C# Coding Test Course, Creating Blu-ray

This course aims to delve into algorithm problems using C#. First, we will introduce the problem of “Making a Blu-ray” and explain the necessary algorithmic approaches and C# code to detail the resolution process.

Problem Description

The “Making a Blu-ray” problem has the following conditions. Several movie titles are given, and your goal is to create the minimum number of discs needed to fit all movies onto Blu-ray discs. Each movie has a specific playback time, and there is a limitation on the capacity of each disc. Finding a way to arrange the movies according to the given conditions is the essence of this problem.

Input

  • Number of movies N (1 ≤ N ≤ 100)
  • Playback time of each movie M[i] (1 ≤ M[i] ≤ 500)
  • Capacity of the disc D (1 ≤ D ≤ 500)

Output

You need to output the minimum number of discs required.

Problem Solving Process

1. Approach to the Problem

To solve the problem, we need to appropriately distribute the list of movies to maximize the utilization of the disc capacity. This problem can be approached with a simple search and conditional statements, and we can consider all cases using a backtracking technique.

2. Algorithm Design

The first thing to do is to calculate the total length of movies that can fit on each disc to achieve maximum efficiency.

The basic idea is as follows:

  • Sort the movies in descending order.
  • Check the current capacity of the disc and add the movie.
  • If the movie cannot be added to the current disc, create a new disc.

3. C# Code Implementation

Now, let’s write the C# code to solve the problem. The code below is based on the approach described above.


using System;
using System.Linq;

class Program
{
    static void Main()
    {
        int N = int.Parse(Console.ReadLine());
        int D = int.Parse(Console.ReadLine());
        int[] M = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();

        Console.WriteLine(CountDiscs(M, D));
    }

    static int CountDiscs(int[] movies, int capacity)
    {
        Array.Sort(movies);
        Array.Reverse(movies);

        int discs = 0;
        int currentCapacity = 0;

        foreach (var movie in movies)
        {
            if (currentCapacity + movie > capacity)
            {
                discs++;
                currentCapacity = movie; // Current capacity of the new disc
            }
            else
            {
                currentCapacity += movie; // Add movie to the disc
            }
        }

        // Count the last used disc as well
        if (currentCapacity > 0)
            discs++;

        return discs;
    }
}
        

4. Explanation of the Code

The main function of the code is to calculate the number of discs through the CountDiscs method. This method sorts the given movie list in descending order and proceeds to add each movie to the discs.

– If the length of the movie exceeds the disc capacity, a new disc is created, and
– if not, the movie is added to the current disc.

Through this process, the final number of discs required is derived.

Conclusion

In this course, we examined the “Making a Blu-ray” problem as an example of coding tests using C#. The process of approaching and solving such problems greatly helps in developing algorithmic thinking. It is important to gain experience by solving various problems.

In the future, continued interest and practice in solving algorithm problems will be necessary, and through this, you can strengthen your preparation for coding tests. Recording and sharing insights gained while solving problems in your own way can also be a good learning method.

C# Coding Test Course, Depth First Search

Hello, everyone! Today, we will delve deeply into the Depth-First Search (DFS) algorithm for preparing coding tests using C#. Depth-First Search is an efficient algorithm used to explore tree or graph structures. In this article, we will explain the basic concepts of DFS, how to implement it, and how to apply it through real problems step by step.

1. What is Depth-First Search (DFS)?

Depth-First Search (DFS) is a search technique that visits all nodes of a graph. This algorithm explores as deeply as possible from one node before returning to the most recently visited node when there are no further nodes to explore, in order to explore other paths.

1.1 Basic Principles

DFS can be implemented using a stack or conveniently handled using a recursive function. The fundamental principles of DFS are as follows:

  • Select a starting node.
  • Keep track of visited nodes.
  • Choose an unvisited adjacent node and recursively call DFS.
  • If there are no more nodes to visit, return to the previous node.

2. Time Complexity of DFS

The time complexity of DFS is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. This is because it visits all nodes once and checks all edges once. The space complexity depends on the stack used and is O(V) in the worst case.

