Minimum Cuts can be made in the Chessboard such that it is not divided into 2 parts

Given M x N Chessboard. The task is to determine the Maximum numbers of cuts that we can make in the Chessboard such that the Chessboard is not divided into 2 parts.

Examples:

Input: M = 2, N = 4
Output: Maximum cuts = 3

Input: M = 3, N = 3
Output: Maximum cuts = 4


Representation:

  1. For M = 2, N = 2 We can only make 1 cut (mark in red). if we make 1 more cut then the chessboard will divide into 2 pieces.

  2. For M = 2, N = 4 We can makes 3 cuts (marks in red). if we make 1 more cut then the chessboard will divide into 2 pieces.

So, it can be observed that no. of cuts = (m-1) * (n-1).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// function that calculates the
// maximum no. of cuts
int numberOfCuts(int M, int N)
{
    int result = 0;
  
    result = (M - 1) * (N - 1);
  
    return result;
}
  
// Driver Code
int main()
{
    int M = 4, N = 4;
  
    // Calling function.
    int Cuts = numberOfCuts(M, N);
  
    cout << "Maximum cuts = " << Cuts;
  
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation of above approach 


  


class GFG { 


      


// function that calculates the 


// maximum no. of cuts 


static int numberOfCuts(int M, int N) 


{ 


    int result = 0; 


  


    result = (M - 1) * (N - 1); 


  


    return result; 


} 


  


// Driver Code 


public static void main(String args[]) 


{ 


    int M = 4, N = 4; 


  


    // Calling function. 


    int Cuts = numberOfCuts(M, N); 


  


    System.out.println("Maximum cuts = " + Cuts); 


  


} 


} 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 implementation of  


# above approach 


  


# function that calculates the 


# maximum no. of cuts 


def numberOfCuts(M, N): 


    result = 0


      


    result = (M - 1) * (N - 1) 


      


    return result 


  


# Driver code 


if __name__=='__main__': 


      


    M, N = 4, 4


      


    # Calling function. 


    Cuts = numberOfCuts(M, N) 


      


    print("Maximum cuts = ", Cuts) 


  


# This code is contributed by 


# Kriti_mangal 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















//C#  implementation of above approach  


using System; 


  


public class GFG{ 


// function that calculates the  


// maximum no. of cuts  


static int numberOfCuts(int M, int N)  


{  


    int result = 0;  


  


    result = (M - 1) * (N - 1);  


  


    return result;  


}  


  


// Driver Code  


      


    static public void Main (){ 


      


    int M = 4, N = 4;  


    // Calling function.  


    int Cuts = numberOfCuts(M, N);  


  


    Console.WriteLine("Maximum cuts = " + Cuts);  


    } 


//This code is contributed by akt_mit     


} 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?php 


// php implementation of above approach 


  


// function that calculates the 


// maximum no. of cuts 


function numberOfCuts($M, $N) 


{ 


    $result = 0; 


  


    $result = ($M - 1) * ($N - 1); 


  


    return $result; 


} 


  


// Driver Code 


$M = 4; 


$N = 4; 


  


// Calling function. 


$Cuts = numberOfCuts($M, $N); 


  


echo "Maximum cuts = ", $Cuts ; 


  


// This code is contributed by ANKITRAI1 


?> 



















                        chevron_right







                        filter_none











Output:


Maximum cuts = 9






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Naman_Garg
Check out this Author's contributed articles.
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.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  Kirti_Mangal, jit_t, AnkitRai01