# Maximum sum of hour glass in matrix

• Difficulty Level : Easy
• Last Updated : 31 Jul, 2022

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 with
maximum sum
1 0 0
4
0 2 4```
Recommended Practice

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 an 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` `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){` `         ``cout << ``"Not possible"` `<< endl;` `          ``exit``(0);` `    ``}`   `    ``// 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

## C

 `/* C program to find the maximum` `    ``sum of hour glass in a Matrix */` `#include ` `#include ` `// 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){` `         ``printf``(``"Not Possible"``);` `          ``exit``(0);` `    ``}`   `    ``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 than 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);` `    ``printf``(``"Maximum sum of hour glass = %d"``, res);` `    ``return` `0;` `}`   `// This code is written by Akshay Prakash` `// Code is modified by Susobhan Akhuli`

## 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``){` `        ``System.out.println(``"Not possible"``);` `        ``System.exit(``0``);` `    ``}`   `    ``// 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 than ` `            ``// 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);` `        ``System.out.println(``"Maximum sum of hour glass = "``+ res);` `    ``}` `    `  `}`   `// This code is contributed by vt_m .` `// Code is modified by Susobhan Akhuli`

## Python3

 `# 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``):` `        ``print``(``"Not possible"``)` `        ``exit()`   `    ``# 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)`   `print``(f``"Maximum sum of hour glass = {res}"``)`   `# This code is written by Akshay Prakash` `# Code is modified by Susobhan Akhuli`

## 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){` `      ``Console.WriteLine(``"Not possible"``);` `      ``Environment.Exit(0);` `    ``}`   `    ``// 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 than ` `            ``// 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);` `        ``Console.WriteLine(``"Maximum sum of hour glass = "``+ res);` `    ``}` `    `  `}`   `// This code is contributed by vt_m .` `// Code is modified by Susobhan Akhuli`

## PHP

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## Javascript

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Output

`Maximum sum of hour glass = 13`

Time complexity: O(R x C).
Auxiliary Space: O(1)

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