3. Implementing DFS

Now, let’s implement DFS using C#. The following code represents a graph in the form of an adjacency list and implements DFS.


using System;
using System.Collections.Generic;

class Graph
{
    private int V; // Number of vertices
    private List[] adj; // Adjacency list

    public Graph(int v)
    {
        V = v;
        adj = new List[V];
        for (int i = 0; i < V; i++)
        {
            adj[i] = new List();
        }
    }

    public void AddEdge(int v, int w)
    {
        adj[v].Add(w); // Add node w to node v
    }

    public void DFS(int v)
    {
        bool[] visited = new bool[V]; // Visited nodes array
        DFSUtil(v, visited); // Recursive call
    }

    private void DFSUtil(int v, bool[] visited)
    {
        visited[v] = true; // Visit current node
        Console.Write(v + " "); // Print current node

        foreach (int neighbor in adj[v])
        {
            if (!visited[neighbor]) // If adjacent node has not been visited
            {
                DFSUtil(neighbor, visited); // Recursive call
            }
        }
    }
}

class Program
{
    static void Main(string[] args)
    {
        Graph g = new Graph(4);
        g.AddEdge(0, 1);
        g.AddEdge(0, 2);
        g.AddEdge(1, 2);
        g.AddEdge(2, 0);
        g.AddEdge(2, 3);
        g.AddEdge(3, 3);

        Console.WriteLine("Depth First Traversal (starting node 2):");
        g.DFS(2);
    }
}

3.1 Code Explanation

The above code has the following structure:

  • Graph Class: Contains the number of vertices and the adjacency list.
  • AddEdge Method: Adds an edge between two nodes.
  • DFS Method: Initializes the visited array for DFS traversal and calls DFSUtil.
  • DFSUtil Method: Recursively performs DFS and prints the visited nodes.

4. Algorithm Problem: Counting Islands

Now, let’s solve a real problem using DFS. The problem is ‘Counting Islands’.

4.1 Problem Description

The given 2D array consists of water (0) and land (1). An island is a collection of land (1), formed by land that is connected to each other. An island is defined as a collection of land that is adjacent either vertically or horizontally. Write a program that counts the number of islands in this 2D array.

4.2 Approach

To solve this problem, we will perform the following steps:

  • Traverse the 2D array and call DFS when land (1) is found.
  • Within DFS, visit all connected lands and increase the count.
  • After exploring all lands, return the number of islands.

4.3 C# Code Implementation


using System;

class IslandCounter
{
    private int[,] grid;
    private int rows, cols;

    public IslandCounter(int[,] grid)
    {
        this.grid = grid;
        rows = grid.GetLength(0);
        cols = grid.GetLength(1);
    }

    public int CountIslands()
    {
        bool[,] visited = new bool[rows, cols];
        int count = 0;

        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                if (grid[i, j] == 1 && !visited[i, j])
                {
                    DFS(i, j, visited); // Call DFS
                    count++;
                }
            }
        }
        return count;
    }

    private void DFS(int i, int j, bool[,] visited)
    {
        if (i >= rows || i < 0 || j >= cols || j < 0 || grid[i, j] == 0 || visited[i, j])
            return;

        visited[i, j] = true; // Mark as visited

        // Call DFS in 8 adjacent directions
        DFS(i + 1, j, visited);
        DFS(i - 1, j, visited);
        DFS(i, j + 1, visited);
        DFS(i, j - 1, visited);
        DFS(i + 1, j + 1, visited);
        DFS(i - 1, j - 1, visited);
        DFS(i + 1, j - 1, visited);
        DFS(i - 1, j + 1, visited);
    }
}

class Program
{
    static void Main(string[] args)
    {
        int[,] grid = new int[,]
        {
            { 1, 1, 0, 0, 0 },
            { 0, 1, 0, 0, 1 },
            { 0, 0, 0, 1, 1 },
            { 0, 0, 0, 0, 0 },
            { 1, 0, 1, 0, 1 }
        };

        IslandCounter counter = new IslandCounter(grid);
        Console.WriteLine("Number of islands: " + counter.CountIslands());
    }
}

4.4 Code Explanation

The above code has the following structure:

  • IslandCounter Class: Takes a 2D array as input and includes functionality to count the number of islands.
  • CountIslands Method: Visits the 2D array and counts the number of islands.
  • DFS Method: Marks the connected lands (1) as visited through depth-first search.

