# 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

It is evident from definition of hour glass that number of rows and number of columns must be equal to 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 the implementation of 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) ` `        ``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

## 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 . `

## 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

 ` `

Output:

```Maximum sum of hour glass = 13
```

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Improved By : vt_m, jit_t, SrinivasaLakkaraju

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