Category: C++ Coding Test Tutorials

October 3, 2023

Author: AI Writer

Introduction

The Kevin Bacon rule of six degrees states that any person is connected to Kevin Bacon within a maximum of six degrees. This is an intriguing concept that explains relationships in social networks and is a popular topic in various fields. In this article, we will transform this rule into an algorithmic problem and detail the solution process using C++.

The problem we will tackle is “Finding the person who can reach Kevin Bacon the fastest.” Through research, we will calculate the degree of connection to Kevin Bacon for each person when given the friendship relations among several people.

Problem Definition

The objective is to calculate each person’s Kevin Bacon number within a given network and find the person with the lowest number. To do this, we need to input the number of people and their friendship relations.

Problem Description

The input is provided in the following format:

• The first line contains the number of people N (1 ≤ N ≤ 100)
• From the second line onward, each line consists of two integers A, B representing the friendship relation (A and B are friends).

The output should be the number of the person closest to Kevin Bacon. If there are multiple, output the smallest number.

Problem Solving Approach

To solve this problem, we will use graph traversal techniques. We will treat people as vertices and friendships as edges, applying BFS (Breadth-First Search). We will calculate the shortest distance from each person to Kevin Bacon and find the minimum among them.

Algorithm

1. Construct the graph as an adjacency list based on the input.
2. Calculate each person’s Kevin Bacon number using BFS.
3. Compare the calculated Kevin Bacon numbers to find the minimum.

C++ Implementation

Now, let’s implement the C++ code based on the algorithm above.

#include <iostream>
#include <vector>
#include <queue>
#include <limits> 
using namespace std;

int main() {
    int N, M;
    cin >> N >> M; // Number of people N, number of friendship relations M
    vector<vector<int>> graph(N + 1); // Adjacency list

    for (int i = 0; i < M; i++) {
        int A, B;
        cin >> A >> B; // Friendship relation A, B
        graph[A].push_back(B);
        graph[B].push_back(A); // Undirected graph
    }

    int min_bacon = numeric_limits<int>::max();
    int answer = 0;

    for (int i = 1; i <= N; i++) {
        vector<int> distance(N + 1, -1); // Initialize distances for each person
        queue<int> q;
        distance[i] = 0; // Distance from self is 0
        q.push(i); // Add self to the queue

        while (!q.empty()) {
            int current = q.front();
            q.pop();
            for (int neighbor : graph[current]) {
                if (distance[neighbor] == -1) {
                    distance[neighbor] = distance[current] + 1; // Calculate distance
                    q.push(neighbor); // Add friend to explore
                }
            }
        }

        int bacon_count = 0;
        for (int d : distance) {
            if (d != -1) {
                bacon_count += d; // Sum Kevins Bacon numbers
            }
        }

        if (bacon_count < min_bacon) {
            min_bacon = bacon_count; // Update minimum
            answer = i; // Record person number
        }
    }

    cout << answer << endl; // Output result
    return 0;
}
            

Code Explanation

The above C++ code includes the following key steps:

1. Input Processing: Read N and M, and store friendship relations in an adjacency list format.
2. BFS Implementation: Perform BFS for each person to calculate distances.
3. Result Calculation: Update the Kevin Bacon number based on calculated distances and ultimately find the minimum.

Complexity Analysis

The time complexity of this algorithm is O(N + M). This is because each person explores their friends through BFS, and each edge is explored only once. This is sufficiently efficient for the input range of the given problem.

Conclusion

In this lecture, we explored how to solve an algorithm problem utilizing the Kevin Bacon rule of six degrees in C++. We analyzed relationships within a network using graph traversal techniques like BFS. Such types of problems frequently appear in coding tests, making it crucial to practice varied problems to enhance proficiency.

Finally, the problem-solving process presented here will not only aid in preparation for coding tests but also help address various issues that may arise in actual programming more systematically.

This course introduces how to solve the cocktail making problem using C++. Algorithmic problems generally require generating a desired output for a specific input. This problem involves making cocktails using various ingredients according to given proportions.

Problem Description

You want to make cocktails with various ingredients. Each ingredient requires a specific amount. You want to use each ingredient to make as much cocktail as possible according to the given proportions. Calculate how many servings of cocktails you can make by properly calculating the amount and requirements of the given ingredients.

Input

• The first line contains the number of types of ingredients N (1 ≤ N ≤ 100).
• The second line contains the amount of each ingredient A[i] (1 ≤ A[i] ≤ 10^5).
• The third line contains the requirement of each ingredient B[i] (1 ≤ B[i] ≤ 10^5).

Output

Print the maximum number of servings of cocktails that can be made.

