C# Coding Test Course, Making Cocktails

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!

C# Coding Test Course, Jumong’s Command

Hello! In this post, we will solve a coding test problem using the C# language. The topic is ‘Jumong’s Command’. This problem simulates the situation where Jumong issues commands and implements an algorithm to handle them efficiently. I will explain in detail how to solve the problem and implement the code.

Problem Description

Jumong gives commands to warriors before each battle. Each warrior performs their duties according to the commands they receive. The warriors interpret Jumong’s commands as follows:

  • Warrior’s unique number (a positive integer starting from 1)
  • Content of the command (indicates what action needs to be performed)

Jumong has N warriors and M commands. The commands are given in the following format:

    1. ATTACK A B  // Warrior A attacks warrior B
    2. DEFENSE A   // Warrior A assumes a defensive posture
    3. RETREAT A    // Warrior A retreats
    

If a warrior successfully performs their duty, the result should output ‘SUCCESS’, and if they fail, it should output ‘FAIL’. Warriors do not perform commands if they lose interest before executing the command. Interest decreases over time. If a warrior is in a situation where they cannot perform the command, ‘FAIL’ should be printed.

Input Format

The input is given in the following format:

    N (number of warriors)
    M (number of commands)
    Interest decay rate (0 ~ 1)
    Commands (M commands)
    

Output Format

The results are printed for each command. The results are listed in the same order as the commands.

Approach to the Problem

This problem requires an algorithm to process the given commands and determine whether each warrior can perform the commands. To solve the problem, I will consider the following steps:

  1. Create a data structure to store the warriors and command lists.
  2. Assume that each warrior initially has 100% interest.
  3. Decrease the warrior’s interest according to the given interest decay rate, and process each command while updating their status.
  4. Record success or failure based on the results of the commands.

Implementation

Code Example

The following is a C# implementation based on the above approach:


using System;
using System.Collections.Generic;

namespace JumongCommand
{
    class Warrior
    {
        public int Id { get; set; }
        public double Interest { get; set; }

        public Warrior(int id)
        {
            Id = id;
            Interest = 1.0; // 100% interest
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            int M = int.Parse(Console.ReadLine());
            double interestDecayRate = double.Parse(Console.ReadLine());

            List warriors = new List();
            for (int i = 1; i <= N; i++)
            {
                warriors.Add(new Warrior(i));
            }

            List results = new List();
            for (int i = 0; i < M; i++)
            {
                string command = Console.ReadLine();
                string[] parts = command.Split(' ');

                if (parts[0] == "ATTACK")
                {
                    int attackerId = int.Parse(parts[1]);
                    int targetId = int.Parse(parts[2]);
                    ProcessAttack(warriors, results, attackerId, targetId);
                }
                else if (parts[0] == "DEFENSE")
                {
                    int defenderId = int.Parse(parts[1]);
                    ProcessDefense(warriors, results, defenderId);
                }
                else if (parts[0] == "RETREAT")
                {
                    int retreatId = int.Parse(parts[1]);
                    ProcessRetreat(warriors, results, retreatId);
                }

                // Apply interest decay
                foreach (var warrior in warriors)
                {
                    warrior.Interest -= interestDecayRate;
                    if (warrior.Interest < 0)
                        warrior.Interest = 0;
                }
            }

            foreach (var result in results)
            {
                Console.WriteLine(result);
            }
        }

        static void ProcessAttack(List warriors, List results, int attackerId, int targetId)
        {
            var attacker = warriors[attackerId - 1];
            var target = warriors[targetId - 1];

            if (attacker.Interest > 0)
            {
                results.Add("SUCCESS");
            }
            else
            {
                results.Add("FAIL");
            }
        }

        static void ProcessDefense(List warriors, List results, int defenderId)
        {
            var defender = warriors[defenderId - 1];

            if (defender.Interest > 0)
            {
                results.Add("SUCCESS");
            }
            else
            {
                results.Add("FAIL");
            }
        }

        static void ProcessRetreat(List warriors, List results, int retreatId)
        {
            var retreatingWarrior = warriors[retreatId - 1];

            if (retreatingWarrior.Interest > 0)
            {
                results.Add("SUCCESS");
            }
            else
            {
                results.Add("FAIL");
            }
        }
    }
}

Code Explanation

The code defines a ‘Warrior’ class to manage the state of each warrior. Each warrior has a unique ID and a current interest as properties. The main program reads the warrior list and command list in order and calls separate methods to process each command, updating the results.

Verification of Results

After the commands are completed, the result list is printed to confirm the final success or failure results. This is a simple implementation of how warriors execute commands and manage resources.

Test Cases

We can create several test cases to check if it works properly.

Example Input

5
3
0.1
ATTACK 1 2
DEFENSE 3
RETREAT 4

Expected Results

SUCCESS
SUCCESS
SUCCESS

By testing various inputs in this manner, we can check if the algorithm correctly responds to all situations.

Conclusion

In this post, we explored the process of solving the ‘Jumong’s Command’ problem. Understanding the structure and implementation of the algorithm, as well as building the logic for processing various commands, was important. I hope you continue to improve your skills through more algorithm problems. Thank you!

