Category: Java Coding Test

In this article, we will solve the problem of binary numbers. Binary numbers are sequences composed of 0s and 1s, which must satisfy certain conditions. This problem is one of the frequently asked questions in Java coding tests and can be solved using dynamic programming.

1. Problem Definition

Problem: Write a program to calculate the number of binary numbers of length N. Binary numbers must satisfy the following conditions:

• The first digit of the number must be 1.
• It must be composed only of 0s and 1s.
• No two consecutive 1s are allowed.

For example, the 3-digit binary numbers are 100, 101, and 110. The 4-digit binary numbers are 1000, 1001, 1010, 1100, and 1101. Let’s explore how to solve this problem.

2. Approach

To solve this problem, we can use a dynamic programming (DP) approach. The ways to create a binary number of length N can be structured as follows:

2.1. State Definition

A binary number can be divided into two cases based on whether the last digit is 0 or 1. Thus, we define the following two DP arrays.

• dp0[i]: the number of binary numbers of length i whose last digit is 0
• dp1[i]: the number of binary numbers of length i whose last digit is 1

At this point, the total number of binary numbers of length N can be expressed as dp0[N] + dp1[N]. By discovering the rules for constructing binary numbers, we can derive the following recurrence relation:

2.2. Recurrence Relation

• dp0[i] = dp0[i-1] + dp1[i-1] (sum of the counts of binary numbers of length i-1 ending with 0 and 1)
• dp1[i] = dp0[i-1] (only binary numbers of length i-1 ending with 0 are possible)

2.3. Initial Conditions

The initial values are as follows:

• dp0[1] = 0 (there are no one-digit numbers ending with 0)
• dp1[1] = 1 (there is one one-digit number ending with 1: 1)

3. Java Code Implementation

Now, let’s implement this in Java using these conditions.

public class BinaryNumbers {
    private static int[] dp0;
    private static int[] dp1;

    public static void main(String[] args) {
        int N = 4; // Input: N digits
        dp0 = new int[N + 1];
        dp1 = new int[N + 1];

        // Set initial values
        dp0[1] = 0;
        dp1[1] = 1;

        // Fill the DP table
        for (int i = 2; i <= N; i++) {
            dp0[i] = dp0[i - 1] + dp1[i - 1];
            dp1[i] = dp0[i - 1];
        }

        // Print the result
        int result = dp0[N] + dp1[N];
        System.out.println("Number of binary numbers of length N: " + result);
    }
}

The above code shows the overall structure of the program to find binary numbers. It fills the DP table at each step and outputs the result at the end.

4. Performance Analysis

The time complexity of this algorithm is O(N), and the space complexity is also O(N). This provides a very efficient way to calculate binary numbers, allowing for quick execution even for large N. The algorithm makes good use of dynamic programming and recurrence relations, and thus can be applied to many other similar problems.

5. Problem Variations

This problem can be modified to create various other problems. For example:

• A program that returns an array of binary numbers of length N
• A program that prints all possible combinations of binary numbers
• A program that checks whether a given sequence is a binary number

6. Conclusion

Today we covered how to find binary numbers. Through this, I hope to enhance the understanding of dynamic programming concepts and improve problem-solving skills using this pattern. It will be useful in future coding tests and algorithm problem-solving.

7. References

• Dynamic Programming – Wikipedia
• Geeks for Geeks – Dynamic Programming Tutorials

Author: [Your Name]

Date: [Date]

1. Introduction to Binary Search Algorithm

Binary Search is an algorithm used to find a specific value in a sorted array. It is implemented by comparing the middle value of the array; if the desired value is less than the middle value, it searches in the left part; if greater, it searches in the right part. Binary Search is highly efficient with a time complexity of O(log n).

To use Binary Search, the array must be sorted; otherwise, a different search method, Linear Search, should be used.

2. Example Problem of Binary Search

Problem: Find the Index of a Specific Value

Given a sorted array of integers nums and an integer target, write a function binarySearch that returns the index of target. If target does not exist, it should return -1.

Example Input

nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
target = 5
            

Example Output

4
            

3. Problem Solving Process

3.1 Problem Analysis

The task is to find the index of target in the given nums array, so it is essential to utilize the fact that the array is sorted. The search will be performed by calculating the middle value and narrowing the search range through comparisons with target.

