Category: C# Coding Test Tutorials

In this course, we will cover C# coding test problems using arrays and lists. Arrays and lists are important parts of data structures and are frequently used in various algorithm problems.

Problem Solving Process

To solve this problem, we can use an array or list to store the numbers from 1 to N and sum them up. However, since there is a simpler mathematical formula, we can also use Sum = N * (N + 1) / 2.

Solution Method

1. Input: Get the natural number N from the user.
2. Calculate the sum: Calculate the sum up to N. Typically, a for loop is used, but we can use the formula mentioned above.
3. Output the result: Print the calculated sum.

C# Code Implementation

using System;

class Program
{
    static void Main(string[] args)
    {
        Console.Write("Please enter a natural number N: ");
        int N = int.Parse(Console.ReadLine());
        
        // Method 1: Calculate the sum using a for loop
        int sum = 0;
        for (int i = 1; i <= N; i++)
        {
            sum += i;
        }

        // Method 2: Calculate the sum using the mathematical formula
        // int sum = N * (N + 1) / 2;
        
        Console.WriteLine($"The sum from 1 to {N} is {sum}.");
    }
}

Code Explanation

The code above first gets the natural number N from the user. It provides both methods:

• In Method 1, it uses a for loop to sequentially add the values from 1 to N.
• Method 2 uses the mathematical formula, simply taking the input for N and calculating the result through arithmetic operations.

Complexity Analysis

Time complexity: O(N) (when using the for loop)

Space complexity: O(1) (when not using additional arrays or lists)

Conclusion

In this course, we covered basic algorithms in C# through a simple problem using arrays and lists. We learned the importance of data management through arrays or lists and the usefulness of using mathematical formulas to solve problems.

In the next course, we will address more complex problems related to arrays and lists. We look forward to your participation!

In this lecture, we will cover a problem that frequently appears in coding tests, the ‘Make the Maximum Number by Grouping Numbers’ problem. This problem involves combining given numbers to create the largest possible number. Additionally, I will explain how to solve this using the C# language step by step.

Problem Description

Given several positive integers, write a function that returns the largest number that can be formed by combining these numbers. For example, if the given numbers are [3, 30, 34, 5, 9], the largest number that can be formed by appropriately combining them is 9534330.

Input Format

An integer array numbers will be provided. The size of the array is between 0 and 100, and the length of each number is between 1 and 10.

Output Format

Return the largest number formed by combining all the numbers as a string.

Problem Approach

To solve this problem, several approaches are necessary.

• Sorting: We need to sort the given numbers to create the largest number. The sorting criteria are important, and we need to create rules based on treating each number as a string.
• String Comparison: When comparing two numbers, we need to determine which combination (where one number comes first or second) results in a larger number.

Code Implementation

Let’s write a code in C# to solve the problem based on the above approaches.

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    public static string Solution(int[] numbers)
    {
        // Convert numbers to strings.
        string[] strNumbers = numbers.Select(n => n.ToString()).ToArray();
        
        // Sort the string array.
        Array.Sort(strNumbers, (x, y) => (y + x).CompareTo(x + y));
        
        // Combine the sorted array into a single string.
        string result = string.Join("", strNumbers);
        
        // If the result starts with "0", return only "0".
        return result[0] == '0' ? "0" : result;
    }

    static void Main(string[] args)
    {
        int[] numbers = { 3, 30, 34, 5, 9 };
        string answer = Solution(numbers);
        Console.WriteLine(answer);  // 9534330
    }
}
    

Code Explanation

Let’s take a closer look at the code above.

1. String Conversion: First, convert each number to a string and store it in an array called strNumbers. This is necessary for rearranging the numbers in string form.
2. Sorting: Use the Array.Sort method to sort the strNumbers array. The sorting criterion combines two numbers x and y to sort them in descending order. By comparing (y + x) and (x + y), we determine which combination is larger.
3. Result Combination: Use string.Join to combine the sorted numbers into a single string.
4. Final Check: If the first character of the result string is ‘0’, it indicates that all numbers are 0, so return only ‘0’.

Complexity Analysis

The time complexity of this algorithm is O(N log N). This is the time spent sorting the given number array. N represents the number of digits. The space complexity is O(N) since it requires an array to store the string conversions of each number.

