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Maximum sum of hour glass in matrix
  • Difficulty Level : Easy
  • Last Updated : 11 Feb, 2021

Given a 2D matrix, the task is to find the maximum sum of an hourglass.

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

Approach:
It is evident from the definition of the hourglass that the number of rows and number of columns must be equal to 3. If we count the total number of hourglasses in a matrix, we can say that the count is equal to the count of possible top left cells in an hourglass. The number of top-left cells in an hourglass is equal to (R-2)*(C-2). Therefore, in a matrix total number of hourglass 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

Consider all top left cells of hourglasses one by one. For every cell, we compute the sum of the hourglass formed by it. Finally, return the maximum sum.
Below is the implementation of the above idea:

C++




// 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;
}


C




/* C program to find the maximum
    sum of hour glass in a Matrix */
#include <stdio.h>
 
// Fixing the size of the matrix
// ( Here it is of the order 6
// x 6 )
#define R 5
#define C 5
 
// Function to find the maximum
// sum of the hour glass
int MaxSum(int arr[R][C])
{
    int i, j, sum;
    if (R < 3 || C < 3)
        return -1;
 
    int max_sum
        = -500000; /* Considering the matrix also contains
                      negative values , so initialized with
                      -50000. It can be any value but very
                      smaller.*/
 
    // int max_sum=0 -> Initialize with 0 only if your
    // matrix elements are positive
 
    // Here loop runs (R-2)*(C-2) times considering
    // different top left cells of hour glasses.
    for (i = 0; i < R - 2; i++) {
        for (j = 0; j < C - 2; j++) {
            // Considering arr[i][j] as top left cell of
            // hour glass.
            sum = (arr[i][j] + arr[i][j + 1]
                   + arr[i][j + 2])
                  + (arr[i + 1][j + 1])
                  + (arr[i + 2][j] + arr[i + 2][j + 1]
                     + arr[i + 2][j + 2]);
 
            // If previous sum is less then current sum then
            // update new sum in max_sum
            if (sum > max_sum)
                max_sum = sum;
            else
                continue;
        }
    }
    return max_sum;
}
 
// Driver Code
int main()
{
    int arr[][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 = MaxSum(arr);
    if (res == -1)
        printf("Not Possible");
    else
        printf("Maximum sum of hour glass = %d", res);
    return 0;
}
 
// This code is written by Akshay Prakash


Java




// Java program to find maximum
// sum of hour glass in matrix
import java.io.*;
 
class GFG {
     
static int R = 5;
static int C = 5;
 
// Returns maximum sum of
// hour glass in ar[][]
static int findMaxSum(int [][]mat)
{
    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 = Integer.MIN_VALUE;
    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 = Math.max(max_sum, sum);
        }
    }
    return max_sum;
}
 
    // Driver code
    static public void main (String[] args)
    {
        int [][]mat = {{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)
            System.out.println("Not possible");
        else
            System.out.println("Maximum sum of hour glass = "
                                + res);
    }
     
}
 
// This code is contributed by vt_m .


Python




# Python 3 program to find the maximum
# sum of hour glass in a Matrix
 
# Fixing the size of the Matrix.
# Here it is of order 6 x 6
R = 5
C = 5
 
# Function to find the maximum sum of hour glass
def MaxSum(arr):
 
    # Considering the matrix also contains
    max_sum = -50000 
     
    # Negative values , so initialized with
    # -50000. It can be any value but very
    # smaller.
    # max_sum=0 -> Initialize with 0 only if your
    # matrix elements are positive
 
    if(R < 3 or C < 3):
        return -1
 
    # Here loop runs (R-2)*(C-2) times considering
    # different top left cells of hour glasses.
    for i in range(0, R-2):
        for j in range(0, C-2):
                     
            # Considering arr[i][j] as top
            # left cell of hour glass.
            SUM = (arr[i][j] + arr[i][j + 1] +
                   arr[i][j + 2]) + (arr[i + 1][j + 1]) +
                                          (arr[i + 2][j] +
                    arr[i + 2][j + 1] + arr[i + 2][j + 2])
 
            # If previous sum is less
            # then current sum then
            # update new sum in max_sum
            if(SUM > max_sum):
                max_sum = SUM
            else:
                continue
 
    return max_sum
 
 
# Driver Code
arr = [[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]]
res = MaxSum(arr)
 
if(res == -1):
    print("Not possible")
else:
    print(f"Maximum sum of hour glass = {res}")
 
# This code is written by Akshay Prakash


C#




// C# program to find maximum
// sum of hour glass in matrix
using System;
 
class GFG {
     
static int R = 5;
static int C = 5;
 
// Returns maximum sum of
// hour glass in ar[][]
static int findMaxSum(int [,]mat)
{
    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.MinValue;
    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 = Math.Max(max_sum, sum);
        }
    }
    return max_sum;
}
 
    // Driver code
    static public void Main(String[] args)
    {
        int [,]mat = {{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)
            Console.WriteLine("Not possible");
        else
            Console.WriteLine("Maximum sum of hour glass = "
                               + res);
    }
     
}
 
// This code is contributed by vt_m .


PHP




<?php
// PHP program to find maximum sum
// of hour glass in matrix
$R = 5;
$C = 5;
 
// Returns maximum sum
// of hour glass in ar[][]
function findMaxSum($mat)
{
    global $R; global $C;
    if ($R < 3 || $C < 3)
        return -1;
 
    // Here loop runs (R-2)*(C-2) times considering
    // different top left cells of hour glasses.
    $max_sum = PHP_INT_MIN;
    for ($i = 0; $i < ($R - 2); $i++)
    {
        for ($j = 0; $j < ($C - 2); $j++)
        {
            // Considering mat[i][j] as
            // top left cell of hour glass.
            $sum = ($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 than current sum
            // then update new sum in max_sum
            $max_sum = max($max_sum, $sum);
        }
    }
    return $max_sum;
}
 
// Driver code
$mat = array(array(1, 2, 3, 0, 0),
             array(0, 0, 0, 0, 0),
             array(2, 1, 4, 0, 0),
             array(0, 0, 0, 0, 0),
             array(1, 1, 0, 1, 0));
$res = findMaxSum($mat);
if ($res == -1)
    echo "Not possible", "\n";
else
    echo "Maximum sum of hour glass = ",
         $res, "\n";
 
// This code is contributed by ajit.
?>


Output: 

Maximum sum of hour glass = 13

Reference : 
http://stackoverflow.com/questions/38019861/hourglass-sum-in-2d-array
This article is contributed by AKSHAY PRAKASH. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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