# Count tiles of dimensions 2 * 1 that can be placed in an M * N rectangular board that satisfies the given conditions

Given two integers **M** and **N**, the task is to find the minimum number of tiles of size **2 * 1** that can be placed on an M * N grid such that the following conditions are satisfied:

- Each tile must completely cover 2 squares of the board.
- No pair of tiles may overlap.
- Each tile lies must be placed entirely inside the board. It is allowed to touch the edges of the board.

If it is not possible to cover the entire board, print **-1**

Input:N = 2, M = 4Output:4Explanation:4 tiles of dimension 2 * 1. Place each tile in one column.

Input:N = 3, M = 3Output:-1

**Approach: **Follow the steps below to solve the problem

- If
**N**is even,**(N / 2) * M**tiles can be placed to cover the entire board. - If
**N**is odd, tiles of 2 * 1 tiles, since the length is odd which can not be expressed as a multiple of 2

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count tiles of dimensions` `// 2 x 1 that can be placed in a grid of` `// dimensions M * N as per given conditions` `int` `numberOfTiles(` `int` `N, ` `int` `M)` `{` ` ` `if` `(N % 2 == 1) {` ` ` `return` `-1;` ` ` `}` ` ` `// Number of tiles required` ` ` `return` `(N * 1LL * M) / 2;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 2, M = 4;` ` ` `cout << numberOfTiles(N, M);` ` ` `return` `0;` `}` |

## Java

`// Java Program to implement` `// the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to count tiles of dimensions` ` ` `// 2 x 1 that can be placed in a grid of` ` ` `// dimensions M * N as per given conditions` ` ` `static` `int` `numberOfTiles(` `int` `n, ` `int` `m)` ` ` `{` ` ` `if` `(n % ` `2` `== ` `1` `) {` ` ` `return` `-` `1` `;` ` ` `}` ` ` `// Number of tiles required` ` ` `return` `(m * n) / ` `2` `;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `n = ` `2` `, m = ` `4` `;` ` ` `System.out.println(` ` ` `numberOfTiles(n, m));` ` ` `}` `}` |

## Python

`# Python Program to implement` `# the above approach` `# Function to count tiles of dimensions` `# 2 x 1 that can be placed in a grid of` `# dimensions M * N as per given conditions` `def` `numberOfTiles(N, M):` ` ` `if` `(N ` `%` `2` `=` `=` `1` `):` ` ` `return` `-` `1` ` ` ` ` `# Number of tiles required` ` ` `return` `(N ` `*` `M) ` `/` `/` `2` `# Driver Code` `N ` `=` `2` `M ` `=` `4` `print` `(numberOfTiles(N, M))` `# This code is contributed by shubhamsingh10` |

## C#

`// C# Program to implement` `// the above approach` `using` `System;` `public` `class` `GFG {` ` ` `// Function to tiles of size 2 x 1` ` ` `// find the number of tiles that can` ` ` `// be placed as per the given conditions` ` ` `static` `int` `numberOfTiles(` `int` `n, ` `int` `m)` ` ` `{` ` ` `if` `(n % 2 == 1) {` ` ` `return` `-1;` ` ` `}` ` ` `// Number of tiles required` ` ` `return` `(m * n) / 2;` ` ` `}` ` ` `// Driver Code` ` ` `static` `public` `void` `Main()` ` ` `{` ` ` `int` `n = 2, m = 4;` ` ` `Console.WriteLine(` ` ` `numberOfTiles(n, m));` ` ` `}` `}` |

## Javascript

`<script>` `// Javascript Program to implement` `// the above approach` `// Function to count tiles of dimensions` `// 2 x 1 that can be placed in a grid of` `// dimensions M * N as per given conditions` `function` `numberOfTiles(n, m)` `{` ` ` `if` `(n % 2 == 1)` ` ` `{` ` ` `return` `-1;` ` ` `}` ` ` ` ` `// Number of tiles required` ` ` `return` `(m * n) / 2;` `}` `// Driver Code` `var` `n = 2, m = 4;` `document.write(numberOfTiles(n, m));` `// This code is contributed by kirti` `</script>` |

**Output:**

4

**Time Complexity: **O(1)**Auxiliary Space:** O(1)