# Maximum sum of cocktail glass in a 2D matrix

• Last Updated : 05 Aug, 2021

Given a 2D matrix mat[][], the task is to find the maximum sum of a cocktail glass.

```A Cocktail glass is made of 6 cells in the following form:
X   X
X
X X X ```

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```Input: mat[][] = {
{1, 0, 4, 0, 0},
{0, 3, 0, 0, 0},
{1, 1, 1, 0, 0},
{0, 0, 0, 0, 0},
{0, 0, 0, 0, 0}}
Output: 11
Below is the cocktail glass with
maximum sum:
1   4
3
1 1 1

Input: mat[][] = {
{0, 3, 0, 6, 0},
{0, 1, 1, 0, 0},
{1, 1, 1, 0, 0},
{0, 0, 2, 0, 1},
{0, 2, 0, 1, 3}}
Output: 12```

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

```For mat[][] = {
{0, 3, 0, 6, 0},
{0, 1, 1, 0, 0},
{1, 1, 1, 0, 0},
{0, 0, 2, 0, 1},
{0, 2, 0, 1, 3}}
Possible cocktail glasses are:
0   0  3   6   0   0
1      1       0
1 1 1  1 1 0   1 0 0

0   1  1   0  1   0
1      1      0
0 0 2  0 2 0  2 0 1

1   1  1   0  1   0
0      2      0
0 2 0  2 0 1  0 1 3```

We consider all top left cells of cocktail glasses one by one. For every cell, we compute the sum of the cocktail glass formed by it. Finally, we return the maximum sum.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``const` `int` `R = 5;``const` `int` `C = 5;` `// Function to return the maximum sum``// of a cocktail glass``int` `findMaxCock(``int` `ar[R][C])``{` `    ``// If no cocktail glass is possible``    ``if` `(R < 3 || C < 3)``        ``return` `-1;` `    ``// Initialize max_sum with the mini``    ``int` `max_sum = INT_MIN;` `    ``// Here loop runs (R-2)*(C-2) times considering``    ``// different top left cells of cocktail glasses``    ``for` `(``int` `i = 0; i < R - 2; i++) {``        ``for` `(``int` `j = 0; j < C - 2; j++) {` `            ``// Considering mat[i][j] as the top left``            ``// cell of the cocktail glass``            ``int` `sum = (ar[i][j] + ar[i][j + 2])``                      ``+ (ar[i + 1][j + 1])``                      ``+ (ar[i + 2][j] + ar[i + 2][j + 1]``                         ``+ ar[i + 2][j + 2]);` `            ``// Update the max_sum``            ``max_sum = max(max_sum, sum);``        ``}``    ``}``    ``return` `max_sum;``}` `// Driver code``int` `main()``{``    ``int` `ar[][C] = { { 0, 3, 0, 6, 0 },``                    ``{ 0, 1, 1, 0, 0 },``                    ``{ 1, 1, 1, 0, 0 },``                    ``{ 0, 0, 2, 0, 1 },``                    ``{ 0, 2, 0, 1, 3 } };` `    ``cout << findMaxCock(ar);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `static` `int` `R = ``5``;``static` `int` `C = ``5``;` `// Function to return the maximum sum``// of a cocktail glass``static` `int` `findMaxCock(``int` `ar[][])``{` `    ``// If no cocktail glass is possible``    ``if` `(R < ``3` `|| C < ``3``)``        ``return` `-``1``;` `    ``// Initialize max_sum with the mini``    ``int` `max_sum = Integer.MIN_VALUE;` `    ``// Here loop runs (R-2)*(C-2) times considering``    ``// different top left cells of cocktail glasses``    ``for` `(``int` `i = ``0``; i < R - ``2``; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < C - ``2``; j++)``        ``{` `            ``// Considering mat[i][j] as the top left``            ``// cell of the cocktail glass``            ``int` `sum = (ar[i][j] + ar[i][j + ``2``])``                    ``+ (ar[i + ``1``][j + ``1``])``                    ``+ (ar[i + ``2``][j] + ar[i + ``2``][j + ``1``]``                        ``+ ar[i + ``2``][j + ``2``]);` `            ``// Update the max_sum``            ``max_sum = Math.max(max_sum, sum);``        ``}``    ``}``    ``return` `max_sum;``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `    ``int` `ar[][] = { { ``0``, ``3``, ``0``, ``6``, ``0` `},``                    ``{ ``0``, ``1``, ``1``, ``0``, ``0` `},``                    ``{ ``1``, ``1``, ``1``, ``0``, ``0` `},``                    ``{ ``0``, ``0``, ``2``, ``0``, ``1` `},``                    ``{ ``0``, ``2``, ``0``, ``1``, ``3` `} };` `    ``System.out.println(findMaxCock(ar));``}``}` `// This code is contributed by mits`