3.2 Algorithm Design

The Binary Search algorithm includes the following steps:

1. Initialize the starting index left and ending index right of the array.
2. Calculate the middle index mid: mid = (left + right) / 2.
3. Compare the middle value nums[mid] with target.
4. If the middle value is equal to target, return mid.
5. If the middle value is less than target, update left = mid + 1; otherwise, update right = mid - 1.
6. Repeat this process until left is greater than right.
7. If it decreases, return -1 indicating that target is not in the array.

3.3 Java Code Implementation

Let’s implement the above algorithm in Java.

public class BinarySearch {
    public static int binarySearch(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] == target) {
                return mid; // Return index if found
            } else if (nums[mid] < target) {
                left = mid + 1; // Search right
            } else {
                right = mid - 1; // Search left
            }
        }

        return -1; // Return -1 if not found
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        int target = 5;
        int result = binarySearch(nums, target);
        System.out.println(result);
    }
}

            

4. Time Complexity Analysis

The time complexity of the Binary Search algorithm is O(log n). This is because the size of the array reduces by half with each search. Therefore, Binary Search performs better with larger amounts of data.

5. Conclusion

In this lecture, we explored the concept of Binary Search and problems utilizing it. Binary Search is an efficient searching method for sorted arrays and is a basic algorithm frequently featured in various coding tests. As preparation for actual coding tests, practice solving various Binary Search problems to assess your skills.

This article will cover the concept of ‘bipartite graph’ and an algorithm problem for determining it. A bipartite graph is a graph where the vertices can be divided into two disjoint sets such that every edge connects a vertex in one set to a vertex in the other set. Such graphs can be applied to various problems and play a crucial role in algorithms like bipartite matching.

1. Problem Definition

Here is an example of the problem of determining a bipartite graph:

Problem: Determine if a graph is bipartite.

Input: 
- The first line contains the number of vertices N and the number of edges M. (1 ≤ N, M ≤ 100,000)
- The next M lines contain two vertices u and v for each edge. (1 ≤ u, v ≤ N)

Output:
- Print "Yes" if it is a bipartite graph, otherwise print "No".

2. What is a Bipartite Graph?

A bipartite graph is a graph structure that allows the vertices to be divided under specific conditions. That is, if any two vertices u and v are connected by an edge, then u and v must belong to different sets. Due to this property, bipartite graphs can be colored with two colors. In other words, all vertices belonging to one set are colored the same, while vertices in the other set are colored differently.

3. Algorithm Approach

The method for determining a bipartite graph is as follows. The graph is explored using DFS (Depth First Search) or BFS (Breadth First Search), coloring each vertex during the process.

1. Represent the graph using an adjacency list.
2. If there are unvisited vertices, start BFS or DFS from that vertex.
3. Assign a color to the starting vertex and assign a different color to its adjacent vertices while exploring.
4. If an adjacent vertex has already been visited and has the same color, it determines that it is not a bipartite graph.
5. After exploring all vertices, determine whether it is a bipartite graph and print the result.

4. Java Code Implementation

Below is the Java code implemented based on the above algorithm:

import java.util.*;

public class BipartiteGraph {
    static List> graph;
    static int[] color;
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        
        graph = new ArrayList<>();
        for (int i = 0; i <= N; i++) {
            graph.add(new ArrayList<>());
        }
        
        for (int i = 0; i < M; i++) {
            int u = sc.nextInt();
            int v = sc.nextInt();
            graph.get(u).add(v);
            graph.get(v).add(u);
        }
        
        color = new int[N + 1];
        boolean isBipartite = true;
        
        for (int i = 1; i <= N; i++) {
            if (color[i] == 0) {
                if (!bfs(i)) {
                    isBipartite = false;
                    break;
                }
            }
        }
        
        System.out.println(isBipartite ? "Yes" : "No");
    }
    
    private static boolean bfs(int start) {
        Queue queue = new LinkedList<>();
        queue.offer(start);
        color[start] = 1; // Color the starting vertex
        
        while (!queue.isEmpty()) {
            int node = queue.poll();
            for (int neighbor : graph.get(node)) {
                if (color[neighbor] == 0) {
                    // If the adjacent vertex has not been visited
                    color[neighbor] = 3 - color[node]; // Color it differently
                    queue.offer(neighbor);
                } else if (color[neighbor] == color[node]) {
                    // If the adjacent vertex has the same color
                    return false;
                }
            }
        }
        return true;
    }
}

5. Code Explanation

The above Java code implements an algorithm for determining a bipartite graph. Let’s examine each part in detail.

