C# Coding Test Course, Finding the Minimum Among Prime & Palindrome Numbers

In this course, we will cover coding test problems using C#.
The topic is “Finding the minimum value among prime and palindrome numbers within a given range.”
This problem emphasizes algorithmic thinking and allows you to learn various approaches to writing efficient code.

Problem Description

Given an integer N,
the task is to find all prime and palindrome numbers among the integers from 1 to N and determine the minimum value among them.
A prime number is a positive integer greater than 1 that has no positive divisors other than 1 and itself,
and a palindrome number is a number that reads the same forward and backward.

Input

– Integer N (1 ≤ N ≤ 1,000,000)

Output

– Print the minimum value among the prime and palindrome numbers.
– If there are no such numbers, you should print -1.

Example Input

31

Example Output

11

Algorithm Analysis

To solve this problem, we can approach it in the following steps:

  1. Prime Check: Implement an algorithm to check if numbers within the given range are prime.
  2. Palindrome Check: Write a function to check if each number is a palindrome.
  3. Finding Minimum: Collect the numbers that are both prime and palindrome, then return the minimum value.

C# Code Implementation

Below is the code to solve the problem using C#.


using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int N = int.Parse(Console.ReadLine());
        int minPrimePalindrome = int.MaxValue;

        for (int i = 2; i <= N; i++)
        {
            if (IsPrime(i) && IsPalindrome(i))
            {
                minPrimePalindrome = Math.Min(minPrimePalindrome, i);
            }
        }

        Console.WriteLine(minPrimePalindrome == int.MaxValue ? -1 : minPrimePalindrome);
    }

    static bool IsPrime(int number)
    {
        if (number < 2) return false;
        for (int i = 2; i <= Math.Sqrt(number); i++)
        {
            if (number % i == 0) return false;
        }
        return true;
    }

    static bool IsPalindrome(int number)
    {
        string str = number.ToString();
        int len = str.Length;
        for (int i = 0; i < len / 2; i++)
        {
            if (str[i] != str[len - 1 - i]) return false;
        }
        return true;
    }
}
    

Code Explanation

1. Main Method: For the input integer N, it iterates through numbers from 2 to N.
It checks if each number is both prime and a palindrome, and finds the minimum value among them.

2. IsPrime Method: Determines whether the given number is prime.
If the number is less than 2, it returns false as it is not prime,
and it checks for divisibility from 2 to √number to determine its primality.

3. IsPalindrome Method: Converts the given number to a string and
checks if it is a palindrome by comparing characters from the front and back.
It returns true if all characters match.

Time Complexity

The time complexity of this algorithm is O(N√N).
This is because checking numbers up to N involves O(√N) time for primality testing for each number.
Therefore, as the input range increases, the computation time is affected, so it is important to consider this when writing the code.

Conclusion

Through the problems covered in this course, we have learned how to determine prime numbers and check for palindromes.
Understanding primes and palindromes is a common topic in coding tests,
so it is advisable to practice thoroughly.
Additionally, I recommend building your skills through a variety of practice problems.

C# Coding Test Course, Finding Numbers That Are Not Perfect Squares

Finding Non-Square Numbers

In this tutorial, we will cover an algorithm problem of finding non-square numbers. The goal is to learn various approaches to solve the problem and enhance problem-solving skills for coding tests.

Problem Description

Given an integer N, the task is to count the number of non-square numbers among the integers from 1 to N. For example, if N is 10, the square numbers are 1, 4, and 9, making the non-square numbers 1, 2, 3, 5, 6, 7, 8, and 10, which totals to 8.

Input

  • An integer N (1 ≤ N ≤ 10^6)

Output

  • Print the count of non-square numbers

Example

    Input Example:
    10

    Output Example:
    8
    

Approach to Solve the Problem

There can be various methods to solve this problem. However, to find the optimal method, we need to generate square numbers and infer the count of non-square numbers based on that. A square number is defined as follows:

  • X is an integer, and Y = X * X

To find square numbers from 1 to N, we need the following logic.

1. Create a List of Square Numbers

Calculating square numbers for numbers from 1 to N has a complexity of O(√N), so we pre-calculate the square numbers and store them in a list.

2. Compare with Integer N

Next, count how many numbers from 1 to N are included in the list of square numbers.

3. Derive the Final Result

Subtracting the count of square numbers from N gives the count of non-square numbers. This process has a time complexity of O(N).

