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Tiling Problem

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  • Difficulty Level : Easy
  • Last Updated : 18 Jul, 2022
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Given a “2 x n” board and tiles of size “2 x 1”, count the number of ways to tile the given board using the 2 x 1 tiles. A tile can either be placed horizontally i.e., as a 1 x 2 tile or vertically i.e., as 2 x 1 tile. 

Examples: 

Input: n = 4

Output: 5

Explanation:

For a 2 x 4 board, there are 5 ways

  • All 4 vertical (1 way)
  • All 4 horizontal (1 way)
  • 2 vertical and 2 horizontal (3 ways)

Input: n = 3

Output: 3

Explanation:

We need 3 tiles to tile the board of size  2 x 3.

We can tile the board using following ways

  • Place all 3 tiles vertically.
  • Place 1 tile vertically and remaining 2 tiles horizontally (2 ways)

 

tilingproblem

 

Implementation – 

Let “count(n)” be the count of ways to place tiles on a “2 x n” grid, we have following two ways to place first tile. 
1) If we place first tile vertically, the problem reduces to “count(n-1)” 
2) If we place first tile horizontally, we have to place second tile also horizontally. So the problem reduces to “count(n-2)” 
Therefore, count(n) can be written as below. 

   count(n) = n if n = 1 or n = 2
   count(n) = count(n-1) + count(n-2)

Here’s the code for the above approach:

C++




// C++ program to count the
// no. of ways to place 2*1 size
// tiles in 2*n size board.
#include <iostream>
using namespace std;
 
int getNoOfWays(int n)
{
    // Base case
    if (n <= 2)
      return n;
 
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
}
 
// Driver Function
int main()
{
    cout << getNoOfWays(4) << endl;
    cout << getNoOfWays(3);
    return 0;
}


Java




/* Java program to count the
 no of ways to place 2*1 size
 tiles in 2*n size board. */
import java.io.*;
 
class GFG {
  static int getNoOfWays(int n)
  {
 
    // Base case
    if (n <= 2) {
      return n;
    }
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
  }
 
  // Driver Function
  public static void main(String[] args)
  {
    System.out.println(getNoOfWays(4));
    System.out.println(getNoOfWays(3));
  }
}
 
// This code is contributed by ashwinaditya21.


Python3




# Python3 program to count the
# no. of ways to place 2*1 size
# tiles in 2*n size board.
def getNoOfWays(n):
   
    # Base case
    if n <= 2:
        return n
 
    return getNoOfWays(n - 1) + getNoOfWays(n - 2)
 
# Driver Code
print(getNoOfWays(4))
print(getNoOfWays(3))
 
# This code is contributed by Kevin Joshi


C#




// C# program to implement
// the above approach
using System;
 
class GFG
{
     
static int getNoOfWays(int n)
  {
  
    // Base case
    if (n <= 2) {
      return n;
    }
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
  }
 
// Driver Code
public static void Main()
{
    Console.WriteLine(getNoOfWays(4));
    Console.WriteLine(getNoOfWays(3));
}
}
 
// This code is contributed by code_hunt.


Javascript




<script>
// JavaScript program to count the
// no. of ways to place 2*1 size
// tiles in 2*n size board.
 
function getNoOfWays(n)
{
    // Base case
    if (n <= 2)
      return n;
       
    return getNoOfWays(n - 1) + getNoOfWays(n - 2);
}
 
// Driver Function
document.write(getNoOfWays(4));
document.write(getNoOfWays(3));
 
// This code is contributed by shinjanpatra
</script>


Output:

5
3

Time Complexity: O(n)

Space Complexity: O(1)
The above recurrence is nothing but Fibonacci Number expression. We can find n’th Fibonacci number in O(Log n) time, see below for all method to find n’th Fibonacci Number. 
https://youtu.be/NyICqRtePVs 
https://youtu.be/U9ylW7NsHlI 
Different methods for n’th Fibonacci Number
Count the number of ways to tile the floor of size n x m using 1 x m size tiles 
This article is contributed by Saurabh Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 


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