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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Improved By : Kirti_Mangal, jit_t, AnkitRai01