Example

Input

3
100 200 300
10 20 30

Output

3

Problem Solving Approach

To solve this problem, you need to calculate how many servings of cocktails can be made based on the amount and requirements of each ingredient. Follow the procedure below:

1. Calculate the possible maximum number of cocktails for each ingredient. This is the quotient of the amount of that ingredient divided by its requirement.
2. Take the minimum of the maximum cocktail numbers calculated for all ingredients. This value will be the final number of cocktails that can be made.

C++ Code Implementation

Now let’s write the C++ code based on the above approach. The code will receive input, calculate the possible maximum number of cocktails for each ingredient, and output the minimum value.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int N;
    cin >> N; // Number of ingredient types
    vector<int> A(N); // Amount of ingredients
    vector<int> B(N); // Requirement of ingredients

    // Input amount of ingredients
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    
    // Input requirement of ingredients
    for (int i = 0; i < N; i++) {
        cin >> B[i];
    }

    int maxCocktails = INT_MAX; // Set initial value to a very large number

    // Calculate maximum number of cocktails for each ingredient
    for (int i = 0; i < N; i++) {
        int cocktailsPossible = A[i] / B[i]; // Possible number of cocktails
        maxCocktails = min(maxCocktails, cocktailsPossible); // Update minimum value
    }

    cout << maxCocktails << endl; // Print the result
    return 0;
}
    

Code Explanation

The above C++ code calculates the maximum number of servings of cocktails through the following steps:

1. Receive the number of ingredient types N and declare two vectors A and B to store the amount and requirement of ingredients.
2. Input the amount of ingredients and input the requirement for each ingredient.
3. Initialize the maximum number of cocktails to ‘INT_MAX’ to set a comparison basis.
4. Calculate the possible number of cocktails for each ingredient and store the smallest value in ‘maxCocktails’.
5. Finally, output the calculated maximum number of cocktails.

Complexity Analysis

The time complexity of this problem is O(N). N is the number of ingredient types, and since we need to check the amount and requirements for each ingredient, it has linear time complexity. The space complexity is O(N) because two vectors are used to store the amount and requirements of ingredients.

Conclusion

In this course, we learned how to solve the cocktail making problem using C++. To solve the given problem, you need to handle input and derive desired results through calculations. This helps develop algorithmic problem-solving skills and can be useful in real coding tests.

Expanding the problem by adding more diverse ingredients and conditions would also be good practice. I want to emphasize once again that understanding the problem and approaching it logically is important, in addition to coding.

In recent years, programming and algorithm problem-solving skills have become a crucial assessment criterion not only in the IT industry but also in various fields. In this course, we will learn basic manipulation processes, algorithm design, and optimization through the problem of ‘Card Sorting’. This problem helps enhance understanding of important sorting algorithms in practical work and coding tests.

Problem Description

In old card games, sorting the numbers on cards was essential. You have N cards, each with an integer written on it. You need to sort these cards in ascending order. However, there are some rules in this process:

• The cards consist of integers and must be arranged in sorted order.
• The number of cards N (1 ≤ N ≤ 10^6).
• The numbers written on the cards are integers within the range of -10^9 ≤ number ≤ 10^9.

Input Format

The first line contains the number of cards N, followed by N lines, each containing the number on each card.

Output Format

After sorting the cards in ascending order, print each card’s number on a new line.

Example

Input:
5
3
1
4
1
5

Output:
1
1
3
4
5

Problem-Solving Strategy

This problem is a sorting problem, and various sorting algorithms can be used. However, considering the given range, it is advisable to use basic sorting algorithms such as Quick Sort or Merge Sort. In this course, we will demonstrate how to efficiently solve the problem using the std::sort function included in the C++ STL.

STL Standard Library’s sort() Function

The standard template library (STL) of C++ includes the sort() function. This function sorts the elements of the container provided as an argument. std::sort has an average time complexity of O(N log N) and is very efficient.

Code Implementation

#include <iostream>
#include <vector>
#include <algorithm> // Include to use sort()

int main() {
    int N;
    std::cin >> N; // Input the number of cards
    std::vector<int> cards(N); // Create a vector to store card numbers

    // Input card numbers
    for (int i = 0; i < N; i++) {
        std::cin >> cards[i];
    }

    // Sort card numbers
    std::sort(cards.begin(), cards.end());

    // Output sorted card numbers
    for (int i = 0; i < N; i++) {
        std::cout << cards[i] << std::endl;
    }

    return 0;
}

Code Analysis

The code above consists of the following processes:

1. Library Inclusion: Includes #include <iostream>, #include <vector>, #include <algorithm> to add basic libraries for input, output, and sorting.
2. Input: Inputs the number of cards N from the user, and then stores the N card numbers in the vector.
3. Sorting: Uses the std::sort() function to sort the card numbers stored in the vector in ascending order. Here, the begin() and end() methods are used to set the start and end positions.
4. Output: Prints the sorted card numbers.

Time Complexity Analysis

The std::sort() function mentioned above operates with an average complexity of O(N log N). This is very efficient for processing large amounts of data. When the input size N is up to 10^6, sorting can be executed within a sufficient timeframe.

