Easy Solution | C++
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```\nclass Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        int n=nums1.size();\n        sort(nums1.begin(),nums1.end());\n        sort(nums2.begin(),nums2.end());\n        \n        int ans=INT_MAX,res=INT_MAX;\n        map<int,int>mp;\n        for(auto x:nums2)mp[x]++;\n        \n        int left=-nums1[n-1]+nums2[0];\n        int right=nums2[n-3]-nums1[0];\n        \n        for(int i=left;i<=right;i++){\n            map<int,int>mp1;\n            for(int j=0;j<n;j++){\n                int x=nums1[j]+i;\n                mp1[x]++;\n                \n            }\n            int cnt=0;\n            for(auto [x,y]:mp){\n                if(mp1[x]<y)cnt++;\n            }\n            if(cnt==0)return i;\n            mp1.clear();\n        }\n            \n        return ans;\n    }\n};\n```
['C++']