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 vertically.
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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.
- Place the first tile horizontally : If we place first tile horizontally, the problem reduces to “count(n-1)”
- 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++ 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 gridimport 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 |
Python3
# 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 aboven = 5print ("Count of ways is"),print (count(n)) |
C#
// C# program to count of ways // to place 1 x 4 tiles on// n x 4 gridusing 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 |
Javascript
<script>// JavaScript program to count of ways to place 1 x 4 tiles// on n x 4 grid // Function that count the number of ways to place 1 x 4 tiles // on n x 4 grid. function count(n) { // Create a table to store results of sub-problems // dp[i] stores count of ways for i x 4 grid. let dp = []; dp[0] = 0; // Fill the table from d[1] to dp[n] for(let 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 let n = 5; document.write("Count of ways is: " + count(n)); // This code is contributed by target_2.</script> |
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)
Efficient approach : Space Optimization
In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1] and dp[i-4]. So to optimize the space we can keep track of previous 4 values and current values which will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Handle the base cases:
- If n is less than or equal to 0, return 0.
- If n is less than or equal to 3, return 1.
- Initialize variables dp1, dp2, dp3, and dp4 with values 1, 1, 1, and 2 respectively.
- Initialize dp to 0.
- Iterate from 5 to n:
- Calculate dp as the sum of dp4 and dp1.
- Update the variables: shift dp1 to dp2, dp2 to dp3, dp3 to dp4, and dp4 to dp.
- Return dp as the count of ways.
Implementation:
C++
// C++ code for the above approach approach #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){ // Base Case if (n <= 0) return 0; if (n <= 3) return 1; // create the dp previous instances int dp1 = 1; int dp2 = 1; int dp3 = 1; int dp4 = 2; int dp = 0; // to store current value // iterate to get the current value from previous computations for (int i = 5; i <= n; i++) { dp = dp4 + dp1; dp1 = dp2; dp2 = dp3; dp3 = dp4; dp4 = dp; } // return final answer return dp;}// Driver codeint main(){ int n = 5; cout << "Count of ways is " << count(n) << endl; return 0;}// -- by bhardwajji |
Java
/*package whatever //do not write package name here */import java.io.*;class Gfg { // Returns count of ways to place 1 x 4 tiles on n x 4 grid. static int count(int n) { // Base Case if (n <= 0) return 0; if (n <= 3) return 1; // Create the dp previous instances int dp1 = 1; int dp2 = 1; int dp3 = 1; int dp4 = 2; int dp = 0; // To store the current value // Iterate to get the current value from previous computations for (int i = 5; i <= n; i++) { dp = dp4 + dp1; dp1 = dp2; dp2 = dp3; dp3 = dp4; dp4 = dp; } // Return the final answer return dp; } // Driver code public static void main(String[] args) { int n = 5; System.out.println("Count of ways is " + count(n)); }}// code is contributed by shinjanpatra |
Output :
Count of ways is 3
Time Complexity : O(n)
Auxiliary Space : O(1)
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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