5.1 Graph Representation

The graph is represented as an adjacency list in the form of List>. It stores the list of each vertex and adds edge information to complete the graph structure.

5.2 Color Array

The color array color manages the colors of each vertex. 0 indicates not visited, and 1 and 2 represent two different colors.

5.3 BFS Exploration

The bfs method uses the BFS algorithm to explore the graph. It adds the starting vertex to the queue and colors the visited vertices. Then it assigns colors to adjacent vertices and checks for conflicts. If a vertex with the same color is found, the graph is not a bipartite graph.

6. Time Complexity

The time complexity of this algorithm is O(N + M). Here, N denotes the number of vertices and M denotes the number of edges. This is because all vertices and edges of the graph are explored once.

7. Other Considerations

This algorithm can handle both connected and disconnected graphs. In the case of a disconnected graph, each component is independently checked to determine if it is bipartite.

8. Conclusion

This article addressed the problem of determining a bipartite graph. Such problems are frequently encountered and are particularly useful for algorithm interview preparation. Understanding bipartite graphs will help solve various graph-related problems in the future. Continuous practice and solving a diverse range of problems are recommended to enhance coding skills.

This blog post will explore the Euclidean algorithm, which frequently appears in Java coding tests. The Euclidean algorithm is an efficient method for finding the greatest common divisor (GCD) of two numbers and requires basic mathematical knowledge. In this article, we will present a problem using the Euclidean algorithm and explain the solution process in detail.

Problem Description

Introduction to the Euclidean Algorithm

The Euclidean algorithm is a method devised by the ancient Greek mathematician Euclid, which is used to find the greatest common divisor of two given numbers. When two numbers A and B are given, the property GCD(A, B) = GCD(B, A % B) is used. This process is repeated until A becomes 0, at which point B’s value becomes the GCD.

The Euclidean Algorithm

1. If A is not 0: GCD(A, B) = GCD(B % A, A)
2. If A is 0: GCD(A, B) = B

Now, let’s apply the above algorithm to solve the problem through coding.

Problem Solving

Step 1: Designing the Function

First, we design a function to find the greatest common divisor of two numbers. We will use a recursive function.

public static int gcd(int a, int b) {
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

The code above takes two integers A and B as arguments and calculates the greatest common divisor recursively. If B is 0, A is the GCD. In other cases, it calls GCD(B, A % B) to continue calculating the GCD.

Step 2: Writing the Main Function

Now, we will write the main function to handle input and call the previously created GCD function.

import java.util.Scanner;

public class GCDExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter two integers (A B): ");
        int a = scanner.nextInt();
        int b = scanner.nextInt();

        int result = gcd(a, b);
        System.out.println("The greatest common divisor of " + a + " and " + b + " is: " + result);
    }

    public static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

The above code takes two numbers from the user to calculate and print the greatest common divisor. It uses the Scanner class to get input and calls the gcd method to obtain the calculation result.

Step 3: Testing the Code

Now, let’s run the program to calculate the greatest common divisor of the two numbers. For example, if we input 48 and 18, the following result is obtained.

Enter two integers (A B): 48 18
The greatest common divisor of 48 and 18 is: 6

Optimization and Additional Considerations

The Euclidean algorithm is a very efficient algorithm, with a time complexity of O(log(min(A, B))). However, in the given problem, we could also consider more diverse applications or performance optimizations.

Using an Array for Greatest Common Divisor

For example, if we have an array of several numbers, we can think of a way to calculate the greatest common divisor for multiple numbers. This can be implemented in the following form.

public static int gcdArray(int[] numbers) {
    int result = numbers[0];
    for (int i = 1; i < numbers.length; i++) {
        result = gcd(result, numbers[i]);
    }
    return result;
}

The above method takes an integer array as an argument and calculates the greatest common divisor of all the numbers in the array.

Conclusion

In this article, we solved the problem of finding the greatest common divisor using the Euclidean algorithm. This algorithm frequently appears in coding tests and serves as a good tool for assessing candidates’ problem-solving skills. We have shown that by appropriately handling input, we can create programs that function reliably for numbers of various sizes.

Organizing and practicing such algorithmic problems systematically will greatly aid you in preparing for coding tests. It is essential to practice various problems to refine your skills and be able to adapt flexibly when faced with difficult problems.

References

• Euclidean Algorithm – Wikipedia