Author: root

Hello, everyone! Today, we will explore a problem that frequently appears in coding tests using C#, Bridging. This problem requires a good understanding of the concepts of combinations and dynamic programming to solve. To better understand the bridging problem we will tackle, I will explain its definition and basic approach.

Problem Definition

The bridging problem is about finding the number of ways to choose k bridges from a given n bridges. Here, k must be less than or equal to n. This problem follows the basic principles of combinations, and can be formally expressed as follows:

        C(n, k) = n! / (k! * (n - k)!)
    

Here, n! represents the product of all integers from 1 to n, and the combination formula C(n, k) indicates the number of ways to choose k items from n.

Example Problem

For example, when choosing 2 bridges from 5, the following combinations are possible:

• Bridge 1, Bridge 2
• Bridge 1, Bridge 3
• Bridge 1, Bridge 4
• Bridge 1, Bridge 5
• Bridge 2, Bridge 3
• Bridge 2, Bridge 4
• Bridge 2, Bridge 5
• Bridge 3, Bridge 4
• Bridge 3, Bridge 5
• Bridge 4, Bridge 5

In other words, there are a total of 10 ways. Based on this concept, we can solve the problem.

Algorithm Approach

To solve the bridging problem, two basic concepts are needed.

1. Combination Formula

This programming problem can be solved using the combination formula. We will tackle the problem based on the C(n, k) formula.

C# Code Implementation

        using System;

        class Program
        {
            static void Main(string[] args)
            {
                Console.WriteLine("Enter the values for n and k (e.g., n k):");
                string[] inputs = Console.ReadLine().Split(' ');
                int n = int.Parse(inputs[0]);
                int k = int.Parse(inputs[1]);

                long result = CalculateCombination(n, k);
                Console.WriteLine($"C({n}, {k}) = {result}");
            }

            static long CalculateCombination(int n, int k)
            {
                if (k > n) return 0;
                if (k == 0 || k == n) return 1;

                long result = 1;
                for (int i = 0; i < k; i++)
                {
                    result = result * (n - i) / (i + 1);
                }
                return result;
            }
        }
        

Code Explanation

The above C# code calculates the number of combinations based on the values of n and k input by the user. The CalculateCombination method executes the logic to compute C(n, k) using the given n and k.

2. Dynamic Programming Method

Another way to calculate the number of combinations is by using dynamic programming, which allows for the use of memoization techniques.

C# Code Implementation: Dynamic Programming

        using System;

        class Program
        {
            static void Main(string[] args)
            {
                Console.WriteLine("Enter the values for n and k (e.g., n k):");
                string[] inputs = Console.ReadLine().Split(' ');
                int n = int.Parse(inputs[0]);
                int k = int.Parse(inputs[1]);

                long[,] dp = new long[n + 1, k + 1];

                for (int i = 0; i <= n; i++)
                {
                    for (int j = 0; j <= Math.Min(i, k); j++)
                    {
                        if (j == 0 || j == i)
                            dp[i, j] = 1;
                        else
                            dp[i, j] = dp[i - 1, j - 1] + dp[i - 1, j];
                    }
                }

                Console.WriteLine($"C({n}, {k}) = {dp[n, k]}");
            }
        }
        

Code Explanation

The dynamic programming code above uses a 2D array dp to store C(n, k). It utilizes nested loops to calculate the number of combinations for each case. This code solves the problem with a time complexity of O(n * k) through the memoization approach.

Input/Output Example

Let’s demonstrate the process where, upon the user’s entry of n and k, the program outputs the count of combinations:

Input

    5 2
    

Output

    C(5, 2) = 10
    

Conclusion

The bridging problem is a good exercise for understanding the concept of combinations and learning how to implement it programmatically. We have provided two methods for calculating combinations using C#, and it’s important to understand the pros and cons of each method. I hope this problem enhances your understanding of combinations and dynamic programming, and helps you develop the ability to solve various algorithmic challenges.

References

• Combination – Wikipedia
• Dynamic Programming – Wikipedia

One of the common problems that appears in coding tests is the ‘Minimum Cost Path’ problem.
This problem requires minimizing costs while satisfying certain conditions,
and can be efficiently solved using various algorithms.
In this article, we will explore how to solve the minimum cost path problem using C# in detail.

Problem Description

In the given graph, each edge has a specific cost.
When tasked with finding the minimum cost path from a starting vertex to all other vertices,
we need to solve this efficiently.

Problem Definition

Let the number of vertices be N and the number of edges be M.
Suppose we are given a graph that satisfies the following conditions:

• N (1 <= N <= 1000): Number of vertices
• M (1 <= M <= 10000): Number of edges
• Costs are integers between 1 and 10000

Input: Information about the starting vertex and the costs between each vertex is provided.
Output: An array of the minimum cost paths from the starting vertex to all other vertices is returned.

Example

Input

        5 7
        1 2 2
        1 3 3
        2 3 1
        2 4 5
        3 4 1
        3 5 5
        4 5 2
        1
        

Output

        0 2 3 7 5
        

In the above example, starting from vertex 1,
it takes a cost of 2 to reach vertex 2, 3 to reach vertex 3, 7 to reach vertex 4, and 5 to reach vertex 5.
Each cost must be minimized while setting the paths correctly.