## Python3

 `# Python 3 implementation of the approach``import` `sys` `R ``=` `5``C ``=` `5` `# Function to return the maximum sum``# of a cocktail glass``def` `findMaxCock(ar):``    ` `    ``# If no cocktail glass is possible``    ``if` `(R < ``3` `or` `C < ``3``):``        ``return` `-``1` `    ``# Initialize max_sum with the mini``    ``max_sum ``=` `-``sys.maxsize ``-` `1` `    ``# Here loop runs (R-2)*(C-2) times considering``    ``# different top left cells of cocktail glasses``    ``for` `i ``in` `range``(R ``-` `2``):``        ``for` `j ``in` `range``(C ``-` `2``):``            ` `            ``# Considering mat[i][j] as the top left``            ``# cell of the cocktail glass``            ``sum` `=` `((ar[i][j] ``+` `ar[i][j ``+` `2``]) ``+``                   ``(ar[i ``+` `1``][j ``+` `1``]) ``+``                   ``(ar[i ``+` `2``][j] ``+` `ar[i ``+` `2``][j ``+` `1``] ``+``                    ``ar[i ``+` `2``][j ``+` `2``]))` `            ``# Update the max_sum``            ``max_sum ``=` `max``(max_sum, ``sum``)` `    ``return` `max_sum;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``ar ``=` `[[``0``, ``3``, ``0``, ``6``, ``0``],``          ``[``0``, ``1``, ``1``, ``0``, ``0``],``          ``[``1``, ``1``, ``1``, ``0``, ``0``],``          ``[``0``, ``0``, ``2``, ``0``, ``1``],``          ``[``0``, ``2``, ``0``, ``1``, ``3``]]` `    ``print``(findMaxCock(ar))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``static` `int` `R = 5;``    ``static` `int` `C = 5;``    ` `    ``// Function to return the maximum sum``    ``// of a cocktail glass``    ``static` `int` `findMaxCock(``int` `[,]ar)``    ``{``    ` `        ``// If no cocktail glass is possible``        ``if` `(R < 3 || C < 3)``            ``return` `-1;``    ` `        ``// Initialize max_sum with the mini``        ``int` `max_sum = ``int``.MinValue;``    ` `        ``// Here loop runs (R-2)*(C-2) times considering``        ``// different top left cells of cocktail glasses``        ``for` `(``int` `i = 0; i < R - 2; i++)``        ``{``            ``for` `(``int` `j = 0; j < C - 2; j++)``            ``{``    ` `                ``// Considering mat[i][j] as the top left``                ``// cell of the cocktail glass``                ``int` `sum = (ar[i,j] + ar[i,j + 2])``                        ``+ (ar[i + 1,j + 1])``                        ``+ (ar[i + 2,j] + ar[i + 2,j + 1]``                            ``+ ar[i + 2,j + 2]);``    ` `                ``// Update the max_sum``                ``max_sum = Math.Max(max_sum, sum);``            ``}``        ``}``        ``return` `max_sum;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``    ` `        ``int` `[,]ar = { { 0, 3, 0, 6, 0 },``                        ``{ 0, 1, 1, 0, 0 },``                        ``{ 1, 1, 1, 0, 0 },``                        ``{ 0, 0, 2, 0, 1 },``                        ``{ 0, 2, 0, 1, 3 } };``    ` `        ``Console.WriteLine(findMaxCock(ar));``    ``}``}` `// This code is contributed by Ryuga`

## PHP

 ``

## Javascript

 ``
Output:
`12`

Time Complexity : O(R * C)
Auxiliary Space: O(1)

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