Category: Java Coding Test

In modern society where online coding tests are increasingly required, the ability to effectively solve algorithm problems is an essential skill for developers. Today, we will deal with an algorithm problem that utilizes mathematical concepts. In this course, we will implement the Euler’s Totient Function and explain the process of solving it using the Java programming language.

What is Euler’s Totient Function?

The Euler’s Totient Function for a given positive integer n represents the number of integers between 1 and n that are coprime with n. In other words, it counts the number of integers whose greatest common divisor with n is 1. For example:

• φ(1) = 1 (The only number coprime with 1 is 1 itself)
• φ(2) = 1 (The only number coprime with 2 among 1 and 2 is 1)
• φ(3) = 2 (The numbers coprime with 3 among 1, 2, and 3 are 1 and 2)
• φ(4) = 2 (The numbers coprime with 4 among 1, 2, 3, and 4 are 1 and 3)

This function is also important in various fields of number theory and cryptography. Now, let’s solve the problem of implementing this function.

Problem Description

Implement a function that calculates Euler’s Totient Function for a given integer n. The input to the function is an integer n (1 ≤ n ≤ 10^6), and the output is the value of φ(n).

Problem Solving Strategy

There are various methods to find Euler’s Totient Function, but we can implement it efficiently using the following algorithm:

1. Prime Factorization: First, find the prime factors of n. Prime factors are the smallest primes that can express an integer as a product.
2. Calculating the Euler’s Totient Function: It can be calculated using the property of π(n). This can be expressed in the following formula:

φ(n) = n * (1 - 1/p1) * (1 - 1/p2) * ... * (1 - 1/pk)

In this formula, p1, p2, ..., pk are the prime factors of n. This formula allows us to compute Euler’s Totient Function quickly.

Java Code Implementation

Now let’s write a Java code to implement the Euler’s Totient Function. The following code snippet shows how to solve this problem:

public class EulerTotient {

    public static int eulerTotient(int n) {
        int result = n; // Initialize the result to n
        for (int p = 2; p * p <= n; p++) {
            // Check if p is a prime factor of n
            if (n % p == 0) {
                // If a prime factor, calculate the result
                while (n % p == 0) {
                    n /= p;
                }
                result -= result / p; // Update the result
            }
        }
        // Handle the last prime factor
        if (n > 1) {
            result -= result / n;
        }
        return result;
    }

    public static void main(String[] args) {
        int n = 10; // Define the input for testing
        System.out.println("φ(" + n + ") = " + eulerTotient(n)); // Output the result
    }
}

The above code performs operations using prime factorization to efficiently calculate the Euler’s Totient Function. For the input value n, π(n) is calculated and the result is printed.

Code Explanation

1. Function Definition

public static int eulerTotient(int n) method calculates and returns the Euler’s Totient Function for the input integer n. The final result is stored in a variable called result, initialized to n.

2. Prime Factor Check

A for loop is used where p ranges from 2 to √n, checking whether p is a prime factor of n. If it is a prime factor, it performs operations on all powers of p dividing n.

3. Result Calculation

When a prime factor p is found, the result is updated considering the relation with n to ultimately calculate the value of φ(n). Lastly, if n is greater than 1, we must also handle the last prime factor.

4. Main Method

The main method defines the test value for n and is structured to output the result.

Result Check

When the above code is executed, you can verify the value of the Euler’s Totient Function for the input value 10. The result is as follows:

φ(10) = 4

There are 4 numbers less than 10 that are coprime with 10, namely 1, 3, 7, and 9. This confirms that the implemented Euler’s Totient Function works correctly.

Review and Conclusion

In this course, we have implemented the Euler’s Totient Function and explored the process of solving algorithm problems. The Euler’s Totient Function is an algorithm based on mathematical reasoning, and it’s beneficial to know it due to its large applicability in solving various problems. The code implemented in Java can efficiently calculate Euler’s Totient values for numbers within a specific range and can be easily applied in various situations.

Continuing to tackle algorithm problems and practicing repeatedly will be important moving forward. This will greatly help in achieving results in coding tests. Thank you for reading!

Problem Description

The Euler’s Totient function (φ(n)) represents the number of positive integers that are coprime to n for a given natural number n. For example, φ(1) = 1, φ(2) = 1, φ(3) = 2, φ(4) = 2, φ(5) = 4. The goal of this problem is to write a program that calculates φ(n) for a given natural number n.

Problem: Calculate φ(n)

When a natural number n is given, write a Java program to calculate φ(n) for n. Note that n is an integer between 1 and 10^6 inclusive.

Input Format

• Natural number n is provided in the first line.

Output Format

• Output the value of φ(n) for n.

Problem Solving Process

To solve the problem, one must understand the definition and properties of Euler’s Totient function accurately and design an algorithm to calculate it efficiently.

Definition of Euler’s Totient Function

The Euler’s Totient function is calculated as follows:

φ(n) = n * (1 - 1/p1) * (1 - 1/p2) * ... * (1 - 1/pk)
            

Here, p1, p2, …, pk are the distinct prime factors of n. That is, the result can be derived from the number of primes that appear in the prime factorization of n and their product.

