Category: Java Coding Test

Hello! In today’s lecture, we will solve an algorithm problem called “Sum of Remainders” using Java. This problem involves finding the remainders when a given integer array is divided by a specified number and then summing those remainders. This problem covers important basic concepts for preparing for coding tests.

Problem Description

Given an integer array arr and an integer m, write a function that calculates the remainder of each element in the array when divided by m and returns the total sum of those remainders.

Input

• The first line contains the size of the array n (1 ≤ n ≤ 100,000).
• The second line contains n integers arr[i] (1 ≤ arr[i] ≤ 1,000,000) separated by spaces.
• The third line contains the integer m (1 ≤ m ≤ 1,000,000).

Output

Print the sum of the remainders after dividing the elements of the array by m.

Example

Input:
5
1 2 3 4 5
3

Output:
15

Problem-Solving Strategy

To solve this problem, follow these steps:

1. After receiving the integer array and integer m, calculate the remainders for all elements in the array using m.
2. Add all those remainder values together.
3. Print the final result.

Implementation

Now, let’s implement the above strategy in Java. First, we will write a method to input the array and calculate the remainder for each element when divided by m. Below is the Java code that solves this problem:

import java.util.Scanner;

public class RemainderSum {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        // 1. Input
        int n = scanner.nextInt(); // size of the array
        int[] arr = new int[n];
        
        // Input array elements
        for (int i = 0; i < n; i++) {
            arr[i] = scanner.nextInt();
        }
        
        int m = scanner.nextInt(); // divisor

        // 2. Calculate sum of remainders
        long remainderSum = 0; // variable to store the result
        for (int i = 0; i < n; i++) {
            remainderSum += arr[i] % m; // add remainders
        }

        // 3. Print result
        System.out.println(remainderSum);
        
        scanner.close();
    }
}

Code Explanation

1. Input Handling: The Scanner class is used to handle input. First, it reads the size of the array n, then stores each element in arr. Finally, it reads the integer m that we want to divide by.
2. Calculating the Sum of Remainders: After initializing the remainderSum variable, a loop is used to add the remainders of each element in the array divided by m. The long type is used to safely store the sum of large numbers.
3. Output Result: Finally, the sum of the remainders is printed.

Complexity Analysis

The time complexity of this programming problem is O(n). Due to the length of the array, we need to iterate through all elements in the worst case. The space complexity is O(1) as there is almost no additional space required.

Additional Example Tests

It is also good to consider a few additional test cases after implementing:

Example 1

Input:
3
5 10 15
4

Output:
6

Example 2

Input:
4
1 100 10000 1000000
1

Output:
0

Example 3

Input:
5
0 10 20 30 40
7

Output:
10

Conclusion

Today, we solved the "Sum of Remainders" problem using Java. Through this example, we learned how to effectively utilize array processing, loops, and remainder calculations. These types of problems can be transformed in various ways, helping to build adaptability. We hope to improve our skills by solving many different algorithm problems in the future!

Thank you!

Many problems in coding tests are based on graph or tree structures. One of the algorithms that is useful in solving these problems is Depth-First Search (DFS). In this tutorial, we will learn about the concept of DFS and how to implement it in Java.

1. What is Depth-First Search (DFS)?

Depth-First Search is a type of graph traversal algorithm that explores nodes as deeply as possible along each branch before backtracking. In other words, it keeps moving down one branch until no further progress can be made, at which point it backtracks to the previous node and explores other branches.

2. Characteristics of DFS

• Uses a stack to keep track of visited nodes.
• Can be implemented through recursion.
• Unlike Breadth-First Search (BFS), it primarily considers the depth of nodes or the path.

3. Problem Definition

Let us assume there is a graph as follows:

        A
       / \
      B   C
     / \   \
    D   E   F
    

We aim to find the path from A to D using DFS. The goal is to find the path and print the nodes visited from A to D.

4. Problem-Solving Process

To solve the problem using DFS, we first need to represent the graph and then implement the DFS algorithm.

4.1 Graph Representation

Graphs can usually be represented by an adjacency list or an adjacency matrix. Here, we will represent the graph using an adjacency list.

Implementing the Graph in Java

    import java.util.*;

    class Graph {
        private Map> adjacencyList;

        public Graph() {
            adjacencyList = new HashMap<>();
        }

        public void addEdge(String source, String destination) {
            adjacencyList.putIfAbsent(source, new ArrayList<>());
            adjacencyList.get(source).add(destination);
        }
        
        public Map> getAdjacencyList() {
            return adjacencyList;
        }
    }
    

4.2 Implementing the DFS Algorithm

The DFS algorithm can be implemented as follows. A Set is used to keep track of visited nodes, and depth exploration is performed through recursive calls.