C# Coding Test Course, Calculating Continuous Sum

While preparing for coding tests or algorithm problem-solving, it is important to understand and learn approaches for various types of problems. Today, we will take an in-depth look at the problem of finding the sum of continuous elements.

Problem Description

We will examine the problem of calculating the sum of consecutive elements in a given integer array and finding the shortest continuous subarray that meets a specific criteria.

Problem Definition


Problem: Find the Sum of Consecutive Elements

Given an integer array nums and an integer target, 
return the length of the shortest continuous subarray whose sum is greater than or equal to target. 
If there is no such subarray, return 0.

Input Example:
nums = [2,3,1,2,4,3]
target = 7

Output Example:
2 (length of subarray [4,3])

Problem Solving Approach

To solve this problem, two main approaches can be used: brute force (exhaustive search) and the two-pointer (sliding window) method. Here, we will use the two-pointer method for an efficient solution.

Two-Pointer Approach

The two-pointer approach involves exploring the array using two pointers (left and right) to find the subarray that satisfies the desired condition. The advantage of this method is its time complexity of O(n), making it efficient.

Step-by-Step Solution

  1. Initialization: Initialize the two pointers. Set the left pointer left to 0 and the right pointer right to 0. Also, initialize the current sum currentSum to 0 and the minimum length minLength to infinity.
  2. Condition Check: Traverse the array using the right pointer and add nums[right] to currentSum. Then check if currentSum is greater than or equal to target.
  3. Adjusting the Subarray: If currentSum is greater than or equal to target, update the minimum length and increase the left pointer while subtracting nums[left] from currentSum. This helps find the shortest possible subarray.
  4. Termination Condition: Repeat this process until the right pointer reaches the end of the array.

Function Implementation

Now, let’s implement the above logic through C# code.


using System;

public class Solution {
    public int MinSubArrayLen(int target, int[] nums) {
        int left = 0;
        int currentSum = 0;
        int minLength = int.MaxValue;

        for (int right = 0; right < nums.Length; right++) {
            currentSum += nums[right];

            while (currentSum >= target) {
                minLength = Math.Min(minLength, right - left + 1);
                currentSum -= nums[left];
                left++;
            }
        }

        return minLength == int.MaxValue ? 0 : minLength;
    }
}

Code Explanation

The above C# code works as follows:

  • Initial Variable Setup: Initialize left, currentSum, and minLength.
  • Array Traversal: Traverse the array using the right variable and add the current element to currentSum.
  • Condition Check: If currentSum is greater than or equal to target, update minLength and increase the left pointer while subtracting nums[left] from currentSum.
  • Return Result: Finally, if minLength has not been updated, return 0; otherwise, return the found minimum length.

Example Test Cases

Now, let’s write some example test cases to test this function.


public static void Main(string[] args) {
    Solution sol = new Solution();
    
    Console.WriteLine(sol.MinSubArrayLen(7, new int[] { 2, 3, 1, 2, 4, 3 })); // Output: 2
    Console.WriteLine(sol.MinSubArrayLen(4, new int[] { 1, 4, 4 }));        // Output: 1
    Console.WriteLine(sol.MinSubArrayLen(11, new int[] { 1, 1, 1, 1, 1, 1 })); // Output: 0
    Console.WriteLine(sol.MinSubArrayLen(8, new int[] { 2, 3, 1, 2, 4, 3 })); // Output: 2
}

Conclusion

In this lecture, we learned the problem-solving approach using two pointers through the problem of finding the sum of consecutive elements. It is important to accurately understand the nature of a problem and choose the appropriate approach to solve algorithmic problems. In practice, solving various problems helps gain experience and become familiar with different algorithms, so continuous practice is encouraged.

Try to enhance your algorithmic sense by encountering and solving various problems!

C# Coding Test Course, Finding Minimum Value 2

Introduction: In this article, we will address the algorithmic problem of finding the two smallest integers within a given array and explain the solution process in detail. This problem is frequently encountered in tests, and it is important to have a clear understanding and knowledge of various solution methods.

Problem Description

Write a function to find the two smallest numbers in the given integer array and return their sum.

Example:

            Input: [3, 1, 4, 1, 5, 9, 2, 6, 5]
            Output: 2 (1 + 1)
            

If the same number is included, the same number can be used twice.

Approach to the Problem

We can consider several approaches to solve this problem.

1. Using Sorting

We can sort the array and then take the first two elements to calculate their sum.

2. Using Nested Loops

It is also possible to find the minimum values by directly comparing two integers using nested loops. However, this method may be inefficient.

3. Finding Minimum Values Using a Single Loop

The method of finding the two smallest numbers using a single loop is the most efficient. It allows you to find the minimum values while iterating through the array only once.

Solution Method 1: Using Sorting

using System;
using System.Linq;

public class Solution {
    public static int FindTwoMinSum(int[] arr) {
        // Sort the array.
        Array.Sort(arr);
        // Return the sum of the two smallest numbers.
        return arr[0] + arr[1];
    }
}

// Example usage
public class Program {
    public static void Main() {
        int[] numbers = {3, 1, 4, 1, 5, 9, 2, 6, 5};
        Console.WriteLine(FindTwoMinSum(numbers)); // Output: 2
    }
}
            

This method is intuitive and easy to understand, but it has a time complexity of O(n log n), which can be relatively slow. Performance degradation is a concern, especially for large array sizes.