Calculating φ(n) through Prime Factorization

The steps to use this equation to calculate φ(n) in Java are as follows:

1. Set the initial value of n received as input.
2. Iterate from 2 to the square root of n to find primes.
3. For each prime p, check if n is divisible by p, and if it is, divide n by p and multiply φ(n) by (1 – 1/p).
4. Finally, repeat until n becomes 1.
5. Output the calculated value of φ(n).

Java Code Example

Below is an example of Java code that calculates φ(n):

import java.util.Scanner;

public class EulerTotient {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int originalN = n;
        double result = n;

        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                while (n % i == 0) {
                    n /= i;
                }
                result *= (1.0 - (1.0 / (double)i));
            }
        }

        if (n > 1) {
            result *= (1.0 - (1.0 / (double)n));
        }

        System.out.println((int)result);
    }
}
            

This code calculates and outputs φ(n) for the value of n entered by the user. It has calculated φ(n) in cases where it can be divided by primes and the response is provided accordingly.

Efficiency Analysis

The time complexity of this algorithm is O(√n) because it needs to iterate up to the square root of n to find primes. This method can handle cases where n is as large as 10^6 without issues. Moreover, the space complexity is O(1), minimizing additional memory usage, making it efficient.

Conclusion

The Euler’s Totient function is one of the very important concepts in number theory and algorithm problems. Through the process of solving this problem, one can enhance their understanding of primes, prime factorization, and basic mathematical principles. Remember that this is a useful problem for preparing for coding tests using Java.

Problem Description

This is a problem to find the number of connected components in a given graph. A connected component refers to a subgraph in which there is a path between any two vertices. In other words, if two vertices are connected, they belong to the same connected component.

Problem Definition

Output the number of connected components in the given undirected graph.

Input

The first line contains the number of vertices n (1 ≤ n ≤ 1000) and the number of edges m (0 ≤ m ≤ 10000). The next m lines provide the two endpoints of each edge u and v. u and v are distinct vertices, represented by integers from 1 to n.

Output

Output the number of connected components.

Example

    Input:
    5 3
    1 2
    2 3
    4 5

    Output:
    2
    

Problem Solving Strategy

To solve the problem, we will follow these steps:

1. Represent the graph in the form of an adjacency list.
2. Use DFS (Depth First Search) or BFS (Breadth First Search) to explore the connected components.
3. Count the number of connected components during the exploration.

1. Graph Representation

We represent the undirected graph as an adjacency list. Each vertex stores a list of its connected vertices. In Java, this can be easily implemented using ArrayList.

2. Exploration Using DFS

After representing the graph, we perform DFS for each vertex to visit the connected vertices. We use an array to keep track of visited vertices to avoid visiting them again.

3. Implementation and Result Derivation

We maintain a counter variable to count the total number of connected components, and we increase the count each time DFS starts at a new vertex.

Java Code Implementation

import java.util.ArrayList;
import java.util.Scanner;

public class ConnectedComponents {
    static ArrayList[] graph;
    static boolean[] visited;
    static int count;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        int n = scanner.nextInt(); // Number of vertices
        int m = scanner.nextInt(); // Number of edges
        
        graph = new ArrayList[n + 1];
        visited = new boolean[n + 1];
        
        for (int i = 1; i <= n; i++) {
            graph[i] = new ArrayList<>();
        }
        
        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt();
            int v = scanner.nextInt();
            graph[u].add(v);
            graph[v].add(u);
        }
        
        count = 0; // Initialize connected components count
        
        for (int i = 1; i <= n; i++) {
            if (!visited[i]) {
                dfs(i); // Execute DFS
                count++; // Increase count when a new connected component is found
            }
        }
        
        System.out.println(count); // Output the result
        scanner.close();
    }

    public static void dfs(int node) {
        visited[node] = true; // Mark the current node as visited
        for (int neighbor : graph[node]) {
            if (!visited[neighbor]) {
                dfs(neighbor); // Explore adjacent nodes
            }
        }
    }
}
    

Code Explanation

The above Java code works as follows:

1. Input the number of vertices n and edges m, and initialize the adjacency list graph.
2. Input the edge information to construct the graph.
3. Perform DFS for each vertex. If the vertex has not been visited, call DFS to visit all connected vertices.
4. Increase the count of connected components each time DFS is called.
5. Finally, output the number of connected components.

Complexity Analysis

The time complexity of this algorithm is O(n + m), where n is the number of vertices and m is the number of edges. This is because all vertices and edges are visited once during the DFS. The space complexity also uses O(n + m) additional space.

Conclusion

The problem of finding the number of connected components can be solved using exploration algorithms such as DFS or BFS. This problem requires a deep understanding of graphs, and it is important to master accurate graph representation and exploration methodologies in the problem-solving process. By doing so, you can build a useful foundational knowledge for coding tests.

Through this tutorial, we have learned the basics of graph theory and how to find the number of connected components. Continue to practice various graph problems to improve your algorithm skills!