Java DFS Method

    class DFS {
        private Set visited;
        private List path;

        public DFS() {
            visited = new HashSet<>();
            path = new ArrayList<>();
        }

        public void depthFirstSearch(Graph graph, String node) {
            if (!visited.contains(node)) {
                visited.add(node);
                path.add(node);
                
                for (String neighbor : graph.getAdjacencyList().get(node)) {
                    depthFirstSearch(graph, neighbor);
                }
            }
        }

        public List getPath() {
            return path;
        }
    }
    

5. Complete Code

Now, let’s combine the complete code for use. Below is the complete code that creates the graph and executes DFS.

    public class Main {
        public static void main(String[] args) {
            Graph graph = new Graph();
            graph.addEdge("A", "B");
            graph.addEdge("A", "C");
            graph.addEdge("B", "D");
            graph.addEdge("B", "E");
            graph.addEdge("C", "F");

            DFS dfs = new DFS();
            dfs.depthFirstSearch(graph, "A");

            System.out.println("DFS Path: " + dfs.getPath());
        }
    }
    

6. Code Execution Result

When you run the above code, you can obtain the following result:

    DFS Path: [A, B, D, E, C, F]
    

As can be seen from the result, DFS starts from A, visits B, and then goes deep to D. The order of exploration in DFS is maintained.

7. Advanced Learning

This tutorial introduced the concept of DFS and basic implementation methods. It is advisable to solve various problems using DFS to build your skills. Typical problems include:

• Maze solving
• Finding connected components
• Cycle detection

8. Conclusion

Through this tutorial, we learned the basic concept of Depth-First Search and how to implement it in Java. Graph problems frequently appear in coding tests, so it is essential to memorize and practice them. In the next tutorial, we will discuss another search algorithm, Breadth-First Search (BFS).

Hello! Today, we will solve algorithm problems related to geometry in the Java coding test course. Geometric problems are often covered in algorithm exams and help in understanding basic shapes like plane geometry, triangles, circles, and polygons. These problems usually provide an opportunity to develop mathematical thinking skills based on the theoretical properties of shapes.

Problem Statement

Let’s solve the following geometric problem:

Problem: Given two points P1(1, 3) and P2(4, 6) on a plane, write a program to calculate the length of the line segment connecting these two points.

The length of the line segment is defined as the distance between the two points, and the distance between two points P1(x1, y1) and P2(x2, y2) can be calculated using the following formula:

distance = √((x2 - x1)² + (y2 - y1)²)

Problem Solving Process

1. Problem Analysis

Analyzing the problem, the coordinates of the two points P1 and P2 are fixed, and the goal is to find the distance between these two points. The mathematical concept that can be used here is the Pythagorean theorem. We can calculate the Euclidean distance using the coordinates of the given two points.

2. Mathematical Judgment

We are given two points P1(1, 3) and P2(4, 6). Based on this, we can calculate the differences in the x-coordinate and y-coordinate to apply the distance formula.

– Difference in x-coordinates: (x2 - x1) = (4 - 1) = 3
– Difference in y-coordinates: (y2 - y1) = (6 - 3) = 3

3. Distance Calculation

Substituting the calculated differences into the distance formula:

distance = √(3² + 3²) = √(9 + 9) = √18 = 3√2 ≈ 4.24

4. Writing Java Code

Now let’s implement this algorithm in Java code. The flow of the algorithm is as follows:

public class DistanceCalculator {
    public static void main(String[] args) {
        double x1 = 1, y1 = 3; // Coordinates of the first point P1
        double x2 = 4, y2 = 6; // Coordinates of the second point P2
        
        double distance = calculateDistance(x1, y1, x2, y2);
        System.out.println("Distance between points P1 and P2: " + distance);
    }

    // Method to calculate the distance between two points
    public static double calculateDistance(double x1, double y1, double x2, double y2) {
        return Math.sqrt(Math.pow((x2 - x1), 2) + Math.pow((y2 - y1), 2));
    }
}

5. Output Result

When the above code is run, the program calculates and outputs the distance between the two points P1 and P2. The result is approximately 4.24.

Conclusion

Through this lecture, we have understood how to calculate the distance between points in plane geometry. The problem-solving process proceeded as follows: problem analysis, mathematical judgment, distance calculation, writing Java code, and outputting the result. These geometric problems can be applied to various algorithm problems, and it is important to solidify the basic concepts.

