O(N^2) ||easy 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```\nclass Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        //remove the 2 elements\n        //sort the array\n        //simple find the array with a same difference as second array between 2 numbers\n        sort(nums1.begin(),nums1.end());\n        sort(nums2.begin(),nums2.end());\n        int mini=INT_MAX;\n        //can i make n^2 ime complexity ofcourse \n        //so selecting numbers will be in n^2 ways\n        //selecting first index to remove\n        for(int i=0;i<nums1.size()-1;i++){\n            for(int j=i+1;j<nums1.size();j++){\n                vector <int>v;\n                for(int k=0;k<nums1.size();k++){\n                    if(k!=i&&k!=j)\n                      v.push_back(nums1[k]);  \n                   }\n                   if(v.size()==1)mini=min(mini,nums2[0]-v[0]);\n                   \n                for(int l=0;l<nums2.size()-1;l++){\n                    if(v[l]-v[l+1]!=nums2[l]-nums2[l+1]){\n                        break;\n                    }\n                  if(l==nums2.size()-2){\n                    mini=min(mini,nums2[0]-v[0]);\n                  }  \n                }\n                \n                }\n            }\n            return  mini;\n    }\n};\n```
['C++']