Maximum sum of hour glass in matrix

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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