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Maximum sum of cocktail glass in a 2D matrix

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  • Last Updated : 09 Jun, 2022
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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 

 

Examples: 
 

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 <bits/stdc++.h>
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




<?PHP
// PHP implementation of the approach
$R = 5;
$C = 5;
 
// Function to return the maximum sum
// of a cocktail glass
function findMaxCock($ar)
{
    global $R, $C;
     
    // If no cocktail glass is possible
    if ($R < 3 || $C < 3)
        return -1;
 
    // Initialize max_sum with the mini
    $max_sum = PHP_INT_MIN;
 
    // Here loop runs (R-2)*(C-2) times considering
    // different top left cells of cocktail glasses
    for ($i = 0; $i < $R - 2; $i++)
    {
        for ($j = 0; $j < $C - 2; $j++)
        {
 
            // 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
$ar = array(array( 0, 3, 0, 6, 0 ),
            array( 0, 1, 1, 0, 0 ),
            array( 1, 1, 1, 0, 0 ),
            array( 0, 0, 2, 0, 1 ),
            array( 0, 2, 0, 1, 3 ));
 
echo(findMaxCock($ar));
 
// This code is contributed by Code_Mech
?>

Javascript




<script>
 
// Javascript implementation of the approach
var R = 5;
var C = 5;
 
// Function to return the maximum sum
// of a cocktail glass
function findMaxCock(ar)
{
 
    // If no cocktail glass is possible
    if (R < 3 || C < 3)
        return -1;
 
    // Initialize max_sum with the mini
    var max_sum = -1000000000;
 
    // Here loop runs (R-2)*(C-2) times considering
    // different top left cells of cocktail glasses
    for (var i = 0; i < R - 2; i++) {
        for (var j = 0; j < C - 2; j++) {
 
            // Considering mat[i][j] as the top left
            // cell of the cocktail glass
            var 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
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 ] ];
                 
document.write(findMaxCock(ar));
 
</script>

Output: 

12

 

Time Complexity : O(R * C)
Auxiliary Space: O(1), since no extra space has been taken.


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