Category: Java Coding Test

Problem Description

Today’s problem is “Building a Bridge”. This problem is one of the frequently asked types in coding interviews, where the goal is to find the optimal solution based on the given information.

Problem Definition

Let’s assume a city needs a bridge with really high stairs. This bridge must be located between two beaches (left and right) and must meet specific requirements.

Input

• n: Length of the beach where the bridge needs to be built (1 ≤ n ≤ 100)
• Height array: The heights of the areas of each beach (1 ≤ elements of height array ≤ 100)

Output

Return the maximum height of the bridge. In other words, you need to output the minimum height at all areas where the bridge passes.

Example Problem

Input Example

    n = 5
    height = [5, 3, 6, 2, 4]
    

Output Example

    3
    

Explanation: The maximum height that can be swum to build the bridge is 3. This is because the second area from the left, Height[1], is 3.

Problem Solving Process

Step 1: Understand the Problem

This problem requires finding the position of the highest bridge to be built. The height of the bridge is restricted by the heights of each regional area. Therefore, the height of the bridge is limited to the minimum of all the higher areas.

Step 2: Design an Approach to the Problem

The approach to solving the problem is simply to find the minimum value of the array. The height of the bridge is determined by the lowest height at each section of the bridge. To achieve this, follow these steps:

1. Find the minimum value in the given array.
2. Set the found minimum value as the maximum height of the bridge.
3. Return the result.

Step 3: Write Java Code

Now, let’s solve the problem with Java code. We will write a simple program to find the optimal height of the bridge.

    import java.util.Arrays;

    public class BridgeBuilder {
        public static void main(String[] args) {
            int n = 5;
            int[] height = {5, 3, 6, 2, 4};
            int maxBridgeHeight = findMaxBridgeHeight(height);
            System.out.println("The maximum height of the bridge is: " + maxBridgeHeight);
        }

        public static int findMaxBridgeHeight(int[] height) {
            // Find the minimum value in the height array
            return Arrays.stream(height).min().orElse(Integer.MAX_VALUE);
        }
    }
    

Step 4: Explain the Code

The code above demonstrates a simple logic to calculate the maximum height of the bridge based on the heights of the given beaches.

• First, we import the java.util.Arrays package to easily handle the array.
• We create a method called findMaxBridgeHeight that returns the minimum value of the given height array.
• To find the minimum value, we used Java 8’s Stream API to create concise code.

Step 5: Analyze Time Complexity

The time complexity of this algorithm is O(n), because we need to check all the elements in the array to find the minimum value. This method is efficient and practical as long as the input size (n) does not become extremely large.

Conclusion

The bridge-building problem can be understood simply as a process of finding the minimum value of an array. This approach is very useful for solving algorithmic problems and is commonly used when dealing with data such as arrays or lists. It teaches us techniques that can be applied to various problems.

Additional Exercise Problems

• If the height of the bridge can be changed, what can be done to make the bridge higher?
• Write a program that dynamically changes the height of the bridge according to the traffic volume of vehicles.

Through these exercise problems, you can become a developer with a higher understanding by solving variations of the bridge-building problem.

Problem Description

The Bridge Building Problem arises in the following context. You want to build N bridges and M piers. The goal of this problem is to calculate the number of possible combinations of bridge placements.
The bridges must not overlap, and each pier serves to connect two bridges. To calculate the number of ways to place the bridges, knowledge of combinations and dynamic programming is required.

Problem Definition

Input

• Size N (1 ≤ N ≤ 30): Number of bridges
• Size M (1 ≤ M ≤ N): Number of piers

Output

• Print the total number of ways to place the bridges as an integer.

Problem Approach

The bridge building problem is essentially a combinatorial problem. We need to find the number of ways to select M bridges from N bridges. The number of combinations is calculated using the following formula:

C(N, M) = N! / (M! * (N-M)!)

However, directly calculating the factorial can be time-consuming, so we can use dynamic programming to calculate it efficiently. We will use two arrays.
The first array will keep track of the number of piers chosen to place the bridges, and the second array will record the number of ways to place the bridges.

Problem Solution

Step 1: Initialize the Dynamic Programming Array

First, we initialize the DP array. dp[i][j] represents the number of ways to place i bridges and j piers. After initializing the array to 0,
we set dp[0][0] = 1. This means there is 1 way when there are no bridges and no piers.

