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:3

Explanation:

For a 2 x 4 board, there are 3 ways

- All 4 vertical
- All 4 horizontal
- 2 vertical and 2 horizontal

Input:n = 3

Output:2

Explanation:We need 2 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.

**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 == 0)` ` ` `return` `0;` ` ` `if` `(n == 1)` ` ` `return` `1;` ` ` `return` `getNoOfWays(n - 1) + getNoOfWays(n - 2);` `}` `// Driver Function` `int` `main()` `{` ` ` `cout << getNoOfWays(4) << endl;` ` ` `cout << getNoOfWays(3);` ` ` `return` `0;` `}` |

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**Output:**

3 2

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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