c++ solution
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:\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```\n\nclass Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        // Sorting both vectors\n        sort(nums1.begin(), nums1.end());\n        sort(nums2.begin(), nums2.end());\n\n        // Counting occurrences of differences between nums2 and nums1 elements\n        unordered_map<int, int> count_x;\n        int min_possible_x = INT_MAX;\n\n        for (int i = 0; i < nums1.size(); ++i) {\n            for (int j = 0; j < nums2.size(); ++j) {\n                int x = nums2[j] - nums1[i];\n                count_x[x]++;\n            }\n        }\n\n        // Checking for each unique difference if it can be the minimum added integer\n        for (auto& [x, count] : count_x) {\n            // Transforming nums1 by adding x to each element\n            vector<int> transformed(nums1.size());\n            transform(nums1.begin(), nums1.end(), transformed.begin(), [&](int num) { return num + x; });\n            sort(transformed.begin(), transformed.end());\n\n            // Checking if transformed nums1 can match nums2 with at most 2 mismatches\n            if (canMatchMinusTwo(transformed, nums2)) {\n                min_possible_x = min(min_possible_x, x);\n            }\n        }\n\n        return min_possible_x;\n    }\n\n    // Function to check if transformed nums1 can match nums2 with at most 2 mismatches\n    bool canMatchMinusTwo(vector<int>& transformed, vector<int>& nums2) {\n        int i = 0, j = 0, skip = 2;\n\n        // Iterating over both vectors, allowing at most 2 mismatches\n        while (i < transformed.size() && j < nums2.size()) {\n            if (transformed[i] == nums2[j]) {\n                i++;\n                j++;\n            } else {\n                i++;\n                skip--;\n                if (skip < 0) {\n                    return false;\n                }\n            }\n        }\n\n        // Returning true if transformed nums1 matches nums2 with at most 2 mismatches\n        return j == nums2.size() && (i - j) <= 2;\n    }\n};\n```
['C++']