Step 2: Find the Recurrence Relation

When adding one bridge, there is an option to place one pier. We can fill the dp array using the following recurrence relation.

dp[i][j] = dp[i-1][j-1] + dp[i-1][j]

Here, dp[i-1][j-1] refers to the case of placing i-1 bridges and j-1 piers, adding a new bridge, while
dp[i-1][j] refers to the case of not adding a new bridge and using i-1 bridges to place j piers.

Step 3: Print the Final Result

After filling all the ranges, we can print dp[N][M] to get the final number of ways to place the bridges.

Java Code Implementation

The Java code to solve the bridge building problem is as follows:

public class BridgeBuilding {
    public static void main(String[] args) {
        int N = 5;  // Number of bridges
        int M = 3;  // Number of piers
        
        System.out.println(countWays(N, M));
    }
    
    public static int countWays(int n, int m) {
        int[][] dp = new int[n + 1][m + 1];

        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= Math.min(i, m); j++) {
                if (j == 0) {
                    dp[i][j] = 1; // The case of placing no piers
                } else if (i > 0) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }
            }
        }
        
        return dp[n][m];
    }
}

Time Complexity

The time complexity of this algorithm is O(N*M). This time arises from filling the DP table using two nested loops. The space complexity is O(N*M), as we need space to store the DP array.

Conclusion

In this lecture, we covered the ‘Bridge Building’ problem. To solve this problem, an understanding of combinatorial theory and dynamic programming is essential.
The provided Java code allows us to solve the problem, and similar approaches can be used when encountering similar types of problems.
In the next lecture, we will discuss another algorithm problem.

Hello! In this lecture, we will address the problem of calculating the area of a polygon. This problem will be implemented using Java, and I will explain the process of solving the problem in detail, applying basic mathematical concepts and algorithms.

Problem Description

Write a program to calculate the area of a polygon based on the coordinates of given points. The points are given in the form of (x1, y1), (x2, y2), ..., (xn, yn), and it is assumed that these points are connected to form the polygon. The area to be computed must satisfy the following conditions:

• The points are given in a clockwise or counterclockwise direction.
• An accurate and efficient algorithm must be implemented.

Input

The first line contains the number of points n. The following n lines contain the coordinates x, y of each point as integers.

Output

Output the area of the polygon as a real number, rounded to two decimal places.

Example Input

4
0 0
4 0
4 3
0 4

Example Output

12.00

Solution Method

There are various algorithms to calculate the area of a polygon, but in this lecture, we will use Schroder’s formula (or the polygon area formula). According to this formula, the area of a polygon A is calculated as follows:

A = 0.5 * |Σ (xiyi+1 - yixi+1)|

Here, (xi, yi) are the coordinates of each point of the polygon, and (xn+1, yn+1) = (x1, y1) to return to the first point.

Step 1: Designing the Algorithm Structure

First, let’s design the basic structure of the program. The program consists of the following four steps:

1. Receive input
2. Calculate area
3. Print result
4. Clean up

Step 2: Receiving Input

In Java, we can receive user input through the Scanner class. We will store the coordinates of each point using an ArrayList. Here is a sample code:

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

public class PolygonArea {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();  // Number of points
        List points = new ArrayList<>();
        
        for (int i = 0; i < n; i++) {
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            points.add(new Point(x, y));
        }
        scanner.close();
    }
    
    static class Point {
        int x, y;

        Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
}

Step 3: Calculating Area

Now it is time to add a method to calculate the area. The area calculation can be done by applying the above formula for each point using a for loop:

public static double calculateArea(List points) {
    double area = 0.0;
    int n = points.size();

    for (int i = 0; i < n; i++) {
        Point p1 = points.get(i);
        Point p2 = points.get((i + 1) % n); // Next point, the last point connects to the first point
        area += p1.x * p2.y - p2.x * p1.y;
    }
    return Math.abs(area) / 2.0;
}

Step 4: Printing the Result

Finally, let’s add the code to print the calculated area in the desired format. Add the following code to the main method:

double area = calculateArea(points);
System.out.printf("%.2f\n", area);

Complete Code

Now, if we organize all the code into one, it looks like this:

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

public class PolygonArea {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();  // Number of points
        List points = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            points.add(new Point(x, y));
        }
        scanner.close();
        
        double area = calculateArea(points);
        System.out.printf("%.2f\n", area);
    }

    static class Point {
        int x, y;

        Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public static double calculateArea(List points) {
        double area = 0.0;
        int n = points.size();

        for (int i = 0; i < n; i++) {
            Point p1 = points.get(i);
            Point p2 = points.get((i + 1) % n); // Next point, the last point connects to the first point
            area += p1.x * p2.y - p2.x * p1.y;
        }
        return Math.abs(area) / 2.0;
    }
}

Conclusion

In this lecture, we solved the problem of calculating the area of a polygon. The algorithm we covered is based on Schroder’s formula, which is frequently used in actual programming competitions. Writing code at each step and implementing the algorithm ourselves must have been a very beneficial experience.

Now, you can also tackle the problem of calculating the area for various types of polygons based on this algorithm. Furthermore, I encourage you to challenge yourself to solve more advanced problems by integrating other ideas and algorithms. In the next lecture, try different algorithm problems!

Understanding various algorithms and data structures is important in coding tests. One of them is Breadth-First Search (BFS). In this article, we will explain the basic concepts of BFS, provide example problems, and outline a step-by-step approach to solving those problems.

1. Overview of Breadth-First Search (BFS)

Breadth-First Search is an algorithm used to explore graph or tree structures, starting from the root node and visiting adjacent nodes in priority order. It is primarily implemented using a queue data structure. BFS is useful for finding the shortest path to a specific node and is suitable for problems involving shortest distances.

1.1 Characteristics of BFS

• Searches all nodes at the same depth
• Guarantees the shortest path
• May use a lot of memory

1.2 Time Complexity of BFS

The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. This is because each node and edge is visited once.

2. Example Problem

Problem Description

This problem involves calculating the shortest distances from a starting node in a given graph. The graph is provided in the form of an adjacency list, and you are required to write a function that returns the shortest distance to all nodes reachable from the starting node.

Input

• n (1 ≤ n ≤ 1000): Number of nodes in the graph
• edges: An array of edge lists, where each edge is of the form [a, b]
• start: The starting node (an integer from 1 to n)

Output

Returns an array containing the shortest distances from the starting node to each node. If a node is unreachable, denote it by -1.

2.1 Example

    Input: 
    n = 5
    edges = [[1, 2], [1, 3], [2, 4], [3, 5]]
    start = 1

    Output: 
    [0, 1, 1, 2, -1]
    

3. Problem Solving Process

To solve this problem, we will proceed through the following steps.

3.1 Convert Input Data to Graph Form

First, we will convert the edge list into an adjacency list representation of the graph. This allows us to easily find adjacent nodes for each node.

3.2 Implement BFS Algorithm

We will perform BFS starting from the starting node using a queue. The current node is added to the queue, allowing us to explore adjacent nodes. We will record the shortest distance for visited nodes in an array for later use.

3.2.1 Progress of BFS Exploration

1. Add the starting node to the queue and set the initial value (0) in the distance array.
2. Poll a node from the queue and check all adjacent nodes to that node.
3. If an adjacent node has not been visited, update the distance and add it to the queue.
4. Repeat this process until all nodes have been visited.

3.3 Generate Final Results

After exploring all nodes, we return the distance array to display the shortest paths. Nodes that are unreachable are indicated by -1.

4. Java Code Example

import java.util.*;

public class BFSDistance {
    public static int[] bfsDistance(int n, int[][] edges, int start) {
        List> graph = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            graph.add(new ArrayList<>());
        }
        
        // Create the graph adjacency list
        for (int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]); // Undirected graph
        }

        int[] distances = new int[n + 1];
        Arrays.fill(distances, -1); // Set initial distance to -1
        distances[start] = 0; // Distance to the starting point is 0
        