5. Conclusion

In this article, we explored the basic concept of the Depth-First Search (DFS) algorithm and its implementation using C#. We also learned how to apply the algorithm to the problem of ‘Counting Islands.’ Depth-First Search is an important aspect of graph theory and a useful tool for solving various problems.

Continue to practice solving a variety of problems to improve your algorithm skills. In the next lesson, we will discuss Breadth-First Search (BFS). Thank you!

C# Coding Test Course, Extended Euclidean Algorithm

Hello, today we will learn about one of the algorithms that frequently appears in coding tests, which is the Extended Euclidean Algorithm. In this article, we will cover the basic principles of the Extended Euclidean Algorithm, examine problems that utilize it, and discuss how to solve these problems using C#.

1. What is the Extended Euclidean Algorithm?

The Extended Euclidean Algorithm is an algorithm that finds the following two things for two integers a and b:

  1. It calculates the greatest common divisor (gcd(a, b)) of the two numbers a and b.
  2. It expresses that greatest common divisor in the form of integers x and y that correspond to the ratio of a and b. In other words, it can be represented as gcd(a, b) = ax + by.

This algorithm is primarily used for solving modular identities or linear equations. Specifically, it plays an important role in cryptography.

2. Need for the Algorithm

The extended version of the Euclidean Algorithm is a variant of the Euclidean algorithm that can find the greatest common divisor of two numbers through remainder calculations. However, its usefulness shines in the fact that it not only finds the greatest common divisor but also derives it by combining the two numbers. The x and y obtained through this process can be very useful tools in specific problems.

3. Implementation of the Algorithm

First, let’s define a basic C# function to implement the Extended Euclidean Algorithm. This function accepts two integers a and b as arguments and returns the corresponding greatest common divisor along with the values of x and y.


using System;

class Program
{
    static void Main(string[] args)
    {
        int a = 252;
        int b = 105;
        
        var result = ExtendedGCD(a, b);
        Console.WriteLine($"gcd: {result.Item1}, x: {result.Item2}, y: {result.Item3}");
    }

    static (int, int, int) ExtendedGCD(int a, int b)
    {
        if (b == 0)
        {
            return (a, 1, 0);
        }

        var (gcd, x1, y1) = ExtendedGCD(b, a % b);
        
        int x = y1;
        int y = x1 - (a / b) * y1;

        return (gcd, x, y);
    }
}
    

3.1 Code Explanation

The main part of this code is to implement a function called ExtendedGCD. This function is called recursively to find the greatest common divisor of two numbers:

  • As a base case, when b is 0, the gcd is a, and it returns x as 1 and y as 0.
  • Otherwise, it calls recursively using a and b, and calculates new x and y using the returned values.

4. Example Problem: Finding the GCD and Coefficients

Now, let’s solve a specific problem using the Extended Euclidean Algorithm. For example, given two integers a = 252 and b = 105, we will find their greatest common divisor and the coefficients x, y.

4.1 Problem Definition

For the given two numbers a and b, find the following:

  • gcd(a, b)
  • Find x and y that satisfy gcd(a, b) = ax + by.

4.2 Solution Process

Substituting a = 252 and b = 105 into the given problem, we apply the Extended Euclidean Algorithm. Below is the process:

  1. Call ExtendedGCD(252, 105)
  2. Here b = 105, calling it recursively with ExtendedGCD(105, 252 % 105)
  3. Continuing to call ExtendedGCD(105, 42) where 252 % 105 = 42
  4. Now calling ExtendedGCD(42, 21)
  5. Finally, when we reach ExtendedGCD(21, 0), it returns the greatest common divisor 21, along with the combination x=1, y=0.

During the return process of recursion, the values of x and y are updated one by one, eventually leading us to find gcd(252, 105) = 21 along with the values of x and y. These values are used to express the greatest common divisor as a linear combination of the given numbers.

