# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```Maximum cuts = 9
```

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Improved By : Kirti_Mangal, jit_t, AnkitRai01