Category: C# Coding Test Tutorials

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:

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

Problem Description

Starting from a certain location on a street and during the journey to a specific point, one encounters various people and hears different stories from them.
These individuals can either tell the truth or lie, and we must determine the truth based on what everyone is saying.
Implement an algorithm to determine your position based on the given data.

Problem Definition

Among N people, each person can either tell the truth or lie about what they know.
Each person should be able to determine their truthfulness based on the information from others.
Based on the given conversation information, calculate the number of possible individuals who can speak the truth at a specific point.

Input Format

• The first line contains the number of people, N.
• The second line contains conversation information between people. Each piece of information is in the format “A B T/F”, where A is the information provider, B is the information receiver, T signifies truth, and F signifies a lie.

Output Format

Output the total number of people who can speak the truth.

Problem Solving Course

1. Understanding the Problem

To understand the problem, let’s analyze the meaning of the provided data.
People communicate with each other, and we need to establish clear rules regarding whether the information from A to B is true or false.
In such cases, we can represent this information in a graph and use algorithms like DFS or BFS to determine truthfulness.

2. Algorithm Design

We need to create a graph to represent the relationships of information transmission and consider all possible combinations of truths and lies that can occur in conversations.
Using the DFS (Depth First Search) algorithm, the process of propagating truths and lies through each person is as follows.

3. C# Code Implementation

        
        using System;
        using System.Collections.Generic;

        class Program
        {
            static void Main(string[] args)
            {
                // Input
                int N = int.Parse(Console.ReadLine());
                List> conversations = new List>();
                string line;
                while ((line = Console.ReadLine()) != null && line.Length > 0)
                {
                    var parts = line.Split(' ');
                    int A = int.Parse(parts[0]);
                    int B = int.Parse(parts[1]);
                    bool isTruth = parts[2].Equals("T");
                    conversations.Add(new Tuple(A, B, isTruth));
                }

                // Graph formation
                Dictionary>> graph = new Dictionary>>();
                foreach (var convo in conversations)
                {
                    if (!graph.ContainsKey(convo.Item1))
                        graph[convo.Item1] = new List>();
                    graph[convo.Item1].Add(new Tuple(convo.Item2, convo.Item3));
                }

                // Perform DFS
                HashSet visited = new HashSet();
                int truthfulCount = 0;

                void DFS(int person)
                {
                    if (visited.Contains(person)) return;
                    visited.Add(person);
                    truthfulCount++;
                    if (graph.ContainsKey(person))
                    {
                        foreach (var neighbor in graph[person])
                        {
                            if (neighbor.Item2) // Case of speaking the truth
                                DFS(neighbor.Item1);
                        }
                    }
                }

                foreach (var person in graph.Keys)
                {
                    if (!visited.Contains(person))
                    {
                        DFS(person);
                    }
                }

                // Output result
                Console.WriteLine(truthfulCount);
            }
        }
        
        

4. Code Explanation

The above C# code illustrates a simple way to solve the problem using the DFS algorithm.
First, it takes the input and constructs a graph based on the conversation information between people.
Then, it executes DFS for each person to calculate the number of individuals who can speak the truth and outputs the result.

5. Time Complexity Analysis

The time complexity of this algorithm is O(V + E).
Here, V represents the number of people, and E represents the number of conversations.
Thus, the time taken to explore all vertices and edges of the graph is generally considered efficient.

6. Code Improvement and Optimization

Additionally, one could consider using BFS instead of DFS for optimization.
Moreover, introducing memoization to store already confirmed results of truths and lies would allow for quicker returns on similar requests in the future.

Conclusion

In this course, we explored how to construct algorithms to solve the given problem using C#.
By utilizing graph algorithms and DFS, we learned the process of calculating the number of individuals who can speak the truth, which equipped us with useful approaches for real coding tests.
I hope this helps improve problem-solving skills by practicing various problem types and enhancing understanding of algorithms.

This article describes how to implement the bubble sort algorithm in C#. Bubble sort is one of the most basic sorting algorithms and often appears in various algorithm tests. In this post, we will take a detailed look at the principle of bubble sort and examine an example implemented in C#.

What is Bubble Sort?

Bubble sort is a simple sorting algorithm that compares adjacent elements of an array to arrange the order of the two elements. One can visualize the largest (or smallest) element ‘bubbling’ up to the end of the array. This algorithm can be implemented simply, but it is not an efficient sorting algorithm. Its time complexity is O(n^2).

Problem Description

Sort the given integer array in ascending order using the bubble sort algorithm.

Input

- Input: Integer array arr = [64, 34, 25, 12, 22, 11, 90]

Output

- Output: Sorted array [11, 12, 22, 25, 34, 64, 90]

Explanation of the Bubble Sort Algorithm

Bubble sort follows these steps:

1. Check the length of the array.
2. Use two nested loops to iterate through all elements of the array.
3. Compare two adjacent elements and swap them if they are in the wrong order.
4. After one full comparison, the largest element moves to the end of the array.
5. Repeat this process N-1 times to sort the array.

