C# Coding Test Course, Radix Sort

Hello! Today we will explore the concept of Radix Sort and how to implement it using C#. Radix Sort is one of the efficient ways to sort unsorted data, particularly useful when the range of integers is limited.

1. What is Radix Sort?

Radix Sort is a method of sorting numbers based on the position of their digits, typically divided into LSD (Least Significant Digit) and MSD (Most Significant Digit) methods. Using L starts the sorting from the lowest digit, while starting with M sorts from the highest digit.

2. How Radix Sort Works

The mechanism of Radix Sort works as follows:

  1. Starting from the least significant digit, sort the data based on each digit.
  2. Repeat the sorting process up to the most significant digit.
  3. In this process, a stable sorting algorithm, such as Counting Sort, is used to sort based on each digit.

3. Time Complexity of Radix Sort

The time complexity of Radix Sort is O(d * (n + k)), where d is the number of digits in the numbers, n is the number of integers to sort, and k is the range of digits from 0 to the maximum digit. One of the biggest advantages is that Radix Sort is stable.

4. Example Problem Solving with Radix Sort

Now, let’s solve a problem by sorting a specific list of numbers using Radix Sort.

Problem Description

Sort the given list of integers in ascending order using Radix Sort.
Input: [170, 45, 75, 90, 802, 24, 2, 66]
Output: [2, 24, 45, 66, 75, 90, 170, 802]

Problem Solving Process

  1. Start from the least significant digit. Sort using the first digit in the units place.
C#
public static void LSDRadixSort(int[] array)
{
    // Find the largest number.
    int max = array.Max();
    // Find the number of digits.
    for (int exp = 1; max / exp > 0; exp *= 10)
    {
        CountingSort(array, exp);
    }
}

private static void CountingSort(int[] array, int exp)
{
    int n = array.Length;
    int[] output = new int[n]; // Sorted array
    int[] count = new int[10]; // Count for digits 0-9

    // Count the number of elements corresponding to the digit.
    for (int i = 0; i < n; i++)
    {
        count[(array[i] / exp) % 10]++;
    }

    // Calculate cumulative count.
    for (int i = 1; i < 10; i++)
    {
        count[i] += count[i - 1];
    }

    // Create the sorted array.
    for (int i = n - 1; i >= 0; i--)
    {
        output[count[(array[i] / exp) % 10] - 1] = array[i];
        count[(array[i] / exp) % 10]--;
    }

    // Update the original array.
    for (int i = 0; i < n; i++)
    {
        array[i] = output[i];
    }
}

In the above code, we define the LSDRadixSort method. This method sorts using CountingSort for each digit.

5. Implementation and Execution

Now, we will run the entire program using the above functions. Below is the complete code for the program:

C#
using System;
using System.Linq;

class Program
{
    public static void Main(string[] args)
    {
        int[] array = {170, 45, 75, 90, 802, 24, 2, 66};
        LSDRadixSort(array);

        Console.WriteLine("Sorted array:");
        Console.WriteLine(string.Join(", ", array));
    }
  
    public static void LSDRadixSort(int[] array)
    {
        int max = array.Max();
        for (int exp = 1; max / exp > 0; exp *= 10)
        {
            CountingSort(array, exp);
        }
    }

    private static void CountingSort(int[] array, int exp)
    {
        int n = array.Length;
        int[] output = new int[n];
        int[] count = new int[10];

        for (int i = 0; i < n; i++)
        {
            count[(array[i] / exp) % 10]++;
        }

        for (int i = 1; i < 10; i++)
        {
            count[i] += count[i - 1];
        }

        for (int i = n - 1; i >= 0; i--)
        {
            output[count[(array[i] / exp) % 10] - 1] = array[i];
            count[(array[i] / exp) % 10]--;
        }

        for (int i = 0; i < n; i++)
        {
            array[i] = output[i];
        }
    }
}

When executing the above code, the following results can be obtained:


Sorted array:
2, 24, 45, 66, 75, 90, 170, 802

6. Advantages and Disadvantages of Radix Sort

The advantages of Radix Sort are as follows:

  • Fast speed: Excellent performance under certain conditions. (i.e., when the range is limited)
  • Stable sort: Maintains relative order of equivalent values.

However, there are also disadvantages:

  • Memory usage: Requires additional memory.
  • Limited data: Difficult to use for data types other than integers.

