Tiling Problem
- Difficulty Level : Easy
- Last Updated : 24 May, 2022
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)
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
(getNoOfWays(
4
))
(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 3The 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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