4.3 Results

The results obtained through this process are as follows:

  • gcd(252, 105) = 21
  • With x = -1 and y = 3, it can be represented as 21 = 252 * (-1) + 105 * 3.

5. Practice Problem

So far, we have looked at the understanding and implementation of the Extended Euclidean Algorithm. Try to implement the Extended Euclidean Algorithm yourself through the following exercise problem.

Exercise Problem

Using the Extended Euclidean Algorithm, find the greatest common divisor and coefficients x, y for the following two integers:

  • a = 119, b = 544

Write your answers and verify them through your implemented C# code!

6. Conclusion

Today, we learned about the Extended Euclidean Algorithm. This algorithm is very useful for finding the greatest common divisor of one or more numbers and can express the result as a linear combination, applying to various mathematical problems. This knowledge is extremely important not only in coding tests but also in actual programming. Before moving on to the next advanced topic, it is recommended to review this algorithm once more and practice through various problems.

C# Coding Test Course, Finding Binomial Coefficients 1

In the process of preparing for coding tests, algorithm problems are one of the essential items that must be solved.
Today, we will solve the problem of calculating the binomial coefficient using C#.
The binomial coefficient is a mathematical methodology for calculating combinations, specifically finding nCk for given n and k.

What is a Binomial Coefficient?

The binomial coefficient represents the number of ways to choose k elements from n given elements.
Mathematically, the binomial coefficient is defined as follows:

            C(n, k) = n! / (k! * (n-k)!)
        

Here, n! denotes n factorial.
Factorial is the product of all positive integers up to the given number n:

            n! = n * (n-1) * (n-2) * ... * 2 * 1
        

For instance, C(5, 2) can be calculated as follows:

            C(5, 2) = 5! / (2! * (5-2)!) = 5 * 4 / (2 * 1) = 10
        

Problem Definition

Write a program to calculate the binomial coefficient C(n, k) for given n and k.
The input consists of two integers n and k (0 ≤ k ≤ n ≤ 10,000).
The output is the value of C(n, k).

Algorithm Approach

To solve this problem, we can use two main approaches.
The first is a recursive method, and the second is an iterative method.
However, since n can be as large as 10,000, the recursive method may lead to stack overflow.
Therefore, we will take an iterative approach.

1. Iterative Method

To calculate the binomial coefficient using the iterative method,
we will create a function to compute factorial and use this function to calculate C(n, k).
However, since the values of n and k can be large, the factorial values can become very large,
we can compute the value of nCk iteratively to reduce memory usage.

2. Optimized Calculation of Binomial Coefficient

Utilizing properties of the binomial coefficient, we can compute it as follows:

            C(n, k) = C(n, n-k)
        

Therefore, if k is greater than n/2, we can switch k to n-k to perform the calculation. This makes the computation more efficient.

C# Code Implementation

Now, let’s implement the C# code in an easy-to-understand way.
Below is the C# code that calculates the binomial coefficient:

        
        using System;

        class Program
        {
            static void Main(string[] args)
            {
                // Receive input
                string[] inputs = Console.ReadLine().Split(' ');
                int n = int.Parse(inputs[0]);
                int k = int.Parse(inputs[1]);

                // Calculate C(n, k)
                long result = BinomialCoefficient(n, k);
                
                // Output the result
                Console.WriteLine(result);
            }

            static long BinomialCoefficient(int n, int k)
            {
                // Optimization: if k is greater than n/2
                if (k > n / 2)
                {
                    k = n - k;
                }

                long result = 1;
                for (int i = 0; i < k; i++)
                {
                    result *= (n - i);
                    result /= (i + 1);
                }

                return result;
            }
        }
        
    

Code Explanation

The operation of the above code works as follows:

  1. It receives inputs n and k from the user.
  2. It calls the BinomialCoefficient method to calculate the binomial coefficient.
  3. It outputs the result.

The BinomialCoefficient method compares the value of k with n/2 and performs optimized calculations.
Then, using a loop, it actually calculates the binomial coefficient.
This method has a time complexity of O(k), making it efficient.