Conclusion

In this lecture, we examined how to solve the ‘Make the Maximum Number by Grouping Numbers’ problem using C#. It is important to understand the problem well and to select appropriate data structures and algorithms to solve algorithmic problems. Practice with various problems!

Additional Practice Problems

I recommend further practicing the following problems:

• Combine the given numbers to generate all possible combinations and find the maximum value among them.
• Write an algorithm to create the maximum number when both positive and negative numbers are mixed.
• Explore methods to prevent performance degradation when the input array size becomes very large.

Reference Materials

There are several reference materials that can help in solving algorithm problems like this. Please check the following links:

• GeeksforGeeks
• LeetCode
• HackerRank

Hello, coding test applicants! Today we will delve deep into the problem of Lowest Common Ancestor (LCA). This problem is related to tree structures and requires a deep understanding of algorithms and data structures. We will also explore how to solve this problem using C#.

Problem Definition

The problem is to find the lowest common ancestor of two nodes in a given binary tree. Given two nodes A and B, our goal is to find the deepest node that is an ancestor of both A and B.

Example

        For instance, let's assume there is a binary tree as follows. 

              3
             / \
            5   1
           / \ / \
          6  2 0  8
            / \
           7   4

        The lowest common ancestor of nodes 5 and 1 is 3.
        The lowest common ancestor of nodes 5 and 2 is 5.
        The lowest common ancestor of nodes 6 and 4 is 5.
    

Approach to the Problem

1. Understanding Tree Structure

   First, it is essential to understand the structure of a binary tree. Each node can have at most two children, and it can be divided into a left subtree and a right subtree. You should know how to traverse the tree to find nodes.

2. Recursive Approach

   One of the main approaches is to search recursively. If the current node is either A or B, return the current node and explore the left and right subtrees to find the other node. If both are found, then the current node is the lowest common ancestor.

3. Writing Tree Traversal Algorithm

   Now, we need to write the code in C# to implement the algorithm. In this process, we will incorporate recursion to navigate and verify nodes through the code.

C# Code Implementation

        // Definition of binary tree node class
        public class TreeNode {
            public int Value;
            public TreeNode Left;
            public TreeNode Right;

            public TreeNode(int value) {
                Value = value;
                Left = null;
                Right = null;
            }
        }

        public class Solution {
            public TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
                // Base case: If root is null or root is p or q, return root
                if (root == null || root == p || root == q) {
                    return root;
                }

                // Search each node in the left and right subtrees
                TreeNode left = LowestCommonAncestor(root.Left, p, q);
                TreeNode right = LowestCommonAncestor(root.Right, p, q);

                // If p and q are in different subtrees, the current node is LCA
                if (left != null && right != null) {
                    return root;
                }

                // If only one is found, return that node
                return left ?? right;
            }
        }
    

Code Explanation

This code is written to find the lowest common ancestor of two nodes in a given binary tree. Below are the descriptions of the key parts of the code:

1. TreeNode Class

   This class represents each node in the tree. It stores the value of each node, the left child node, and the right child node.

2. LowestCommonAncestor Method

   This method is called recursively and checks if the current node is null or if it is p or q. It searches for p and q in the left and right subtrees, and if both are found in different subtrees, it returns the current node.

Complexity Analysis

The time complexity of this algorithm is O(N), where N represents the number of nodes in the tree. It is proportional to the need to traverse all nodes. The space complexity is O(H), where H represents the height of the tree. This corresponds to the depth of the recursion call stack.

Example Execution

        // Example tree creation and finding LCA
        TreeNode root = new TreeNode(3);
        root.Left = new TreeNode(5);
        root.Right = new TreeNode(1);
        root.Left.Left = new TreeNode(6);
        root.Left.Right = new TreeNode(2);
        root.Right.Left = new TreeNode(0);
        root.Right.Right = new TreeNode(8);
        root.Left.Right.Left = new TreeNode(7);
        root.Left.Right.Right = new TreeNode(4);

        Solution solution = new Solution();
        TreeNode p = root.Left; // Node 5
        TreeNode q = root.Left.Right; // Node 2
        TreeNode lca = solution.LowestCommonAncestor(root, p, q);
        Console.WriteLine($"Lowest Common Ancestor: {lca.Value}"); // Outputs 5
    

Conclusion

Today, we explored how to solve the lowest common ancestor problem using C#. This problem serves as a foundation for tackling various algorithmic challenges and requires an understanding of tree structure and recursive thinking. Through practice, I hope you will enhance your coding skills and increase your understanding of C# during this opportunity.

