# 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.
```

## We strongly recommend that you click here and practice it, before moving on to the solution.

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)”

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))
```

Output :
`Count of ways is 3`

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

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