Code Implementation

Now, based on the above logic, let’s write the code in C#.


using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int N = int.Parse(Console.ReadLine());
        HashSet squareNumbers = new HashSet();

        // Add square numbers to the HashSet
        for (int i = 1; i * i <= N; i++)
        {
            squareNumbers.Add(i * i);
        }

        int nonSquareCount = N - squareNumbers.Count; // Count of non-square numbers

        Console.WriteLine(nonSquareCount);
    }
}
    

Code Explanation

  1. HashSet squareNumbers: A HashSet to store square numbers. HashSet is used for fast lookup.
  2. for (int i = 1; i * i <= N; i++): Starts from 1 to generate square numbers.
  3. squareNumbers.Add(i * i): Adds the generated square number to the HashSet.
  4. nonSquareCount = N - squareNumbers.Count: Subtracts the count of square numbers from N to get the count of non-square numbers.
  5. Console.WriteLine(nonSquareCount): Prints the final result.

Complexity Analysis

Analyzing the time complexity of this algorithm:

  • Generating square numbers: O(√N) (up to √N square numbers generated)
  • Counting non-square numbers: O(1) (constant time as it retrieves the size of the HashSet)

Therefore, the overall time complexity is O(√N). This algorithm operates very efficiently within the given constraints.

Conclusion

In this tutorial, we explored how to solve the algorithm problem of finding non-square numbers. We hope to enhance problem-solving skills through the thought process behind solving the problem. Continuous practice with various algorithm problems is essential in the future.

C# Coding Test Course, Finding the Fastest Bus Route

In modern society, public transportation plays an important role, and especially when using buses, many people struggle to find the optimal routes. In this lecture, we will look at the process of solving the problem of finding the fastest bus route. To solve this problem, we will use C# to employ algorithms and data structures.

Problem Definition

The problem is to find the fastest path from a starting point to a destination based on several given bus routes and the time required for each route.
The input consists of the following values:

  • Vertex: Bus stops
  • Edge: Bus routes
  • Weight: Time required

Example Input

        5 6
        1 2 3
        1 3 2
        2 3 1
        1 4 1
        2 4 5
        3 5 1
        

The first line indicates the number of vertices and edges, and each subsequent line shows the relationship between stops and the time required.

Problem Solving Process

Step 1: Graph Modeling

To solve the above problem, we need to construct a graph where the bus stops are vertices and the bus routes and times are edges.
To do this, we will create an adjacency list that contains information about each stop and its edges.

Step 2: Selecting the Shortest Path Algorithm

There are several algorithms to find the shortest path, but when there are weighted edges, it is appropriate to use Dijkstra’s Algorithm.
Dijkstra’s Algorithm is an effective way to calculate the shortest distance to each vertex.

Step 3: Implementing C# Code

Now, let’s write C# code based on the above content. The code below demonstrates how to implement Dijkstra’s Algorithm to find the shortest path in the given graph.

        using System;
        using System.Collections.Generic;

        class Program
        {
            static void Main(string[] args)
            {
                int vertexCount = 5; // Number of vertices
                int[,] graph = new int[vertexCount + 1, vertexCount + 1];
                for (int i = 1; i <= vertexCount; i++)
                    for (int j = 1; j <= vertexCount; j++)
                        graph[i, j] = (i == j) ? 0 : int.MaxValue;

                // Add edges (time required)
                graph[1, 2] = 3;
                graph[1, 3] = 2;
                graph[2, 3] = 1;
                graph[1, 4] = 1;
                graph[2, 4] = 5;
                graph[3, 5] = 1;

                // Call Dijkstra's Algorithm
                int[] shortestPath = Dijkstra(graph, vertexCount, 1);

                // Output results
                Console.WriteLine("Shortest distance from vertex 1 to other vertices:");
                for (int i = 1; i <= vertexCount; i++)
                {
                    Console.WriteLine($"Vertex 1 -> Vertex {i}: {(shortestPath[i] == int.MaxValue ? "Unreachable" : shortestPath[i].ToString())}");
                }
            }

            static int[] Dijkstra(int[,] graph, int vertexCount, int start)
            {
                bool[] visited = new bool[vertexCount + 1];
                int[] distance = new int[vertexCount + 1];

                for (int i = 1; i <= vertexCount; i++)
                {
                    distance[i] = (i == start) ? 0 : int.MaxValue;
                }

