Simple C++ solution | Brute Force.
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 * m)\n   \n\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n\n        int n = nums1.size();\n        int a = nums2[0];\n        vector<int> v;\n\n        for(int i = 0; i < n; i++){\n            v.push_back(a-nums1[i]);\n        }\n\n        sort(v.begin(), v.end());\n        int ans = 0;\n\n        for(int i = 0; i < v.size(); i++){\n           unordered_set<int> st;\n\n            int j = 0;\n            for(; j < nums2.size(); j++){\n\n                int e = nums2[j] - v[i];  //Element to be searched in nums1.\n             \n             int k = 0;\n              for(; k < nums1.size(); k++){\n                if(nums1[k] == e && st.find(k) == st.end()){\n                    st.insert(k);\n                    break;\n                }\n              }\n              if(k == nums1.size()) break;\n            }\n            if(j == nums2.size()){\n                ans = v[i];\n                return ans;\n            }\n        }\n\n        return -1;\n    }\n};\n```
['C++']