7. Conclusion

Today, we examined the basic concept of Radix Sort, its operation principle, time complexity, and problem examples. This algorithm is very useful for efficiently handling collections that contain integers. It is also good to try Radix Sort during actual coding tests.

See you in the next lecture! Good luck with your coding test preparation!

C# Coding Test Course, Finding the Longest Increasing Subsequence

Problem Description

Let’s solve the problem of finding the length of the longest increasing subsequence in a given array. An increasing subsequence is a subarray of elements from the array that have different indices and are sorted in ascending order. For example, for the array [10, 22, 9, 33, 21, 50, 41, 60, 80], the longest increasing subsequence is [10, 22, 33, 50, 60, 80], which has a length of 6.

Problem Definition

Write a method that returns the length of the longest increasing subsequence in a given integer array arr.

Input

  • The first line contains the length of the array n. (1 ≤ n ≤ 1000)
  • The second line contains n integers. Each integer is -10^6 ≤ arr[i] ≤ 10^6.

Output

Print the length of the longest increasing subsequence.

Example

Input

9
10 22 9 33 21 50 41 60 80

Output

6

Solution Method

Among the various algorithms for finding the longest increasing subsequence, the most commonly used method is Dynamic Programming. This approach breaks the problem into smaller subproblems, solving each one and using the results to solve the overall problem.

Step-by-Step Explanation

  1. Initialization of Array: Assume the length of the given array arr is n. We initialize a dynamic programming array dp to store the length of the longest increasing subsequence up to each index i. Initially, we set the value to 1 for each element, as each element can form a subsequence by itself.
  2. Nested Loops: Use nested loops to compare all pairs of indices i and j. Here, i should be greater than j, and if arr[i] is greater than arr[j] (i.e., satisfying the increasing condition), we update the value of dp[i]. Specifically, we set dp[i] = max(dp[i], dp[j] + 1). This means adding 1 to the length of the sequence that includes j.
  3. Finding the Maximum Length: After calculating all sequence lengths, we find the maximum value in the dp array and return it.

C# Implementation

using System;

class Program
{
    static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        int[] arr = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);

        int[] dp = new int[n];
        for (int i = 0; i < n; i++)
        {
            dp[i] = 1; // Each element can form at least 1 subsequence
        }

        for (int i = 1; i < n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                if (arr[i] > arr[j] && dp[i] < dp[j] + 1)
                {
                    dp[i] = dp[j] + 1;
                }
            }
        }

        int maxLength = dp[0];
        for (int i = 1; i < n; i++)
        {
            if (dp[i] > maxLength)
            {
                maxLength = dp[i];
            }
        }

        Console.WriteLine(maxLength);
    }
}

Time Complexity

The time complexity of this algorithm is O(n^2). This is because it uses two nested loops. Both the outer and inner loops run for each length of the array, leading to a total of n * n operations in the worst case. The space complexity is O(n), which is the space needed to store the dp array.

Conclusion

In this tutorial, we implemented an algorithm to find the longest increasing subsequence in C#. This problem is frequently encountered in coding tests and is very useful for learning and mastering the basics of dynamic programming. Through consistent practice, one can develop the thinking and approaches needed to solve a variety of problems.

C# Coding Test Course, Delivering Gifts

Problem Description

You are planning an event to deliver Christmas surprise gifts to your friends.
There are N friends, and each friend has a list of gifts they would like to receive.
You need to ensure that as many friends as possible receive their desired gifts.
Each friend specifies one gift they want, and there is only one gift.
However, the gift can only be delivered to one friend.

Input Format

  • The first line contains the number of friends N.
  • From the second line up to the Nth line, the gifts each friend wants are listed.

Output Format

Output the maximum number of friends who can receive their desired gifts.

Constraints

  • 1 ≤ N ≤ 100
  • Gifts are represented by uppercase letters, with a maximum of 26 types.

Approach to the Problem

This problem is fundamentally similar to the maximum matching problem in graph theory.
Each friend can be represented as a vertex, and the gifts they desire can be represented as edges.
To solve this problem, we will take the friends’ gift requests as input,
create a list of friends requesting each gift, and then use an algorithm to find the maximum matching.

Algorithm Overview

– Construct a graph based on the number of friends and the requested gifts.
– Explore the graph to find assignable gifts for each friend.
– Use the DFS (Depth First Search) algorithm to attempt possible matchings.