                for (int count = 0; count < vertexCount - 1; count++)
                {
                    int u = MinDistance(distance, visited, vertexCount);
                    visited[u] = true;

                    for (int v = 1; v <= vertexCount; v++)
                    {
                        if (!visited[v] && graph[u, v] != int.MaxValue &&
                            distance[u] != int.MaxValue &&
                            distance[u] + graph[u, v] < distance[v])
                        {
                            distance[v] = distance[u] + graph[u, v];
                        }
                    }
                }

                return distance;
            }

            static int MinDistance(int[] distance, bool[] visited, int vertexCount)
            {
                int min = int.MaxValue;
                int minIndex = -1;

                for (int v = 1; v <= vertexCount; v++)
                {
                    if (!visited[v] && distance[v] <= min)
                    {
                        min = distance[v];
                        minIndex = v;
                    }
                }
                return minIndex;
            }
        }
        

Step 4: Time Complexity Analysis

The time complexity of Dijkstra’s Algorithm is O(V^2), where V represents the number of vertices.
Therefore, when there are many vertices, you may consider a more efficient method, which is the improved Dijkstra’s Algorithm using a priority queue.

Conclusion

In this lecture, we implemented Dijkstra’s Algorithm to solve the problem of finding the fastest bus route using C#.
Similar problems may arise in actual coding tests or job interviews, so it is important to understand and practice this algorithm thoroughly.

I hope you continue to study various algorithm problems and solutions in the future. If you have any additional questions or algorithm-related inquiries, feel free to ask!

C# Coding Test Course, Getting the Sum of an Interval 1

Hello! Today, we will address one of the frequently asked problems in coding tests using C#, which is the “Calculate Interval Sum” problem. Through this problem, we aim to enhance our understanding of arrays and interval sums, and learn algorithms to solve this efficiently.

Problem Description

The interval sum problem can be defined as follows:

Given an integer array arr and several queries (starting index and ending index), write a program to calculate the sum of the respective interval for each query.

The input consists of the size of the array n, the elements of the array, the number of queries q, and the starting index l and ending index r for each query.

Input Example

        5
        1 2 3 4 5
        3
        1 3
        2 4
        1 5
        

The above input is as follows:

  • Array size: 5
  • Array elements: [1, 2, 3, 4, 5]
  • Number of queries: 3
  • Queries: (1, 3), (2, 4), (1, 5)

Output Example

        6
        9
        15
        

The output is the interval sum for each query:

  • Interval sum for (1, 3): 1 + 2 + 3 = 6
  • Interval sum for (2, 4): 2 + 3 + 4 = 9
  • Interval sum for (1, 5): 1 + 2 + 3 + 4 + 5 = 15

Solving the Problem

Now, let’s go through the steps to solve this problem. We will start with a simple approach and then progress to a more efficient method.

1. Simple Approach

The most intuitive method is to directly calculate the sum according to the starting and ending indices of each query. However, this method can be inefficient. For example, in the worst case, if the size of the array is n and the number of queries is q, the time complexity can reach O(n * q).

C#
        using System;

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

                for (int i = 0; i < q; i++)
                {
                    var query = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
                    int l = query[0] - 1; // 0-based index
                    int r = query[1] - 1; // 0-based index

                    int sum = 0;
                    for (int j = l; j <= r; j++)
                    {
                        sum += arr[j];
                    }
                    Console.WriteLine(sum);
                }
            }
        }
        

This code calculates the sum for each query based on the input array elements provided by the user. However, this method is inefficient for large datasets.

2. Efficient Approach: Cumulative Sum Array

An efficient way is to use a cumulative sum array. The cumulative sum pre-calculates the sum up to each index in the array, reducing the processing time for queries.

That is,

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

To calculate the interval sum:

C#
        interval_sum = sum[r] - sum[l-1]
        

This reduces the query processing time to O(1). The overall time complexity becomes O(n + q).

C#
        using System;

        class Program
        {
            static void Main()
            {
                int n = int.Parse(Console.ReadLine());
                int[] arr = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
                int q = int.Parse(Console.ReadLine());
                
                long[] sum = new long[n + 1];
                
                for (int i = 1; i <= n; i++)
                {
                    sum[i] = sum[i - 1] + arr[i - 1];
                }

                for (int i = 0; i < q; i++)
                {
                    var query = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
                    int l = query[0] - 1; // 0-based index
                    int r = query[1] - 1; // 0-based index

                    long result = sum[r + 1] - sum[l];
                    Console.WriteLine(result);
                }
            }
        }
        

This code first computes the cumulative sum array and then efficiently calculates the interval sum for each query using the starting and ending indices.

