Given a 2D matrix, the task is that we find maximum sum of a hour glass.

An hour glass is made of 7 cells in following form. A B C D E F G

Examples:

Input : 1 1 1 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 Output : 7 Below is the hour glass with maximum sum: 1 1 1 1 1 1 1 Input : 0 3 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 2 4 4 0 0 0 2 4 Output : 11 Below is the hour glass wuth maximum sum 1 0 0 4 0 2 4

It is evident from definition of hour glass that number of rows and number of columns must be greater than 3. If we count total number of hour glasses in a matrix, we can say that the count is equal to count of possible top left cells in hour glass. Number of top-left cells in a hour glass is equal to (R-2)*(C-2). Therefore, in a matrix total number of hour glass is ** (R-2)*(C-2) **

mat[][] = 2 3 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 2 4 4 0 0 0 2 0 Possible hour glass are : 2 3 0 3 0 0 0 0 0 1 0 0 1 1 1 1 1 0 1 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 2 0 2 4 2 4 4 1 1 1 1 1 0 1 0 0 0 2 4 0 0 0 0 0 2 0 2 0

We consider all top left cells of hour glasses one by one. For every cell, we compute sum of hour glass formed by it. Finally we return maximum sum.

Below is C++ implementation of above idea.

// C++ program to find maximum sum of hour // glass in matrix #include<bits/stdc++.h> using namespace std; const int R = 5; const int C = 5; // Returns maximum sum of hour glass in ar[][] int findMaxSum(int mat[R][C]) { if (R<3 || C<3) return -1; // Here loop runs (R-2)*(C-2) times considering // different top left cells of hour glasses. int max_sum = INT_MIN; for (int i=0; i<R-2; i++) { for (int j=0; j<C-2; j++) { // Considering mat[i][j] as top left cell of // hour glass. int sum = (mat[i][j]+mat[i][j+1]+mat[i][j+2])+ (mat[i+1][j+1])+ (mat[i+2][j]+mat[i+2][j+1]+mat[i+2][j+2]); // If previous sum is less then current sum then // update new sum in max_sum max_sum = max(max_sum, sum); } } return max_sum; } // Driver code int main() { int mat[][C] = {{1, 2, 3, 0, 0}, {0, 0, 0, 0, 0}, {2, 1, 4, 0, 0}, {0, 0, 0, 0, 0}, {1, 1, 0, 1, 0}}; int res = findMaxSum(mat); if (res == -1) cout << "Not possible" << endl; else cout << "Maximum sum of hour glass = " << res << endl; return 0; }

Output:

Maximum sum of hour glass = 13

**Reference : **

http://stackoverflow.com/questions/38019861/hourglass-sum-in-2d-array

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