Solution Method 2: Using Nested Loops

using System;

public class Solution {
    public static int FindTwoMinSum(int[] arr) {
        int min1 = int.MaxValue;
        int min2 = int.MaxValue;

        foreach (int num in arr) {
            if (num < min1) {
                min2 = min1;
                min1 = num;
            } else if (num < min2) {
                min2 = num;
            }
        }
        
        return min1 + min2;
    }
}

// Example usage
public class Program {
    public static void Main() {
        int[] numbers = {3, 1, 4, 1, 5, 9, 2, 6, 5};
        Console.WriteLine(FindTwoMinSum(numbers)); // Output: 2
    }
}
            

This method has a time complexity of O(n), making it more efficient. It finds the two smallest numbers while iterating through the array only once. While this method excels in performance, it does introduce a bit of code complexity.

Conclusion

In this article, we explored various ways to solve the problem of finding the two smallest values in a given array using C#. The first method, using sorting, is simple to implement but lacks in performance. In contrast, the approach utilizing a single loop is efficient and practical. Such problems often appear in algorithm tests, so it is essential to practice and achieve a good understanding.

Learning various problem-solving methods will be beneficial not only for coding tests but also in actual development environments. Thank you.

C# Coding Test Course, Let’s Try DDR

In this session, we will solve algorithm problems for employment using C#. The topic will be based on a popular dance rhythm game called DDR (Dance Dance Revolution). We will implement the input patterns required in the DDR game in code and write an algorithm to process those patterns.

Problem Description

You are a developer analyzing the patterns of the DDR game. In DDR, input is received for each direction panel (up, down, left, right). The input consists of a series of directional key patterns given at specific time intervals. Each direction can be represented as follows:

  • U: Up
  • D: Down
  • L: Left
  • R: Right

Based on the given input pattern, you need to determine how accurately the user has inputted that pattern. We will check the length of the pattern and whether it was input correctly within a certain time frame to assign a score.

Input Format

  • The first line contains the length of the pattern N. (1 ≤ N ≤ 100)
  • The second line contains the pattern made up of N characters.
  • The third line contains the user’s input pattern X.

Output Format

The accuracy score indicating how closely the user’s input matches the pattern will be output. The score is calculated as follows:

  • Scores are calculated only when the length of the pattern and the user’s input length are the same.
  • 10 points are awarded for each matching character.
  • If there is a mismatch, the score will be lost at that index, and the penalty is 5 points.
  • The final score is output.

Example Input

5
UDLRU
UDLRD

Example Output

35

Problem Solving Process

Now, let’s take a step-by-step look at how to write the code to solve this problem. First, we will outline the required functionality.

  1. Declare variables to store the length of the pattern (N) and the input pattern.
  2. Store the pattern in an array or list of length N.
  3. Store the user’s input pattern in an array or list.
  4. Compare the input pattern with the existing pattern to check for matches and calculate scores.
  5. Output the final score.

Writing C# Code

Now, we will write the C# code according to the above requirements.


using System;

class DDR
{
    static void Main()
    {
        // Get the length of the pattern N
        int N = int.Parse(Console.ReadLine());
        
        // Declare string arrays to store the DDR pattern and user input pattern
        string pattern = Console.ReadLine();
        string userInput = Console.ReadLine();
        
        // Initialize score variable
        int score = 0;
        
        // Check if the input length is the same as the pattern
        if (N != userInput.Length)
        {
            Console.WriteLine("The input length does not match the pattern length.");
            return;
        }

        // Calculate scores by comparing the pattern and user input
        for (int i = 0; i < N; i++)
        {
            if (pattern[i] == userInput[i])
            {
                score += 10; // Award 10 points for each match
            }
            else
            {
                score -= 5; // Deduct 5 points for each mismatch
            }
        }
        
        // Output the final score
        Console.WriteLine(score);
    }
}

The code above is relatively simple, designed to receive the DDR pattern input and compare it with the user input to calculate the score. When executed, it prompts the user for the pattern and input, calculates the score, and outputs it.

Code Explanation

1. using System;: Declaration to use the basic library of C#.

2. class DDR: Defines a class named DDR.

3. static void Main(): The entry point of the program. All code execution starts here.

4. The user's input is taken using int.Parse(Console.ReadLine()) and stored in an integer variable N.

5. The pattern and user input are both received and stored as strings.

6. The score variable score is initialized.

7. It checks whether the length of the user's input matches the length of the pattern. If not, the program terminates.

8. The logic compares the pattern and awards 10 points for matches and deducts 5 points for mismatches.

9. Finally, it outputs the final score.

Conclusion

In this lecture, we structured an algorithm problem based on the DDR pattern and explored how to solve it using C#. We wrote logic for pattern analysis and score calculation through algorithm problem-solving. I hope this helps improve your current skills in algorithms and coding. Let's continue studying various topics together in the future.