Written on: October 30, 2023

Problem Description

This is a problem of finding the permutation of a given number through an input with issues and determining its order in the sequence. For example, if the given array is [1, 2, 3], the permutations of this array are as follows:

• [1, 2, 3]
• [1, 3, 2]
• [2, 1, 3]
• [2, 3, 1]
• [3, 1, 2]
• [3, 2, 1]

In this case, if the given order of the array is [2, 3, 1], it corresponds to the 4th position in total.

Input and Output Format

Input

The first line contains two integers N (1 ≤ N ≤ 9) and K (1 ≤ K ≤ N!). This represents the total number of integers in the array and the specific order of the permutation to find.

Output

Print the K-th permutation that is to be found.

Problem Solving

To solve this problem, the following steps should be followed:

1. Problem Analysis

Using the given N and K, we need to create an array consisting of N numbers and find the K-th permutation of that array. By utilizing the concept of permutations, we can create several combinations that reveal the order of all numbers. For example, when N is 3 and the array is [1, 2, 3], finding the permutations of this array is very intuitive.

2. Understanding Permutation Generation Algorithms

There are various approaches to generating permutations. One of them is a recursive method. However, since the problem is to find a specific K-th permutation, rather than using exhaustive search, we will use the properties of permutations to directly calculate the K-th permutation. To do this, we can implement a method that recursively generates possible number combinations to check for the K-th combination.

3. C# Implementation

Now, let’s implement the above approach in C#. Below is the code written in C#:

        using System;
        using System.Collections.Generic;

        class Program
        {
            static void Main(string[] args)
            {
                // Read N and K
                string[] input = Console.ReadLine().Split();
                int N = int.Parse(input[0]);
                int K = int.Parse(input[1]);

                // Initialize the numbers list
                List numbers = new List();
                for (int i = 1; i <= N; i++)
                {
                    numbers.Add(i);
                }

                // Result list to store the sequence
                List result = new List();
                bool[] used = new bool[N];

                // Recursive call to find the K-th permutation
                FindKthPermutation(numbers, used, result, K, 0);
            }

            static void FindKthPermutation(List numbers, bool[] used, List result, int K, int depth)
            {
                // Base condition: if depth is N
                if (depth == numbers.Count)
                {
                    // If we found the K-th permutation, print it
                    K--;
                    if (K == 0)
                    {
                        Console.WriteLine(string.Join(" ", result));
                    }
                    return;
                }

                for (int i = 0; i < numbers.Count; i++)
                {
                    if (!used[i])
                    {
                        // Mark as used
                        result.Add(numbers[i]);
                        used[i] = true;

                        // Recursive call
                        FindKthPermutation(numbers, used, result, K, depth + 1);

                        // Backtrack
                        used[i] = false;
                        result.RemoveAt(result.Count - 1);
                    }
                }
            }
        }
        

4. Code Explanation

The above C# code works on the following principles:

Input Handling

The code reads the values of N and K from the user. It then generates a list containing numbers from 1 to N.

Recursive Permutation Generation

The FindKthPermutation method generates permutations recursively. It continues until the current depth equals N, adding unused numbers to the list and making recursive calls with these values.

K-th Permutation Check

If K becomes 0, it means we have found the K-th permutation in the current list, so we print this list. Afterward, it backtracks to restore the unused numbers in the process.

5. Time Complexity

The base time complexity of this algorithm is O(N!). This is because it can consider all N! permutations through recursive calls. However, we can reduce the implementation time in many cases because we are directly finding the K-th permutation.

6. Conclusion

In this tutorial, we learned about the principles of generating permutations and how to find the K-th permutation among them. We verified the process of solving the problem by invoking functions recursively using C#. The method of finding a specific order in coding tests can be applied to various problems. It is essential to continuously practice to master these strategies.

Hello! In this post, we will discuss the coding test problem ‘Determining whether line segments intersect’ using C#.
This problem is a geometric problem, which determines whether the two given line segments intersect.
It is a fundamental problem that is utilized in various fields, making it useful for improving algorithm problem-solving skills and mathematical reasoning.

