C# Coding Test Course, Dictionary Search

Problem Description

Given a list of words and a word to search for, the task is to check whether the search word is included in the word list. This problem can
be solved using a dictionary data structure. This can be easily implemented in C# using the Dictionary class.

Algorithm Problem

Problem


        Given a list of words words, write the following checkWords method. 
        
The checkWords method takes a list of strings and a string as parameters, and it should return true if the string is in the list, and false if it is not.
Example
Input: words = ["apple", "banana", "cherry"], searchWord = "banana"
Output: true

Solution

One of the most efficient methods to solve this problem is by using a hash table. By using a hash table to store the list of words
beforehand, we can find words with O(1) time complexity during searches. Utilizing C#’s Dictionary class allows
us to implement this, so we will solve the problem using the following steps.

Step 1: Convert the Word List to a Dictionary

First, we will convert the given word list into a Dictionary. Each word will be used as a key and the value will be
an unnecessary empty string.


        // C# code
        using System;
        using System.Collections.Generic;

        public class DictionarySearch
        {
            private Dictionary wordDictionary;

            public DictionarySearch(List words)
            {
                wordDictionary = new Dictionary();
                foreach (var word in words)
                {
                    wordDictionary[word] = true; // Value is not needed, set to true
                }
            }
        }
        

Step 2: Write the Word Search Method

Now, we will write the checkWords method. This method will check whether the word being searched exists in the dictionary.


        public bool CheckWords(string searchWord)
        {
            return wordDictionary.ContainsKey(searchWord);
        }
        

Step 3: Implement the Final Class and Method

We will combine the above steps to complete the class as a final solution.


        using System;
        using System.Collections.Generic;

        public class DictionarySearch
        {
            private Dictionary wordDictionary;

            public DictionarySearch(List words)
            {
                wordDictionary = new Dictionary();
                foreach (var word in words)
                {
                    wordDictionary[word] = true;
                }
            }

            public bool CheckWords(string searchWord)
            {
                return wordDictionary.ContainsKey(searchWord);
            }
        }

        class Program
        {
            static void Main(string[] args)
            {
                List words = new List { "apple", "banana", "cherry" };
                DictionarySearch dictionarySearch = new DictionarySearch(words);

                Console.WriteLine(dictionarySearch.CheckWords("banana")); // Output: true
                Console.WriteLine(dictionarySearch.CheckWords("grape")); // Output: false
            }
        }
        

Complexity Analysis

– Time complexity: O(1) – Using a hash table to store and search for words is optimally performed in O(1) time.
– Space complexity: O(n) – Proportional to the size n of the given word list.

Key Points

– Using the Dictionary class is key to solving the problem. It ensures fast search speeds.
– In string comparisons, the results may differ based on case sensitivity, so caution is needed.
– After running the overall code, it is necessary to validate it through various test cases.

Conclusion

In this tutorial, we explained how to solve the dictionary search problem using C#. We demonstrated that a data structure approach
utilizing a hash table can efficiently satisfy the problem’s given conditions. The use of such data structures is frequently encountered
in actual coding tests, so practicing in advance is important.

C# coding test course, Ax + By = C

The coding test is a competition where programmers are evaluated on their skills in order to find suitable candidates for companies. Today, we are going to solve a frequently encountered topic in coding tests: equation problems. In particular, we will take a detailed look at how to solve problems in the form of ‘Ax + By = C’ using C#.

Problem Description

Given integers A, B, and C, write a program to find integer solutions (x, y) that satisfy the equation Ax + By = C. If solutions exist, all solutions must be printed, and if there are no solutions or if there are an infinite number of solutions, the corresponding messages should be printed.

Input Format

  • On one line, three integers A, B, C are provided. (−109 ≤ A, B, C ≤ 109)

Output Format

  • If there are no solutions, print “There are no solutions.”
  • If there are an infinite number of solutions, print “There are infinitely many solutions.”
  • If there are a finite number of solutions, all solutions must be printed.

