Easy solution -beats 100%
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:$$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {\n        int count1 = 0 ;\n        int count2 = 0;\n        int n = nums1.size() ;\n        int m = nums2.size() ;\n        for (int i=0; i<n; i++){\n            for (int j=0; j<m; j++){\n                if (nums1[i] == nums2[j]){\n                    count1++ ;\n                    break ;\n                }\n            }\n        }\n        for (int j=0; j<m; j++){\n            for (int i=0; i<n; i++){\n                if (nums2[j] == nums1[i]){\n                    count2++ ;\n                    break ;\n                }\n            }\n        }\n        vector<int> answer ; \n        answer.push_back(count1) ;\n        answer.push_back(count2) ;\n        return answer ;\n    }\n};\n```
['C++']