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++

// 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;
}

Java

// 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);

}
}

Python3

# 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

C#

//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
}

PHP

<?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
?>

Javascript

<script>

// Javascript implementation of above approach

// function that calculates the
// maximum no. of cuts
function numberOfCuts(M, N)
{
    var result = 0;

    result = (M - 1) * (N - 1);

    return result;
}

// Driver Code
var M = 4, N = 4;

// Calling function.
var Cuts = numberOfCuts(M, N);
document.write( "Maximum cuts = " + Cuts);

</script>

Output:

Maximum cuts = 9

Time Complexity: O(1)

Auxiliary Space: O(1)