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++

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// 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;
}

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Java














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// 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); 


  


} 


} 



















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Python3














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# 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 



















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C#














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//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     


} 



















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PHP














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<?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 


?> 



















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Output:


Maximum cuts = 9






          
          
          
          

            

    

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