Test Cases

Below are test cases to verify the operation of the code:

        
        Input: 5 2
        Output: 10

        Input: 10 3
        Output: 120

        Input: 10 7
        Output: 120 // Same as C(10, 3)
        
        Input: 10000 5000
        Output: Approximate value (this case may take time to compute.)
        
    

Conclusion

Today, we solved the problem of calculating the binomial coefficient using C#.
We were able to efficiently compute the binomial coefficient through the iterative method and optimized calculation approach.
I hope this article helps in your preparation for coding tests.
I plan to continue a series of lectures covering various algorithm problems, so please stay tuned.

C# Coding Test Course, Finding the Minimum Number of Matrix Multiplication Operations

Problem Definition

This is a problem of finding the minimum number of operations required to multiply a given N matrices.
The number of operations for matrix multiplication is calculated as follows:
For two matrices A and B, they can be multiplied only if the number of columns in A is equal to the number of rows in B.
The number of operations is calculated as:

Operations count = A's row count * A's column count * B's column count

N matrices can be multiplied in sequence, but the total number of operations varies depending on the order of multiplication.
Therefore, we must find the optimal multiplication order, for which we will use Dynamic Programming techniques.

Input Format

The first line contains the number of matrices N (1 ≤ N ≤ 30).
The second line contains N integers M1, M2, …, MN (1 ≤ Mi ≤ 100) representing the size of each matrix.
Each integer represents the number of rows and columns of the matrix.

Output Format

Output the minimum number of operations required to multiply the matrices.

Example Input

3
10 20 30

Example Output

6000

Approach to Problem Solving

This problem can be solved using Dynamic Programming.
The problem of determining the order of matrix multiplication has the following recursive relationship.

1. Definition of Dynamic Programming

We define a DP array dp[i][j] that represents the minimum number of operations required to multiply the matrices from i to j.
Therefore, the goal is to compute dp[0][N-1].

2. Recursive Relationship

dp[i][j] = min(dp[i][k] + dp[k+1][j] + M[i] * M[k+1] * M[j+1]) (i ≤ k < j)
k can be set as another matrix between i and j, thereby considering all possible combinations to find the optimal multiplication order.

C# Code Implementation

Now we will implement the entire algorithm in C#.
The code below reads the matrix sizes through input and calculates the minimum number of operations using Dynamic Programming.

using System;

    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int[] M = new int[N + 1];
            string[] dimensions = Console.ReadLine().Split();

            for (int i = 0; i < N; i++)
            {
                M[i] = int.Parse(dimensions[i]);
                if (i != 0) M[i + 1] = int.Parse(dimensions[i]);
            }

            int[,] dp = new int[N, N];
            for (int len = 2; len <= N; len++)
            {
                for (int i = 0; i <= N - len; i++)
                {
                    int j = i + len - 1;
                    dp[i, j] = int.MaxValue;
                    for (int k = i; k < j; k++)
                    {
                        int q = dp[i, k] + dp[k + 1, j] + M[i] * M[k + 1] * M[j + 1];
                        dp[i, j] = Math.Min(dp[i, j], q);
                    }
                }
            }

            Console.WriteLine(dp[0, N - 1]);
        }
    }

Execution Explanation

The above C# code first receives input for the number of matrices and their sizes.
After initializing the dp[i][j] array, it sets indices i and j for all combinations of matrices through two nested loops.
It also considers all possible splits for k to calculate the minimum number of operations.
Finally, dp[0][N-1] gives the minimum multiplication operations.

Time Complexity and Space Complexity

The time complexity of this algorithm is O(N^3).
Due to the presence of two nested loops and one internal loop, it has a worst-case complexity of O(N^3).
The space complexity is O(N^2), as it requires memory space to store the DP array.

Conclusion

The problem of finding the minimum number of operations for matrix multiplication can be effectively solved using Dynamic Programming.
I hope the algorithm and code described above help you in solving coding test problems.
Through practice, I hope you can solve more problems and become familiar with various algorithmic techniques.

References

– For more information on algorithm problems, check platforms like Google, Baekjoon, and LeetCode.
– For the basics of Dynamic Programming and various patterns, please refer to related books.