Plan for Solving the Problem

The basic idea to solve the problem is as follows:

  1. If both A and B are 0:
    • If C is also 0, the equation is always true, so there are infinitely many solutions.
    • If C is not 0, there are no solutions.
  2. If A is 0:
    • In this case, the equation becomes By = C. If B is 0, there is a solution only when C is 0; otherwise, there are no solutions.
    • If B is not 0, there is a solution y = C/B, and x can be chosen arbitrarily, so there are infinitely many solutions.
  3. If B is 0:
    • In this case, the equation becomes Ax = C. If A is 0, there is a solution only when C is 0; otherwise, there are no solutions.
    • If A is not 0, there is a solution x = C/A, and y can be chosen arbitrarily, so there are infinitely many solutions.
  4. If both A and B are not 0:
    • If we choose an integer x arbitrarily, y becomes (C – Ax) / B.
    • To confirm y is an integer, C – Ax must be divisible by B. We will find and print the value of y for that x value.

C# Code Implementation


using System;

class Program
{
    static void Main()
    {
        string[] input = Console.ReadLine().Split(' ');
        int A = int.Parse(input[0]);
        int B = int.Parse(input[1]);
        int C = int.Parse(input[2]);

        if (A == 0 && B == 0)
        {
            if (C == 0)
                Console.WriteLine("There are infinitely many solutions.");
            else
                Console.WriteLine("There are no solutions.");
        }
        else if (A == 0)
        {
            if (B == 0)
            {
                if (C == 0)
                    Console.WriteLine("There are infinitely many solutions.");
                else
                    Console.WriteLine("There are no solutions.");
            }
            else
            {
                Console.WriteLine($"y = {C / B}, x can be chosen arbitrarily.");
            }
        }
        else if (B == 0)
        {
            if (A == 0)
            {
                if (C == 0)
                    Console.WriteLine("There are infinitely many solutions.");
                else
                    Console.WriteLine("There are no solutions.");
            }
            else
            {
                Console.WriteLine($"x = {C / A}, y can be chosen arbitrarily.");
            }
        }
        else
        {
            for (int x = -100; x <= 100; x++)
            {
                if ((C - A * x) % B == 0)
                {
                    int y = (C - A * x) / B;
                    Console.WriteLine($"x = {x}, y = {y}");
                }
            }
        }
    }
}

Code Explanation

The above code is implemented to check various cases based on the given values of A, B, and C to derive the solution. It checks whether solutions exist and whether the number of solutions is infinite or finite.

In particular, when A is not equal to 0, the code iterates over a specific range to find y values that satisfy the condition for different x values. The range for x is limited to -100 to 100 here, but in reality, a broader range could be considered. However, in cases where there are infinitely many solutions, x and y values can be printed for specific ranges.

Conclusion

This problem illustrates the necessity of considering various cases to find solutions to an equation. We explored a modular problem-solving approach through basic mathematical principles and C# coding. In future coding tests, it would be beneficial to reference this approach when encountering similar problems.

Finally, validate this code with various test cases and consider the extensibility of solving different problems with the same algorithm. Practicing algorithms is never an easy task, but remember that perseverance and consistency are crucial!

C# Coding Test Course, Game Development

Problem Description

In a fictional game, there are enemy characters and player characters. The enemy character starts from the (x, y) coordinate and needs to approach the player character. The enemy character must find the optimal path to move closest to the player’s position every turn. To solve this problem, we need to calculate the optimal path for the enemy to approach the player.

Problem Definition

The following inputs are provided:

  • Enemy’s initial position (X1, Y1)
  • Player’s position (X2, Y2)

The output should return the optimal path for the enemy to reach the player (including all intermediate coordinates).

Input Example

    0 0
    3 4
    

Output Example

    (0,0) → (1,1) → (2,2) → (3,3) → (3,4)
    

Solution Strategy

This problem involves basic distance calculation and path tracking. We need to update the enemy’s position each turn and calculate the next position to approach the player. To do this, we use the Euclidean distance formula to calculate the distance and direction between two points, finding the optimal path the enemy can take each turn.

1. Distance Calculation

    dist(X1, Y1, X2, Y2) = sqrt((X2 - X1)² + (Y2 - Y1)²)
    

2. Movement Logic Implementation

By allowing the enemy to approach the player each turn, we adjust the X and Y coordinates.

