easy 2 steps solution using nested loop
Intuition\n we have to go through every index of nums1 vector and check weather nums1[i] exist in nums2 or not, and vice verca \n\n# Approach\n to do that, we w
# Intuition\n<!-- we have to go through every index of nums1 vector and check weather nums1[i] exist in nums2 or not, and vice verca -->\n\n# Approach\n<!-- to do that, we will use nested loop, first nested loop will search for nums1[i] in nus2 and the second nested loop will search nums2[i] in nums1 -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\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        vector<int> answer1;\n        vector<int> answer2;\n        int n = nums1.size();\n        int m = nums2.size();\n        int count1 = 0, count2 = 0;\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        return {count1, count2};\n    }\n};\n```
['C++']