Conclusion

The interval sum problem frequently appears in coding tests, and we can see that using a cumulative sum array is an efficient solution. This method is particularly advantageous when the size of the array and the number of queries are large. Using appropriate data structures to enhance the efficiency of an algorithm and reduce query processing time is a critical factor in solving algorithm problems.

Based on what we've learned today, try practicing various interval sum problem-solving exercises. Thank you!

C# Coding Test Course, Making an Integer into 1

This posting will detail how to solve the problem of reducing a given integer to 1 using C#. This problem is a common type in coding tests and is very useful for developing problem-solving skills and algorithm design abilities.

Problem Description

There are several operations to reduce the given integer x to 1. Each operation is as follows:

  • Subtracting 1 (x => x - 1)
  • Dividing by 2 (x => x / 2, possible when x is even)
  • Dividing by 3 (x => x / 3, possible when x is divisible by 3)

The goal is to reduce x to 1 with the minimum number of operations. For example, if x = 10, the following operations can be performed:

Example:

10 → 5 (divide 10 by 2) → 4 (subtract 1 from 5) → 2 (divide 4 by 2) → 1 (subtract 1 from 2)

Problem Solving Strategy

There are various approaches to solve this problem, but one of the most efficient methods is to use BFS (Breadth-First Search). BFS is advantageous for finding the shortest path and solves the problem in a step-by-step manner.

BFS Algorithm Overview

BFS is an algorithm that stores nodes to be explored in a queue and explores them in the order they were added. In this problem, each state (integer) is treated as a node, and the new states reachable through each operation are added to the queue. This allows us to find the minimum number of operations to reach 1.

Implementation

Now, let’s write a code to solve the problem in C#. The code below is a function that calculates the minimum number of operations to reduce the given integer x to 1.


using System;
using System.Collections.Generic;

public class Solution {
    public int MinOperations(int x) {
        // A set to track visited nodes
        HashSet visited = new HashSet();
        // A queue for BFS exploration
        Queue> queue = new Queue>();
        
        // Initial state (x, 0) - current value and operation count
        queue.Enqueue(new Tuple(x, 0));
        visited.Add(x); // Mark the initial node as visited
        
        while (queue.Count > 0) {
            var current = queue.Dequeue();
            int currentValue = current.Item1;
            int operationsCount = current.Item2;
            
            // If we have reached the goal state of 1
            if (currentValue == 1) {
                return operationsCount; // Return the minimum operation count
            }
            
            // Perform possible operations
            // x - 1
            if (currentValue - 1 >= 1 && !visited.Contains(currentValue - 1)) {
                visited.Add(currentValue - 1);
                queue.Enqueue(new Tuple(currentValue - 1, operationsCount + 1));
            }
            // x / 2
            if (currentValue % 2 == 0 && !visited.Contains(currentValue / 2)) {
                visited.Add(currentValue / 2);
                queue.Enqueue(new Tuple(currentValue / 2, operationsCount + 1));
            }
            // x / 3
            if (currentValue % 3 == 0 && !visited.Contains(currentValue / 3)) {
                visited.Add(currentValue / 3);
                queue.Enqueue(new Tuple(currentValue / 3, operationsCount + 1));
            }
        }
        
        return -1; // In case it is not reachable
    }
}

Code Explanation

Looking at the C# code above, the MinOperations function takes x as input and returns the minimum number of operations. This function performs all necessary operations to reduce the integer to 1 through BFS exploration.

Main Components

  • Visit Record (visited): Tracks nodes (integers) that have already been explored.
  • Queue (queue): Stores the current node and operation counts to proceed with BFS.
  • Conditional Statements: Check if each operation results in a valid new state (>= 1) and whether it has been visited.

Time Complexity Analysis

The time complexity of this algorithm is O(x) in the worst case. Since BFS is performed for each integer, in the worst case, it may need to explore all possibilities. However, it generally performs quickly.

Conclusion

In this posting, we explored how to solve the problem of reducing an integer to 1 using BFS in C#. By applying this method, you can efficiently find the minimum number of operations, and it greatly helps in developing algorithmic thinking. Solving additional practice problems is also a good way to expand this knowledge.

Based on this problem, try challenging various modified versions. Keep pondering problem-solving to gain a deeper understanding!