3. Path Storage

We store the coordinates after each move in a list to finally output the path.

C# Implementation Example

    using System;
    using System.Collections.Generic;

    class Program
    {
        static void Main(string[] args)
        {
            // Enemy's initial position
            var enemyPosition = (X: 0, Y: 0);
            // Player's position
            var playerPosition = (X: 3, Y: 4);
            var path = new List<(int, int)>();

            while (enemyPosition != playerPosition)
            {
                path.Add(enemyPosition);
                enemyPosition = MoveTowardsPlayer(enemyPosition, playerPosition);
            }
            path.Add(playerPosition);

            Console.WriteLine("Path taken:");
            foreach (var pos in path)
                Console.WriteLine($"({pos.Item1}, {pos.Item2})");
        }

        static (int, int) MoveTowardsPlayer((int X, int Y) enemy, (int X, int Y) player)
        {
            int newX = enemy.X + (player.X > enemy.X ? 1 : (player.X < enemy.X ? -1 : 0));
            int newY = enemy.Y + (player.Y > enemy.Y ? 1 : (player.Y < enemy.Y ? -1 : 0));
            return (newX, newY);
        }
    }
    

Conclusion

This problem helps us understand pathfinding and interactions between objects in game development. Solving coding test problems using C# will be beneficial for future real game development projects. We can see that we can calculate the optimal movement path for the enemy to reach the player based on the given positions.

C# Coding Test Course, Understanding Friend Relationships

Problem Description

You are developing an email service that analyzes friendship relationships. Each user has a friend list, and friends form a close relationship with each other. Your task is to determine whether two users are friends based on the given friendship data and to compute the indirect closeness between them, including friends of friends.

The input consists of a friend list and the IDs of the users to check. The friend list is in the form of a list, where each element consists of a pair of user IDs that are friends with each other. Based on this information, write a program to determine if two users are indirectly friends or not.

Input Example

    Friend List: [(1, 2), (2, 3), (3, 4), (1, 5)]
    User ID1: 1
    User ID2: 4
    

Output Example

    1 and 4 are friends or indirect friends.
    

Solution Method

To solve this problem, you need to use the structure of a graph. Model the friendship relationships as vertices and edges, and explore the relationship between the two users using the BFS (Breadth-First Search) or DFS (Depth-First Search) algorithm. This process will allow you to confirm both direct and indirect friendships.

1. Graph Modeling

Create an undirected graph based on the friend list. Each user is represented by their user ID, and the friendship relationship is represented as an edge between two users.

2. Implementing BFS or DFS

Now that the graph is created, find the indirect friendship relationship between the given two users using BFS or DFS. During this process, keep track of visited nodes and stop the search if a node is already visited.

3. Code Example


using System;
using System.Collections.Generic;

class FriendNetwork
{
    private Dictionary> graph = new Dictionary>();

    public void AddFriendship(int user1, int user2)
    {
        if (!graph.ContainsKey(user1))
            graph[user1] = new List();
        if (!graph.ContainsKey(user2))
            graph[user2] = new List();

        graph[user1].Add(user2);
        graph[user2].Add(user1);
    }

    public bool AreFriends(int user1, int user2)
    {
        if (user1 == user2) return true;
        HashSet visited = new HashSet();
        Queue queue = new Queue();
        
        queue.Enqueue(user1);
        visited.Add(user1);

        while (queue.Count > 0)
        {
            int current = queue.Dequeue();
            foreach (var friend in graph[current])
            {
                if (friend == user2)
                    return true;

                if (!visited.Contains(friend))
                {
                    visited.Add(friend);
                    queue.Enqueue(friend);
                }
            }
        }
        return false;
    }
}

class Program
{
    static void Main(string[] args)
    {
        FriendNetwork network = new FriendNetwork();
        
        // Add friendships
        network.AddFriendship(1, 2);
        network.AddFriendship(2, 3);
        network.AddFriendship(3, 4);
        network.AddFriendship(1, 5);

        // Check the relationship between two users
        int user1 = 1;
        int user2 = 4;

        if (network.AreFriends(user1, user2))
            Console.WriteLine($"{user1} and {user2} are friends or indirect friends.");
        else
            Console.WriteLine($"{user1} and {user2} are not friends.");
    }
}

4. Code Explanation

The code above demonstrates how to manage friendship relationships and verify if two users are friends. The FriendNetwork class provides a graph to store the friend list and methods for adding friendships. The AreFriends method uses BFS to check the connectivity between the two users.