Implementing Bubble Sort in C#

Now, let’s implement the above process in C# code.

using System;

class Program
{
    static void Main()
    {
        int[] arr = { 64, 34, 25, 12, 22, 11, 90 };
        Console.WriteLine("Original array:");
        PrintArray(arr);
        
        BubbleSort(arr);
        
        Console.WriteLine("Sorted array:");
        PrintArray(arr);
    }

    static void BubbleSort(int[] arr)
    {
        int n = arr.Length;
        for (int i = 0; i < n - 1; i++)
        {
            for (int j = 0; j < n - i - 1; j++)
            {
                if (arr[j] > arr[j + 1])
                {
                    // Swap arr[j] and arr[j + 1]
                    int temp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = temp;
                }
            }
        }
    }

    static void PrintArray(int[] arr)
    {
        foreach (var item in arr)
        {
            Console.Write(item + " ");
        }
        Console.WriteLine();
    }
}

Explanation of the Code

The above code is an example of implementing bubble sort using a simple C# console application.

• BubbleSort(int[] arr): A method that sorts the given array. This method uses two nested loops to iterate through the array, comparing and swapping adjacent elements.
• PrintArray(int[] arr): A helper method that prints the elements of the array, separating each element with a space.

Conclusion

Bubble sort is an easy-to-learn and understand sorting algorithm, but it is advisable to use other algorithms with better performance in practice. However, understanding the concepts of basic sorting algorithms is important in preparing for coding tests.

I hope this article helps with your future coding test preparations. If you have any questions or points for discussion, please leave a comment!

1. Problem Description

In this lecture, we will address the problem of representing sets using C#. Sets are collections of elements that do not allow duplicates, and they are useful in certain situations. For example, sets can be effectively used when handling lists of unique user IDs or various product codes.

Problem: Representation of Sets

Given two arrays A and B, implement a function that calculates the union, intersection, and difference of the two arrays. The function is defined in the following format:

public static void CalculateSets(int[] A, int[] B)

2. Solution Process

To solve this problem, it’s essential to understand the definitions of each set operation and how to utilize sets in C#.

2.1 Definition of Sets

• Union: The set that includes all elements from both sets.
• Intersection: The set consisting of elements common to both sets.
• Difference: The set obtained by excluding elements of the second set from the first set.

2.2 C# Set Class

In C#, sets can be easily implemented using the HashSet class. A HashSet does not allow duplicates and can represent an unordered set.

2.3 Function Implementation

Now we will implement the required function CalculateSets. Below is the code to implement each set operation.

using System;
using System.Collections.Generic;

public class SetOperations
{
    public static void CalculateSets(int[] A, int[] B)
    {
        HashSet setA = new HashSet(A);
        HashSet setB = new HashSet(B);

        // Union
        HashSet union = new HashSet(setA);
        union.UnionWith(setB);
        Console.WriteLine("Union: " + string.Join(", ", union));

        // Intersection
        HashSet intersection = new HashSet(setA);
        intersection.IntersectWith(setB);
        Console.WriteLine("Intersection: " + string.Join(", ", intersection));

        // Difference
        HashSet difference = new HashSet(setA);
        difference.ExceptWith(setB);
        Console.WriteLine("Difference: " + string.Join(", ", difference));
    }
}

3. Code Explanation

Let’s explain the above code step by step.

3.1 Creating HashSets

First, we create a HashSet for each of the given arrays A and B. This way, we obtain a set with duplicate elements removed.

HashSet setA = new HashSet(A);
HashSet setB = new HashSet(B);

3.2 Calculating Union

To calculate the union, we first make a copy of setA and then use the UnionWith method to add elements from setB.

HashSet union = new HashSet(setA);
union.UnionWith(setB);
Console.WriteLine("Union: " + string.Join(", ", union));

3.3 Calculating Intersection

The intersection is obtained by calling the IntersectWith method on a copy of setA with setB.

HashSet intersection = new HashSet(setA);
intersection.IntersectWith(setB);
Console.WriteLine("Intersection: " + string.Join(", ", intersection));

3.4 Calculating Difference

The difference is calculated by removing elements of setB using the ExceptWith method.

HashSet difference = new HashSet(setA);
difference.ExceptWith(setB);
Console.WriteLine("Difference: " + string.Join(", ", difference));

4. Full Code

Finally, the full code that includes all content is as follows:

using System;
using System.Collections.Generic;

public class SetOperations
{
    public static void CalculateSets(int[] A, int[] B)
    {
        HashSet setA = new HashSet(A);
        HashSet setB = new HashSet(B);

        // Union
        HashSet union = new HashSet(setA);
        union.UnionWith(setB);
        Console.WriteLine("Union: " + string.Join(", ", union));

        // Intersection
        HashSet intersection = new HashSet(setA);
        intersection.IntersectWith(setB);
        Console.WriteLine("Intersection: " + string.Join(", ", intersection));

        // Difference
        HashSet difference = new HashSet(setA);
        difference.ExceptWith(setB);
        Console.WriteLine("Difference: " + string.Join(", ", difference));
    }
}

class Program
{
    static void Main()
    {
        int[] A = { 1, 2, 3, 4, 5 };
        int[] B = { 4, 5, 6, 7, 8 };
        SetOperations.CalculateSets(A, B);
    }
}

5. Testing and Results

To test the above code, I defined two arrays in the Main method and called the CalculateSets function.

