C# Coding Test Course, 2 N Tile Filling

Hello! Today, we will explore a detailed problem description and solution process for the 2*N tile filling problem for everyone preparing for a C# coding test. This problem is closely related to Dynamic Programming and is very useful for developing problem-solving skills.

Problem Description

We have a rectangular grid of size 2*N. Our goal is to calculate the number of ways to fill this rectangle using 2*1 size tiles. We can rotate the tiles, so each tile can be placed either vertically or horizontally.

Input

Integer N (1 ≤ N ≤ 30) – the width of the rectangle.

Output

Print the number of ways to fill the 2*N rectangle with tiles.

Understanding the Problem

To solve this problem, we need to consider several approaches. The number of ways to fill the rectangle with tiles can be thought of as follows:

1. Establishing the Recurrence Relation

The ways to fill a 2*N rectangle can be divided into two cases:

Using 2 tiles horizontally: In this case, the size of the remaining rectangle is 2*(N-1).
Using 1 tile vertically: In this case, the size of the remaining rectangle is 2*(N-2).

Therefore, this can be expressed mathematically as:

    F(N) = F(N-1) + F(N-2)

Here, F(N) represents the number of ways to fill a rectangle of size N.

2. Setting Initial Conditions

The initial conditions are as follows:

F(1) = 1 (A rectangle of size 2*1 can be filled with either a 2*1 tile or a 1*2 tile.)
F(2) = 2 (A rectangle of size 2*2 can be filled with either 2 horizontal tiles or 2 vertical tiles.)

Implementing C# Code

Now, based on the recurrence relation that we established above, let’s implement the algorithm using C#.

    
    using System;

    class Program
    {
        static void Main()
        {
            int N = int.Parse(Console.ReadLine());
            Console.WriteLine(TileFilling(N));
        }

        static int TileFilling(int N)
        {
            if (N == 1) return 1;
            if (N == 2) return 2;

            int[] dp = new int[N + 1];
            dp[1] = 1;
            dp[2] = 2;

            for (int i = 3; i <= N; i++)
            {
                dp[i] = dp[i - 1] + dp[i - 2];
            }

            return dp[N];
        }
    }

Code Explanation

In the above code, we use a dynamic programming array called dp to store the number of ways to fill the tiles up to N. After setting the cases for lengths 1 and 2 through initial conditions, we perform calculations based on the recurrence relation with a loop.

Main Functions Explanation

Main(): The starting point of the program that receives the N value and outputs the number of ways to fill the tiles.
TileFilling(int N): Calculates and returns the number of ways to fill the tiles for a given N.

Complexity Analysis

The time complexity of this algorithm is O(N), and the space complexity is O(N). This means that the execution time increases in proportion to N, and it allows for efficient memory usage by storing previous results in an array.

Test Cases

Let's check if the algorithm works well for various N values.

Input: N = 1 → Output: 1
Input: N = 2 → Output: 2
Input: N = 3 → Output: 3
Input: N = 4 → Output: 5
Input: N = 5 → Output: 8
Input: N = 6 → Output: 13

Conclusion

Today, we learned how to solve the 2*N tile filling problem using C#. This problem is a typical example that can be solved using dynamic programming and will greatly help in developing algorithmic thinking. I hope this process of solving the problem has been very helpful in your coding test preparation.

Next time, I will bring another interesting algorithm problem. If you have a topic you would like to see, please let me know in the comments! Thank you.