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:
- If N is even, then the task is to place m rows of (N/2) number of tiles to cover the whole floor.
- 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.
- 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)
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