Maximum number of tiles required to cover the floor of given size using 2×1 size tiles

Given a floor of size MxN size and tiles of size 2×1, the task is to find the maximum number of tiles required to cover the floor as much as possible of size MxN size.
Note: A tile can either be placed horizontally or vertically and no two tiles should overlap.

Examples:

Input: M = 2, N = 4
Output: 4
Explanation:
4 tiles are needed to cover the floor.

Input: M = 3, N = 3
Output: 4
Explanation:
4 tiles are needed to cover the floor.

Approach:

1. If N is even, then the task is to place m rows of (N/2) number of tiles to cover the whole floor.
2. Else if N is odd, then cover M rows till N – 1 (even) columns in the same way as discussed in the above point and put (M/2) number of tiles in the last column. If both M and N are odd, then one cell of the floor remains uncovered.
3. Therefore, the maximum number of tiles is floor((M * N) / 2).

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
void maximumTiles(int n, int m)
{
    // Print the answer
    cout << (m * n) / 2 << endl;
}

// Driver Code
int main()
{
    // Given M and N
    int M = 3;
    int N = 4;

    // Function Call
    maximumTiles(N, M);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
static void maximumTiles(int n, int m)
{

    // Print the answer
    System.out.println((m * n) / 2);
}

// Driver code
public static void main (String[] args)
{

    // Given M and N
    int M = 3;
    int N = 4;

    // Function call
    maximumTiles(N, M);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to find the maximum number
# of tiles required to cover the floor
# of size m x n using 2 x 1 size tiles
def maximumTiles(n, m):

    # Print the answer
    print(int((m * n) / 2));

# Driver code
if __name__ == '__main__':

    # Given M and N
    M = 3;
    N = 4;

    # Function call
    maximumTiles(N, M);

# This code is contributed by sapnasingh4991

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
static void maximumTiles(int n, int m)
{

    // Print the answer
    Console.WriteLine((m * n) / 2);
}

// Driver code
public static void Main(String[] args)
{

    // Given M and N
    int M = 3;
    int N = 4;

    // Function call
    maximumTiles(N, M);
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// JavaScript program for the above approach

// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
function maximumTiles(n, m)
{

    // Print the answer
    document.write((m * n) / 2);
}

// Driver Code

     // Given M and N
    let M = 3;
    let N = 4;

    // Function call
    maximumTiles(N, M);

</script>

Output:

6

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