        Queue queue = new LinkedList<>();
        queue.add(start);
        
        while (!queue.isEmpty()) {
            int currentNode = queue.poll();
            for (int neighbor : graph.get(currentNode)) {
                if (distances[neighbor] == -1) { // If not visited
                    distances[neighbor] = distances[currentNode] + 1;
                    queue.add(neighbor);
                }
            }
        }
        
        return Arrays.copyOfRange(distances, 1, distances.length); // Return from 1 to n
    }

    public static void main(String[] args) {
        int n = 5;
        int[][] edges = {{1, 2}, {1, 3}, {2, 4}, {3, 5}};
        int start = 1;

        int[] result = bfsDistance(n, edges, start);
        System.out.println(Arrays.toString(result)); // [0, 1, 1, 2, -1]
    }
}

5. Conclusion

In this lecture, we demonstrated the concept and application of Breadth-First Search (BFS). BFS is a highly effective algorithm for finding the shortest path and a powerful tool for exploring graphs and trees. We encourage you to practice and apply this technique in coding tests. In the next lecture, we will cover Depth-First Search (DFS)!

6. Additional Practice Problems

We provide additional problems for practicing Breadth-First Search. Try solving the problems below.

Problem: Finding Shortest Path

Given an unweighted graph, determine the shortest paths from the starting node to all other nodes. Additionally, learn how to use Dijkstra’s algorithm to find the shortest path in weighted graphs.

Reference Materials

• BFS – Wikipedia
• BFS Implementation – GeeksforGeeks

Problem Description

Unlike typical sorting problems, this problem requires you to sort each digit of the given number in descending order and output the newly formed number.
It is assumed that the number input from the user is a non-negative integer, and each digit consists of numbers from 0 to 9. For example, if 321 is input,
it is sorted in descending order and 321 is output as is, and if 2143 is input, 4321 is output.

Input Format and Conditions

• The input is a single integer that includes each digit.
• The input value must be within the range of a 32-bit integer, which is greater than or equal to 0.
• If the input value is 0, 0 should be output.

Example Test Cases

Input	Output
2143	4321
321	321
1110	1110
0	0

Approach to the Problem

The approach to solving the problem is as follows.
1. **Input Value String Processing**: First, convert the input integer to a string to separate each digit.
2. **Digit Array Formation**: Store each digit of the string-converted number into an array.
3. **Sorting**: Sort each digit in the array in descending order.
4. **Output Format**: Concatenate the sorted digits into one string and output it.

Detailed Algorithm Process

Let’s take a closer look at the algorithm step by step.

Step 1: Input Processing

Convert the number input by the user to a string.
The value entered by the user can be retrieved through Java’s Scanner class.
For example:

Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();

Step 2: Array of Digits

To construct an array of each character in the input string, you can use the toCharArray() method of the String class.
This method converts each character that makes up the string into a character array.

char[] digits = input.toCharArray();

Step 3: Sorting

You can use the Arrays.sort method for sorting, but to sort in descending order,
a Comparator method is needed. The code is as follows:

Arrays.sort(digits, Collections.reverseOrder());

Step 4: Output Format

Convert the sorted character array back into a string and output the final result.
You can process it with the following code:

String result = new String(digits);
System.out.println(result);

Final Code Implementation

The final code that integrates all steps is as follows:

import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class DescendingOrder {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter a number: ");
        String input = scanner.nextLine();

        if (input.equals("0")) {
            System.out.println("0");
            return;
        }

        char[] digits = input.toCharArray();
        Arrays.sort(digits, Collections.reverseOrder());
        
        String result = new String(digits);
        System.out.println(result);
    }
}

Code Explanation and Additional Optimization

The above code meets the basic requirements. However, some parts of the code can be further optimized.
For example, **negative number handling** and clear **input validation** can be added. It may be necessary to check
whether the string received from the user is actually a number.

if (!input.matches("\\d+")) {
            System.out.println("Invalid input.");
            return;
        }

Testing and Validation

Once the final code is complete, it needs to be tested with various inputs to check if it works properly.
You should try values such as 0, negative numbers, and numbers of various lengths.
The program can be validated with test cases such as:

• Input: 98765 → Output: 98765
• Input: 10203 → Output: 32100
• Input: 9001 → Output: 9100

Conclusion

In this tutorial, we have explained in detail how to sort the digits of a given number in descending order using Java.
We hope you understand the process of solving the problem through a step-by-step approach with the final code.
In real practice, you may encounter more complex problems, so it is important to improve your problem-solving abilities
through regular practice. Thank you.