C++||Easy||Different Approach
To be honest i dont know how this approach came to my mind.\nBut it serves the purpose.\n\n\n    vector<vector<int>> largestLocal(vector<vector<int>>& grid) {\n
To be honest i dont know how this approach came to my mind.\nBut it serves the purpose.\n\n```\n    vector<vector<int>> largestLocal(vector<vector<int>>& grid) {\n        int n=grid.size();\n        int it=0;\n        while(it<2){\n        for(int i=n-1;i>it;i--){\n            for(int j=n-1;j>it;j--){\n              grid[i][j]=max(grid[i][j],max(grid[i][j-1],max(grid[i-1][j],grid[i-1][j-1])));\n                \n                \n            }\n        }\n            it++;\n    }\n        \n     \n       vector<vector<int>> ans( n-2 , vector<int> (n-2)); \n        for(int i=2;i<n;i++){\n            for(int j=2;j<n;j++){\n                ans[i-2][j-2]=grid[i][j];\n            }\n        }\n        return ans;\n    }\n```
['C']