5. Complexity Analysis

The time complexity of this implementation, which uses an adjacency list for the graph, is O(V + E), where V is the number of vertices and E is the number of edges. The space complexity is also O(V + E).

Conclusion

In this way, graph and search algorithms can be appropriately utilized to solve friendship relationship problems. In C#, various data structures and algorithms can elegantly address such issues, showcasing strong problem-solving skills in job interviews and coding tests.

C# Coding Test Course, I Will Become the President of the Residents’ Association

Problem Description

The given problem is as follows. You need to write an algorithm to become the head of the residents’ association for the number of floors in an apartment and the number of residents on each floor. The goal is to implement an algorithm that calculates the number of residents on a specific floor.

The apartment has floors from 0 to N (0 ≤ N ≤ 14), and there are k residents living in apartment k on the n-th floor. For example, there are 3 residents living in apartment 3 on the 3rd floor. The number of residents on each floor is calculated as follows.

– 1st floor: 1 resident in apartment 1, 2 residents in apartment 2, 3 residents in apartment 3…
– 2nd floor: 1 resident in apartment 1, 3 residents in apartment 2, 6 residents in apartment 3…
– The number of residents in apartment k on the n-th floor is stored as the sum of the number of residents in apartment k-1 on the n-th floor and the number of residents in apartment k on the n-th floor.

Input Format

The first line contains the number of test cases T (1 ≤ T ≤ 100). In the following T lines, each test case provides integers N (0 ≤ N ≤ 14) and K (1 ≤ K ≤ 14). Here, N represents the floor number, and K represents the apartment number.

Output Format

For each test case, you should output the number of residents in apartment k on the n-th floor.

Example Input


2
1 3
2 3

Example Output


3
6

Problem Solving Process

To solve this problem, we will use Dynamic Programming (DP) techniques. We can use the following relationship to find the number of residents in apartment k on the n-th floor.


residents(n, k) = residents(n-1, 1) + residents(n, k-1)

Here, we initialize residents(0, k) = k and residents(n, 0) = 0. Below are the basic steps for storing the number of residents in a table.

Step 1: Initialization

First, declare a 2-dimensional array to store the number of residents and set the initial values.


int[,] dp = new int[15, 15]; // Array for 15 floors and 15 apartments
for (int i = 0; i <= 14; i++) {
dp[0, i] = i; // k residents live in apartment k on the 0th floor.
}

Step 2: Fill the DP Table

We fill in the dp table using a nested loop. We consider all cases for n-th floor and k-th apartment as follows.


for (int n = 1; n <= 14; n++) {
for (int k = 1; k <= 14; k++) {
dp[n, k] = dp[n - 1, k] + dp[n, k - 1];
}
}

Step 3: Output Results

Calculate and output the number of residents for all test cases. Below is an example of the final implementation.


using System;
class Program {
static void Main(string[] args) {
int[,] dp = new int[15, 15];
for (int i = 0; i <= 14; i++) {
dp[0, i] = i;
}
for (int n = 1; n <= 14; n++) {
for (int k = 1; k <= 14; k++) {
dp[n, k] = dp[n - 1, k] + dp[n, k - 1];
}
}
int T = int.Parse(Console.ReadLine());
for (int t = 0; t < T; t++) {
string[] input = Console.ReadLine().Split();
int n = int.Parse(input[0]);
int k = int.Parse(input[1]);
Console.WriteLine(dp[n, k]);
}
}
}

Conclusion

Through this problem, we were able to understand the basic concepts of dynamic programming and enhance our problem-solving skills. This problem is one of the types frequently asked in algorithm tests, and there are various variations. To apply this in actual coding tests, it is important to practice DP techniques repeatedly.