Code Implementation

Below is the solution code for the gift delivery problem written in C#.

                
using System;
using System.Collections.Generic;

class Program
{
    static int N;                         // Number of friends
    static List friends;          // List of gifts friends want
    static Dictionary gifts; // Matching status of each gift
    static bool[] visited;                // Record of DFS visitation

    static void Main(string[] args)
    {
        N = int.Parse(Console.ReadLine());
        friends = new List();

        for (int i = 0; i < N; i++)
        {
            friends.Add(Console.ReadLine());
        }

        gifts = new Dictionary();
        int totalMatched = 0;

        foreach (var friend in friends)
        {
            visited = new bool[26];
            if (DFS(friend[0], ref totalMatched))
            {
                totalMatched++;
            }
        }

        Console.WriteLine(totalMatched);
    }

    static bool DFS(char gift, ref int totalMatched)
    {
        int index = gift - 'A';

        if (visited[index]) return false; // Gift already visited
        visited[index] = true;

        if (!gifts.ContainsKey(gift.ToString()) || gifts[gift.ToString()] == -1)
        {
            gifts[gift.ToString()] = totalMatched; // Gift matching
            return true;
        }

        return false;
    }
}
                
            

Code Explanation

The above code solves the gift delivery problem through the following procedures.

  • Input Reading: Read the number of friends N and input the desired gifts for each friend.
  • DFS Function Implementation: Check if the requested gift is available and attempt matching.
  • Visitation Record: Manage visitation records to accommodate requests for each gift only once.
  • Output Result: Output the maximum matching result.

Time Complexity

This algorithm has a time complexity of O(N^2).
Since DFS is called for each friend, and the number of gifts is limited to a maximum of 26,
it is efficient enough within the problem’s constraints.

Conclusion

The “Gift Delivery” problem is an interesting problem of distributing gifts among friends.
It demonstrates that it can be implemented through the principles of DFS and maximum matching in graphs.
This principle can be applied to solve similar problems in the future.

I hope this article helps you prepare for C# coding tests.
Please practice algorithms through various problems.

C# Coding Test Course, Calculating Interval Sum 3

Many people preparing for coding tests need systematic learning for algorithm problem solving. Today, we will discuss the problem of calculating the range sum. In particular, the ‘Range Sum Query 3’ problem is an interesting problem that can utilize various strategies. In this article, I will explain the definition of the problem, the solution method, implementation in C# code, and time complexity analysis in detail.

Problem Definition

Given an integer array arr and multiple queries query, each query represents the start and end of a range. Our goal is to efficiently calculate and return the range sum for each query.

Example

Array: arr = [1, 2, 3, 4, 5]
Query: query = [[1, 3], [0, 2], [2, 4]]
Output: [9, 6, 12]

In the above example, the first query means the range from index 1 to 3, so 2 + 3 + 4 = 9. The remaining queries can be handled in the same way.

Approach to the Problem

There are several ways to solve the problem, but when the number of elements in the array is large, performance can be an important factor. Therefore, an efficient algorithm is essential.

1. Traditional Method

The most basic approach is to directly sum the values of the array for each query. However, this method has the following drawbacks:

  • The time complexity is O(n), making it inefficient when the number of queries increases.
  • It is very inefficient when the number of elements in the array is large.

2. Using Prefix Sum

To solve the problem more efficiently, the exposed method is to use Prefix Sum. The prefix sum array is defined as follows:

sum[i] = arr[0] + arr[1] + ... + arr[i]

Using this calculated prefix sum array, we can quickly compute the given range sum. The sum of the range [left, right] is obtained as sum[right] - sum[left - 1]. This way, the time complexity can be reduced to O(1).

3. C# Code Implementation

Now, let’s implement the code to calculate the prefix sum using the C# language based on the given problem.


using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        // Define array and queries
        int[] arr = { 1, 2, 3, 4, 5 };
        int[][] queries = { new int[] { 1, 3 }, new int[] { 0, 2 }, new int[] { 2, 4 } };

        // List to store the range sums
        List results = new List();

        // Create prefix sum array
        int[] prefixSum = new int[arr.Length];
        prefixSum[0] = arr[0];

        for (int i = 1; i < arr.Length; i++)
        {
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        }

        // Calculate range sum for each query
        foreach (var query in queries)
        {
            int left = query[0];
            int right = query[1];

            if (left == 0)
            {
                results.Add(prefixSum[right]);
            }
            else
            {
                results.Add(prefixSum[right] - prefixSum[left - 1]);
            }
        }

        // Output results
        Console.WriteLine(string.Join(", ", results));
    }
}

Code Analysis and Explanation

The above code is designed to convert the array into a prefix sum in advance so that the range sum for each query can be calculated in O(1) time. The following steps are taken:

  1. Create Prefix Sum Array: After initializing the first element of the array, a loop is used to calculate the prefix sum for each element.
  2. Execute Queries: Iterate through each query to calculate the range sum. If the left boundary is 0, the prefix sum can be retrieved directly; otherwise, the range sum is obtained by finding the difference of two elements.
  3. Output Results: Each result is printed.

