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 Functionint 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 Codeprint(getNoOfWays(4))print(getNoOfWays(3))# This code is contributed by Kevin Joshi |
C#
// C# program to implement// the above approachusing System;class GFG{ static int getNoOfWays(int n) { // Base case if (n <= 2) { return n; } return getNoOfWays(n - 1) + getNoOfWays(n - 2); }// Driver Codepublic 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 Functiondocument.write(getNoOfWays(4));document.write(getNoOfWays(3));// This code is contributed by shinjanpatra</script> |
5 3
Time Complexity: O(2^n)
Auxiliary Space: 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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