Conclusion

In this course, we learned the process of solving the card sorting problem. We were able to write simple yet efficient code utilizing the STL’s sort() function to solve the problem. Such sorting algorithms are frequently used in many programming problems and real situations, so it is important to master and apply them.

In the next session, we will further enhance your coding test skills through additional algorithm problems.

Thank you!

This article discusses the process of solving an algorithm problem that identifies friendship relationships using C++. The problem involves applying graph theory based on the number of friends and their relationships to understand the friendship network. In this context, we will explore the fundamental concepts of data structures and algorithms, and examine the implementation methods in C++ in detail.

Problem Description

Problem: There are N students. Each student has the potential to have friends. The friendship relationship is bidirectional, and given the friendship relationships, implement an algorithm to determine how many friends a specific student has. Additionally, it should be able to check whether two students are friends with each other.

Input:

• The first line contains the number of students N (1 ≤ N ≤ 1000).
• The second line contains M (0 ≤ M ≤ N*(N-1)/2) friendship relationships.
• Each friendship relation consists of A and B, where A is a friend of B and B is a friend of A.
• Finally, queries are given to check the friendship relationships.

Problem Solving Process

To solve the problem, we will go through the following steps:

1. Graph Modeling: Model the friendship relationships as a graph. The vertices represent students, and the edges represent friendship relationships.
2. Constructing an Adjacency List: Create an adjacency list representing friends for each vertex.
3. Count Friends: Implement a function to calculate the number of friends of a specific student.
4. Check Friendship Relationships: Implement a function to check the friendship relationship between two students.

1. Graph Modeling

To solve the given problem, we need to first model the friendship relationships in the form of a graph. In this case, the vertices of the graph represent students, and the edges represent friendship relationships between those students.

Therefore, assuming there are N students, we can represent them with integers ranging from 0 to N-1. For example, student A can be represented as 0, and student B as 1.

2. Constructing an Adjacency List

We use an adjacency list to represent the graph. We can use a hashmap or vector where each student is a key and the list of students who are friends with that student is the value. Below is a method for constructing the adjacency list in C++:

#include 
#include 
#include 
#include 

using namespace std;

unordered_map > friends;

void addFriendship(int a, int b) {
    // Add friendship (bidirectional)
    friends[a].push_back(b);
    friends[b].push_back(a);
}
    

In the above code, the addFriendship function adds a friendship relationship between two students.

3. Counting Friends

To count the number of friends of a specific student, we implement a function that returns the size of that student’s adjacency list. Below is an example implementation:

int countFriends(int student) {
    return friends[student].size();
}
    

This code returns the size of the list of friends for a specific student from the adjacency list.

4. Checking Friendship Relationships

We also need a function to check whether two students are friends. This function searches through the adjacency list to see if the two students are in the same list of friends. Below is an example implementation:

bool areFriends(int a, int b) {
    // Get A's list of friends
    for (int friend_id : friends[a]) {
        if (friend_id == b) {
            return true; // Friendship exists
        }
    }
    return false; // Friendship does not exist
}
    

The above function checks if student A has student B in their list of friends.

Complete Code

Now, I will write the complete code incorporating all the explanations given so far:

#include 
#include 
#include 

using namespace std;

unordered_map > friends;

void addFriendship(int a, int b) {
    friends[a].push_back(b);
    friends[b].push_back(a);
}

int countFriends(int student) {
    return friends[student].size();
}

bool areFriends(int a, int b) {
    for (int friend_id : friends[a]) {
        if (friend_id == b) {
            return true;
        }
    }
    return false;
}

int main() {
    int N, M;
    cout << "Enter the number of students (N) and the number of friendship relations (M): ";
    cin >> N >> M;

    cout << "Enter friendship relations (e.g., A B): " << endl;
    for (int i = 0; i < M; i++) {
        int a, b;
        cin >> a >> b;
        addFriendship(a, b);
    }

    int query;
    cout << "Enter the number of the student to check the number of friends: ";
    cin >> query;
    cout << "Student " << query << " has " << countFriends(query) << " friends." << endl;

    int a, b;
    cout << "Enter the numbers of the two students to check their friendship relationship: ";
    cin >> a >> b;
    if (areFriends(a, b)) {
        cout << "Student " << a << " and student " << b << " are friends." << endl;
    } else {
        cout << "Student " << a << " and student " << b << " are not friends." << endl;
    }

    return 0;
}
    

Conclusion

In this tutorial, we closely examined the process of solving an algorithm problem that identifies friendship relationships using the C++ language. We learned how to model friendship relationships using important concepts from graph theory and data structures, and how to solve the problem through this modeling. Skills in solving such algorithmic problems are crucial for coding tests for employment, so I encourage you to strengthen your abilities through a variety of problems.