C++ || Well-Commented || O(N*N*M) || std::unordered_map || Explanation MaximumSumSubmatrix
In a nutshell the approach is to choose all pairs of rows and shrink the bundle (chosen rows  alongwith rows in between them) into an array (summing up entries 
In a nutshell the approach is to choose all pairs of rows and shrink the bundle (chosen rows  alongwith rows in between them) into an array (summing up entries in each column together) and now **finding the number of subarrays in this array whose sum is equal to the target in O(M).**\n\nThe overall complexity becomes O(N * N * M).\n\n\n```\nclass Solution {\npublic:\n    \n    //Another classical problem of finding number of subarrays having sum equal to target in the array\n    int numSubarraySumTarget(vector<int> &arr,int target){\n        int num_of_subarray=0;\n        unordered_map<int,int> mp;\n        mp.insert({0,1});\n        int curr_sum=0;\n        for(int i=0;i<arr.size();i++){\n            curr_sum+=arr[i];\n            if(mp.find(curr_sum-target)!=mp.end()) num_of_subarray+=mp[curr_sum-target];\n            mp[curr_sum]++;\n        }\n        return num_of_subarray;\n    }\n    \n    \n    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {\n        \n        int n=matrix.size();        //number of rows in the matrix\n        if(n==0) return 0;          //If there are no rows in a matrix then there is no submatrix for any target \n        int m=matrix[0].size();     //number of columns in the matrix\n        \n        int number_of_submatrices=0;\n        \n        \n        //choosing two rows adding there elements of same columns and creating the array out of it / Simply shrinking the bundle of rows to a 1-D array\n        for(int row_x1=0;row_x1<n;row_x1++){\n            vector<int> sum_array(m,0);\n            for(int row_x2=row_x1;row_x2<n;row_x2++){\n                \n                for(int i=0;i<m;i++) sum_array[i]+=matrix[row_x2][i];\n                \n                //Find the number of subarrays that equal to the target in this shrinked array. \n                number_of_submatrices+=numSubarraySumTarget(sum_array,target);\n                \n            }\n            \n        }\n        \n        return number_of_submatrices;\n    }\n};\n```\n\n***There is another variation of the same problem where we find the submatrix with the maximum sum.This can be done by following the first part of choosing rows and bundelling up similarly and then applying the kaden\'s algorithm of the array in O(M).***
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