The results for the given arrays A and B are as follows:

Union: 1, 2, 3, 4, 5, 6, 7, 8
Intersection: 4, 5
Difference: 1, 2, 3

6. Conclusion

In this lecture, we learned how to effectively represent sets using C# and perform operations such as union, intersection, and difference. Such set operations are very useful not only for coding tests but also for solving various algorithm problems. Additionally, utilizing the various methods of C#’s HashSet class allows for easy manipulation of sets. It would be beneficial to revisit the concepts and applications of sets as you prepare for future coding tests.

In this course, we will address a problem frequently found in actual coding tests with the theme of ‘Making Cocktails’. This problem requires various algorithmic thinking and will be solved using the C# programming language. Problems related to making cocktails may include combinations, divide and conquer, optimization problems, etc., and in this course, we will select a specific problem to solve step by step.

Problem Description

You have N ingredients. Each ingredient contains a specific amount of alcohol. Additionally, each ingredient is assigned an index from 1 to N. You wish to select M (1 ≤ M ≤ N) ingredients to make a cocktail. The sum of the alcohol amounts from the selected ingredients must equal K (K is a given integer). Your goal is to find various combinations to calculate the number of possible cocktails you can make.

Input Format

• The first line contains two integers N (number of ingredients) and M (number of ingredients to select).
• The second line contains N integers representing the alcohol amounts of each ingredient.
• The third line contains the target alcohol amount K.

Output Format

Print the number of possible cocktail combinations modulo 1,000,000,007.

Example Input

5 3
1 2 3 4 5
5

Example Output

5

Problem Solving Process

To solve this problem, you need to make various combinations of ingredients and check the sum of their alcohol amounts. In C#, you can use recursive calls to find combinations. Below is a basic algorithm design to solve this problem.

Step 1: Input Parsing

First, parse the input data and store N, M, the alcohol amounts, and K in variables.

using System;
using System.Collections.Generic;

class Program
{
    static void Main(string[] args)
    {
        var input = Console.ReadLine().Split();
        int N = int.Parse(input[0]);
        int M = int.Parse(input[1]);

        var alcohols = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        int K = int.Parse(Console.ReadLine());
        
        // We will find combinations through recursive calls later.
    }
}

Step 2: Implementing the Function to Find Combinations

We will implement a recursive function to find combinations. This function will take the current index, the number of selected ingredients, and the current alcohol amount as parameters.

static int CountCombinations(int[] alcohols, int N, int M, int K, int index, int count, int currentSum)
{
    if(count == M)
    {
        return currentSum == K ? 1 : 0;
    }
    
    if(index >= N)
    {
        return 0;
    }

    // Selecting the current ingredient and making a recursive call
    int includeCurrent = CountCombinations(alcohols, N, M, K, index + 1, count + 1, currentSum + alcohols[index]);
    // Not selecting the current ingredient and making a recursive call
    int excludeCurrent = CountCombinations(alcohols, N, M, K, index + 1, count, currentSum);

    return includeCurrent + excludeCurrent;
}

Step 3: Main Function and Calling Combinations

Now, we will call the combination function defined above in the main function to output the result. Additionally, we will output the result modulo 1,000,000,007.

static void Main(string[] args)
{
    var input = Console.ReadLine().Split();
    int N = int.Parse(input[0]);
    int M = int.Parse(input[1]);

    var alcohols = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
    int K = int.Parse(Console.ReadLine());

    long result = CountCombinations(alcohols, N, M, K, 0, 0, 0);
    const int MOD = 1000000007;

    Console.WriteLine(result % MOD);
}

Complete Source Code

using System;
using System.Collections.Generic;

class Program
{
    static void Main(string[] args)
    {
        var input = Console.ReadLine().Split();
        int N = int.Parse(input[0]);
        int M = int.Parse(input[1]);

        var alcohols = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        int K = int.Parse(Console.ReadLine());

        long result = CountCombinations(alcohols, N, M, K, 0, 0, 0);
        const int MOD = 1000000007;

        Console.WriteLine(result % MOD);
    }

    static int CountCombinations(int[] alcohols, int N, int M, int K, int index, int count, int currentSum)
    {
        if(count == M)
        {
            return currentSum == K ? 1 : 0;
        }
        
        if(index >= N)
        {
            return 0;
        }

        // Selecting the current ingredient and making a recursive call
        int includeCurrent = CountCombinations(alcohols, N, M, K, index + 1, count + 1, currentSum + alcohols[index]);
        // Not selecting the current ingredient and making a recursive call
        int excludeCurrent = CountCombinations(alcohols, N, M, K, index + 1, count, currentSum);

        return includeCurrent + excludeCurrent;
    }
}

Conclusion

This algorithm problem can be solved with a simple approach utilizing combinations and recursion. It is a useful way to create specific values by combining multiple ingredients. Since it frequently appears in coding tests, it is highly recommended to practice this problem.

We will continue to cover various algorithms and problem-solving methods, so we appreciate your interest!