Algorithm Explanation

We will use Dijkstra’s algorithm to solve the problem.
Dijkstra’s algorithm is an algorithm that finds the shortest path in a weighted graph and proceeds as follows:

1. Set the starting vertex and initialize the distance of this vertex to 0.
2. Initialize the distance values of all other vertices to infinity.
3. Select the vertex with the least cost and update the distance values to its neighboring vertices.
4. Repeat the above process for all vertices. Ultimately, the minimum cost for each vertex is calculated.

C# Implementation

Now, let’s implement Dijkstra’s algorithm in C#.
First, we will reference the necessary libraries and use an adjacency list to represent the graph.

        using System;
        using System.Collections.Generic;

        class Program
        {
            static void Main(string[] args)
            {
                int n = 5; // Number of vertices
                int m = 7; // Number of edges
                List> edges = new List>()
                {
                    Tuple.Create(1, 2, 2),
                    Tuple.Create(1, 3, 3),
                    Tuple.Create(2, 3, 1),
                    Tuple.Create(2, 4, 5),
                    Tuple.Create(3, 4, 1),
                    Tuple.Create(3, 5, 5),
                    Tuple.Create(4, 5, 2)
                };
                int start = 1; // Starting vertex

                var result = Dijkstra(n, m, edges, start);
                
                Console.WriteLine(string.Join(" ", result));
            }

            static int[] Dijkstra(int n, int m, List> edges, int start)
            {
                // Initialize graph
                List>> graph = new List>>(n + 1);
                for (int i = 0; i <= n; i++)
                {
                    graph.Add(new List>());
                }

                // Add edges
                foreach (var edge in edges)
                {
                    graph[edge.Item1].Add(Tuple.Create(edge.Item2, edge.Item3));
                }

                // Initialize distances
                int[] distance = new int[n + 1];
                bool[] visited = new bool[n + 1];
                for (int i = 1; i <= n; i++) distance[i] = int.MaxValue;

                // Starting vertex
                distance[start] = 0;
                PriorityQueue> pq = new PriorityQueue>();
                pq.Enqueue(Tuple.Create(0, start));

                while (pq.Count > 0)
                {
                    var current = pq.Dequeue();
                    int dist = current.Item1;
                    int node = current.Item2;

                    if (visited[node]) continue;
                    visited[node] = true;

                    foreach (var neighbor in graph[node])
                    {
                        int newDist = dist + neighbor.Item2;
                        if (newDist < distance[neighbor.Item1])
                        {
                            distance[neighbor.Item1] = newDist;
                            pq.Enqueue(Tuple.Create(newDist, neighbor.Item1));
                        }
                    }
                }
                return distance[1..]; // Return distances excluding start vertex
            }
        }

        // Define priority queue class
        public class PriorityQueue
        {
            List elements = new List();

            public int Count => elements.Count;

            public void Enqueue(T item)
            {
                elements.Add(item);
                var index = elements.Count - 1;
                while (index > 0)
                {
                    var parentIndex = (index - 1) / 2;
                    if (Compare(elements[index], elements[parentIndex]) >= 0) break;
                    Swap(index, parentIndex);
                    index = parentIndex;
                }
            }

            public T Dequeue()
            {
                var lastIndex = elements.Count - 1;
                var firstElement = elements[0];
                elements[0] = elements[lastIndex];
                elements.RemoveAt(lastIndex);

                --lastIndex;

                int index = 0;
                while (true)
                {
                    var leftChildIndex = index * 2 + 1;
                    var rightChildIndex = index * 2 + 2;
                    int smallestChildIndex;

                    if (leftChildIndex > lastIndex) break;
                    if (rightChildIndex > lastIndex) smallestChildIndex = leftChildIndex;
                    else smallestChildIndex = Compare(elements[leftChildIndex], elements[rightChildIndex]) < 0 ? leftChildIndex : rightChildIndex;

                    if (Compare(elements[index], elements[smallestChildIndex]) <= 0) break;

                    Swap(index, smallestChildIndex);
                    index = smallestChildIndex;
                }

                return firstElement;
            }

            void Swap(int indexA, int indexB)
            {
                var temp = elements[indexA];
                elements[indexA] = elements[indexB];
                elements[indexB] = temp;
            }

            int Compare(T a, T b)
            {
                // Custom comparison logic needs to be implemented here
                // For example, for Tuple
                return a.Item1.CompareTo(b.Item1);
            }
        }
        

Algorithm Analysis

Using the above algorithm, the time complexity is O((N + M) log N).
Here, N is the number of vertices, and M is the number of edges.
Dijkstra’s algorithm is generally efficient,
and is effective in finding the minimum cost in many cases.
However, it should be noted that all edge weights must not be negative.

Conclusion

In this tutorial, we examined Dijkstra’s algorithm to solve the minimum cost path problem using C#.
We defined the problem, explained the algorithm, and gradually implemented the actual code.
Since such problems frequently appear in coding tests, it is important to practice and understand them thoroughly.
I hope this article helps in your preparation for coding tests.