Time Complexity Analysis

The time complexity of this algorithm is as follows:

  • Time to create the prefix sum array: O(n)
  • Time to calculate the range sum for each query: O(1) (proportional to the number of queries)

Thus, the overall time complexity can be represented as O(n + m), where m is the number of queries. This is an efficient method when dealing with many queries.

Conclusion

Today, we explored how to design and implement an efficient algorithm through the "Range Sum Query 3" problem. Algorithm problems may be simple concepts, but finding solutions through various approaches is a very important skill for coding tests. I hope this course helps you in writing your own code and solving problems.

Through the utilization of code and improvement of algorithms, I hope to elevate your coding test skills to the next level. See you in the next course!

C# Coding Test Course, Finding the Number of Chase

1. Introduction

Coding tests are conducted by many companies to evaluate applicants’ algorithmic thinking and problem-solving skills. Among the various problems with different topics and difficulties, “Finding Binary Numbers” is a good example where the DP (Dynamic Programming) technique can be utilized. In this article, we will explain the problem of finding binary numbers in detail and proceed step by step to solve the problem using C#.

2. Problem Description

A Binary Number is a number made up of 0s and 1s,
which does not have two consecutive 1s.
For example, examples of binary numbers consisting of 0s and 1s are:
0, 1, 010, 001, 100, 0101, 1010, 1001, 0001, etc.
On the other hand, 11, 00000, or 1111 are not binary numbers.
For instance, given N, the problem is to determine the count of binary numbers of length N.

2.1. Input Format

– An integer N is given on the first line. (1 ≤ N ≤ 1,000)

2.2. Output Format

– Output the count of binary numbers of length N.

2.3. Example

    Input:
    3
    
    Output:
    3
    

In this case, the binary numbers are {001, 010, 100}, totaling 3.

3. Problem Solving Approach

To solve this problem, the following approach is necessary.

3.1. Define DP Array

We can solve this problem using Dynamic Programming.
First, we define a DP array to store the count of binary numbers of length N.
– DP[i] stores the count of binary numbers of length i.

3.2. State Transition Formula

From the current state, we can derive the following state transition formula:
– A binary number of length i can be appended from a binary number of length (i-1) if the last digit is 0,
– or it can be appended from a binary number of length (i-2) if the last digit is 1.

Therefore, it can be expressed as:
– DP[i] = DP[i-1] + DP[i-2]

3.3. Initial Conditions

– DP[1] = 2 (Binary Numbers: 0, 1)
– DP[2] = 3 (Binary Numbers: 00, 01, 10)

4. C# Code Implementation

Based on the above logic, let’s implement the C# code.

    using System;

    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            long[] dp = new long[N + 1];
            
            // Initial conditions
            dp[1] = 2; // Binary Numbers: 0, 1
            if (N > 1)
            {
                dp[2] = 3; // Binary Numbers: 00, 01, 10
            }

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

            // Output result
            Console.WriteLine(dp[N]);
        }
    }
    

5. Code Explanation

In the above code, we read N from the user, initialize the DP array, and then fill the DP array using a loop.
Finally, we can check the count of binary numbers of length N by printing DP[N].

5.1. Time Complexity

The time complexity of this algorithm is O(N). Even when N is at its maximum of 1,000,
it operates quickly to handle large amounts of data.

5.2. Space Complexity

The space complexity is also O(N) because we use the DP array,
which increases memory requirements. However, since N is small,
the memory usage is not a significant issue.

6. Conclusion

In this tutorial, we covered the problem of "Finding Binary Numbers."
We explained the step-by-step process of solving the problem using the techniques of Dynamic Programming
and implemented the code in C#.
Such types of problems are frequently encountered in actual coding tests, so it is advisable to practice and become familiar with them.

While preparing for coding tests, try to solve more problems,
and contemplate various approaches and optimizations for each problem.
Thank you!