<int> st2(nums2.begin(), nums2.end());

    vector<vector<int>> answer(2);
    for(auto n1 : st1) {
      if(!st2.count(n1)) {
        answer[0].push_back(n1);
      }
    }
    for(auto n2 : st2) {
      if(!st1.count(n2)) {
        answer[1].push_back(n2);
      }
    }

    return answer;
  }
};

// Accepted
// 202/202 cases passed (60 ms)
// Your runtime beats 63.81 % of cpp submissions
// Your memory usage beats 53.88 % of cpp submissions (32.7 MB)