Problem Description

Given line segment 1 with two points A(x1, y1) and B(x2, y2), and line segment 2 with two points C(x3, y3) and D(x4, y4),
write a program to determine whether the two segments intersect.

Input

• Coordinates of the two points A and B of the first segment: (x1, y1), (x2, y2)
• Coordinates of the two points C and D of the second segment: (x3, y3), (x4, y4)

Output

Output true if the two segments intersect, otherwise output false.

Approach to Solution

To solve this problem, it is necessary to understand the position relationships of the segments. For the segments to intersect, both of the following conditions must be satisfied:

1. The endpoints of segment AB must lie on opposite sides of segment CD.
2. The endpoints of segment CD must lie on opposite sides of segment AB.

Mathematical Background

Given the points, we can determine if two points are on opposite sides using the cross product of the vectors.
If the signs of the cross product of the two vectors v1 = B - A and v2 = C - A are different, it means point C is on one side of segment AB,
and point D is on the opposite side. This method can be used to determine the intersection of the segments.

Implementation Method

The basic logic to solve the problem is as follows:

1. Define the direction of each segment using the four given points A, B, C, and D.
2. Create a cross product function to calculate the cross product of the two vectors.
3. Compare the signs of the cross product based on the two cases and return whether they intersect.

C# Code Implementation

using System;

class Program
{
    static void Main()
    {
        // Input coordinates of points A, B, C, D
        int[] A = { 1, 1 }; // (x1, y1)
        int[] B = { 4, 4 }; // (x2, y2)
        int[] C = { 1, 4 }; // (x3, y3)
        int[] D = { 4, 1 }; // (x4, y4)

        // Check intersection
        bool isCross = DoSegmentsIntersect(A, B, C, D);
        Console.WriteLine(isCross ? "true" : "false");
    }

    static bool DoSegmentsIntersect(int[] A, int[] B, int[] C, int[] D)
    {
        // Check intersection using cross product
        int d1 = CrossProduct(A, B, C);
        int d2 = CrossProduct(A, B, D);
        int d3 = CrossProduct(C, D, A);
        int d4 = CrossProduct(C, D, B);

        // Must be on opposite sides to intersect
        if (d1 * d2 < 0 && d3 * d4 < 0)
            return true;

        // Also consider the case where the segments touch at endpoints
        if (d1 == 0 && IsOnSegment(A, B, C)) return true;
        if (d2 == 0 && IsOnSegment(A, B, D)) return true;
        if (d3 == 0 && IsOnSegment(C, D, A)) return true;
        if (d4 == 0 && IsOnSegment(C, D, B)) return true;

        return false;
    }

    static int CrossProduct(int[] A, int[] B, int[] C)
    {
        // Calculate cross product
        return (B[0] - A[0]) * (C[1] - A[1]) - (B[1] - A[1]) * (C[0] - A[0]);
    }

    static bool IsOnSegment(int[] A, int[] B, int[] P)
    {
        // Check if point P lies on segment AB
        return Math.Min(A[0], B[0]) <= P[0] && P[0] <= Math.Max(A[0], B[0]) &&
               Math.Min(A[1], B[1]) <= P[1] && P[1] <= Math.Max(A[1], B[1]);
    }
}

Code Explanation

The above code can solve the line segment intersection problem. The detailed explanation of each function is as follows.

• Main: The entry point of the program, which sets the points of each segment and checks for intersection.
• DoSegmentsIntersect: The core logic to determine whether the two segments intersect is implemented in this function.
• CrossProduct: This function calculates the cross product using the given three points. The result of the cross product is used to determine the direction.
• IsOnSegment: This function checks whether a specific point lies on the segment.

Test Cases

To test the given code, several test cases can be considered.

Test Number	Point A	Point B	Point C	Point D	Expected Result
1	(1, 1)	(4, 4)	(1, 4)	(4, 1)	true
2	(1, 1)	(2, 2)	(2, 1)	(3, 1)	false
3	(0, 0)	(3, 3)	(0, 3)	(3, 0)	true
4	(0, 0)	(5, 5)	(1, 1)	(5, 5)	true

Conclusion

In this article, we implemented an algorithm to determine whether line segments intersect using C#.
Understanding geometric problems helps in grasping these problem-solving techniques and lays the foundation for learning advanced algorithms.
Let’s continue solving various algorithm problems together. Thank you!