Count number of ways to fill a “n x 4” grid using “1 x 4” tiles

Given a number n, count number of ways to fill a n x 4 grid using 1 x 4 tiles.

Examples:

Input : n = 1
Output : 1

Input : n = 2
Output : 1
We can only place both tiles horizontally

Input : n = 3
Output : 1
We can only place all tiles horizontally.

Input : n = 4
Output : 2
The two ways are : 
  1) Place all tiles horizontally 
  2) Place all tiles vertically.

Input : n = 5
Output : 3
We can fill a 5 x 4 grid in following ways : 
  1) Place all 5 tiles horizontally
  2) Place first 4 vertically and 1 horizontally.
  3) Place first 1 horizontally and 4 horizontally.

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This problem is mainly an extension of this tiling problem

Let “count(n)” be the count of ways to place tiles on a “n x 4” grid, following two cases arise when we place the first tile.

  1. Place the first tile horizontally : If we place first tile horizontally, the problem reduces to “count(n-1)”
  2. Place the first tile vertically : If we place first tile vertically, then we must place 3 more tiles vertically. So the problem reduces to “count(n-4)”

grid



Therefore, count(n) can be written as below.

   count(n) = 1 if n = 1 or n = 2 or n = 3   
   count(n) = 2 if n = 4
   count(n) = count(n-1) + count(n-4) 

This recurrence is similar to Fibonacci Numbers and can be solved using Dynamic programming.

C/C++

// C++ program to count of ways to place 1 x 4 tiles
// on n x 4 grid.
#include<iostream>
using namespace std;
  
// Returns count of count of ways to place 1 x 4 tiles
// on n x 4 grid.
int count(int n)
{
    // Create a table to store results of subproblems
    // dp[i] stores count of ways for i x 4 grid.
    int dp[n+1];
    dp[0] = 0;
  
    // Fill the table from d[1] to dp[n]
    for (int i=1; i<=n; i++)
    {
        // Base cases
        if (i >= 1 && i <= 3)
            dp[i] = 1;
        else if (i==4)
            dp[i] = 2 ;
  
        else 
            // dp(i-1) : Place first tile horizontally
            // dp(n-4) : Place first tile vertically
            //           which means 3 more tiles have
            //           to be placed vertically.
            dp[i] = dp[i-1] + dp[i-4];
    }
  
    return dp[n];
}
  
// Driver program to test above
int main()
{
    int n = 5;
    cout << "Count of ways is " << count(n);
    return 0;
}

Java

// Java program to count of ways to place 1 x 4 tiles
// on n x 4 grid
import java.io.*;
  
class Grid 
    // Function that count the number of ways to place 1 x 4 tiles
    // on n x 4 grid.
    static int count(int n)
    {
        // Create a table to store results of sub-problems
        // dp[i] stores count of ways for i x 4 grid.
        int[] dp = new int[n+1];
        dp[0] = 0;
        // Fill the table from d[1] to dp[n]
        for(int i=1;i<=n;i++)
        {
            // Base cases
            if (i >= 1 && i <= 3)
                dp[i] = 1;
            else if (i==4)
                dp[i] = 2 ;
  
            else
            {
                    // dp(i-1) : Place first tile horizontally
                    // dp(i-4) : Place first tile vertically
                    //         which means 3 more tiles have
                    //         to be placed vertically.
                    dp[i] = dp[i-1] + dp[i-4];
            }
        }
        return dp[n];
    }
      
    // Driver program
    public static void main (String[] args)
    {
        int n = 5;
        System.out.println("Count of ways is: " + count(n));
    }
}
  
// Contributed by Pramod Kumar

Python

# Python program to count of ways to place 1 x 4 tiles
# on n x 4 grid.
  
# Returns count of count of ways to place 1 x 4 tiles
# on n x 4 grid.
def count(n):
  
    # Create a table to store results of subproblems
    # dp[i] stores count of ways for i x 4 grid.
    dp = [0 for _ in range(n+1)]
  
    # Fill the table from d[1] to dp[n]
    for i in range(1,n+1):
  
        # Base cases
        if i <= 3:
            dp[i] = 1
        elif i == 4:
            dp[i] = 2
        else:
            # dp(i-1) : Place first tile horizontally
            # dp(n-4) : Place first tile vertically
            #           which means 3 more tiles have
            #           to be placed vertically.
            dp[i] = dp[i-1] + dp[i-4]
  
    return dp[n]
  
# Driver code to test above
n = 5
print ("Count of ways is"),
print (count(n))

C#

// C# program to count of ways 
// to place 1 x 4 tiles on
// n x 4 grid
using System;
  
class GFG
{
      
    // Function that count the number 
    // of ways to place 1 x 4 tiles
    // on n x 4 grid.
    static int count(int n)
    {
          
        // Create a table to store results
        // of sub-problems dp[i] stores
        // count of ways for i x 4 grid.
        int[] dp = new int[n + 1];
        dp[0] = 0;
          
        // Fill the table from d[1]
        // to dp[n]
        for(int i = 1; i <= n; i++)
        {
              
            // Base cases
            if (i >= 1 && i <= 3)
                dp[i] = 1;
            else if (i == 4)
                dp[i] = 2 ;
  
            else
            {
                  
                // dp(i-1) : Place first tile
                // horizontally dp(i-4) : 
                // Place first tile vertically
                // which means 3 more tiles have
                // to be placed vertically.
                dp[i] = dp[i - 1] + 
                        dp[i - 4];
            }
        }
        return dp[n];
    }
      
    // Driver Code
    public static void Main ()
    {
        int n = 5;
        Console.WriteLine("Count of ways is: "
                           + count(n));
    }
}
  
// This code is contributed by Sam007

PHP

<?php
// PHP program to count of ways to 
// place 1 x 4 tiles on n x 4 grid.
  
// Returns count of count of ways 
// to place 1 x 4 tiles
// on n x 4 grid.
function countt($n)
{
      
    // Create a table to store
    // results of subproblems
    // dp[i] stores count of 
    // ways for i x 4 grid.
    $dp[$n + 1] = 0;
    $dp[0] = 0;
  
    // Fill the table
    // from d[1] to dp[n]
    for ($i = 1; $i <= $n; $i++)
    {
          
        // Base cases
        if ($i >= 1 && $i <= 3)
            $dp[$i] = 1;
        else if ($i == 4)
            $dp[$i] = 2 ;
  
        else
            // dp(i-1) : Place first tile horizontally
            // dp(n-4) : Place first tile vertically
            //             which means 3 more tiles have
            //             to be placed vertically.
            $dp[$i] = $dp[$i - 1] + $dp[$i - 4];
    }
  
    return $dp[$n];
}
  
    // Driver Code
    $n = 5;
    echo "Count of ways is " , countt($n);
  
// This code is contributed by nitin mittal.
?>

Output :

Count of ways is 3

Time Complexity : O(n)
Auxiliary Space : O(n)

This article is contributed by Rajat Jha. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : Sam007, nitin mittal

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