Cleanest solution so far || very simple
Intuition\nsort both arrays . after transformation , \nnums1[0] or nums1[1] or nums1[2]  are only possible options who could have transformed to nums2[0] ,becau
# Intuition\nsort both arrays . after transformation , \nnums1[0] or nums1[1] or nums1[2]  are only possible options who could have transformed to nums2[0] ,because atmost only 2 elements can be deleted . so only 3 possible x are there.\nCheck for each x if they could tranform nums1 to nums2 under given conditions .\n\n\n# Code\n```\nclass Solution {\npublic:\n    int n , m;\n    bool isok(vector<int>& nums1,vector<int>& nums2,int x){\n        int i = 0 , j = 0;\n        while(i<n && j<m){\n            if(nums1[i]+x == nums2[j]) j++;\n            i++;\n        }\n        if(j==m) return true;\n        return false;\n    }\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        n = nums1.size(),m = nums2.size();\n        sort(nums1.begin(),nums1.end());\n        sort(nums2.begin(),nums2.end());\n        int ans = INT_MAX;\n        for(int i=0;i<3;i++) {\n            if(isok(nums1,nums2,nums2[0]-nums1[i])){\n                ans=min(ans,(nums2[0]-nums1[i]));\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']