We look forward to covering more geometric problems and algorithms in the future! Thank you.

Hello! Today we will learn about Radix Sort. Radix Sort is an algorithm that sorts integers or strings based on each digit, and works very efficiently under certain conditions. In this article, we will explain the concept of Radix Sort, how it operates, and how to implement it in Java. I hope that through Radix Sort, you can enhance your algorithmic problem-solving abilities.

1. Concept of Radix Sort

Radix Sort is not a typical comparison-based sorting algorithm, but rather a non-comparative sorting method that uses the digits of the keys for sorting. Generally, Radix Sort goes through the following main steps:

• Separate all numbers by digit
• Start sorting from the least significant bit (LSB) on the far right
• Fill the positions to ensure strict ordering
• Repeat for all digits until final sorting is complete

The time complexity of Radix Sort is O(n*k), where n is the number of data points and k is the number of digits. Therefore, Radix Sort is efficient when the number of digits is low, or when the dataset is not very large.

2. How Radix Sort Works

To understand how Radix Sort operates, let’s explain with a simple example. Let’s assume we are sorting the following array using Radix Sort:

[170, 45, 75, 90, 802, 24, 2, 66]

Radix Sort proceeds through the following steps:

Step 1: Sort by LSB (Least Significant Bit)

Sort the array based on the lowest digit (1’s place):

[170, 90, 802, 24, 2, 45, 75, 66]

This will place smaller numbers in the 1’s place first.

Step 2: Sort by the 10’s place

This time, sort based on the 10’s place:

[170, 802, 24, 2, 45, 75, 90, 66]

This process continues sequentially for each digit.

Step 3: Sort by the 100’s place

Finally, sorting by the 100’s place gives:

[2, 24, 45, 66, 75, 90, 170, 802]

Now we can see that the array is completely sorted.

3. Implementing Radix Sort in Java

Now let’s implement Radix Sort in Java. You can use the following code to perform Radix Sort:

    public class RadixSort {
        // Method to find and return the maximum value in the given array
        static int getMax(int[] arr, int n) {
            int max = arr[0];
            for (int i = 1; i < n; i++) {
                if (arr[i] > max) {
                    max = arr[i];
                }
            }
            return max;
        }

        // Method to perform counting sort on a specific digit
        static void countingSort(int[] arr, int n, int exp) {
            int[] output = new int[n]; // Array to hold the sorted result
            int[] count = new int[10]; // Count for digits 0-9

            for (int i = 0; i < n; i++) {
                count[(arr[i] / exp) % 10]++;
            }

            for (int i = 1; i < 10; i++) {
                count[i] += count[i - 1];
            }

            for (int i = n - 1; i >= 0; i--) {
                output[count[(arr[i] / exp) % 10] - 1] = arr[i];
                count[(arr[i] / exp) % 10]--;
            }

            for (int i = 0; i < n; i++) {
                arr[i] = output[i];
            }
        }

        // Radix sort method
        static void radixSort(int[] arr) {
            int n = arr.length;
            int max = getMax(arr, n);

            for (int exp = 1; max / exp > 0; exp *= 10) {
                countingSort(arr, n, exp);
            }
        }

        // Main method
        public static void main(String[] args) {
            int[] arr = {170, 45, 75, 90, 802, 24, 2, 66};
            radixSort(arr);

            System.out.println("Sorted array: ");
            for (int num : arr) {
                System.out.print(num + " ");
            }
        }
    }
    

This code sorts each digit of the numbers and ultimately outputs an array sorted in ascending order. The methods getMax, countingSort, and radixSort implement the roles of each step, making the working principle of Radix Sort easy to understand.

4. Advantages and Disadvantages of Radix Sort

Advantages

• With a time complexity of O(n * k), it is very efficient for sorting regular data.
• Its predictable access pattern makes it advantageous in systems like databases.

Disadvantages

• Additional memory usage may make it unsuitable for large datasets.
• It is difficult to apply to floating-point numbers.

5. Radix Sort Application Problem

Now, let’s solve a simple problem where we can utilize Radix Sort.

Problem Description

Given an array of integers, use Radix Sort to sort the array in ascending order. The maximum length of the array is 1000, and every element is an integer between 0 and 10000.

Constraints

• You must use Radix Sort and cannot use other sorting algorithms.
• You should solve it without using unnecessary variables.

Solution Process

The problem can be solved by directly applying the Radix Sort algorithm as described above.

Example Input

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

Example Output

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

Code Execution

    public class RadixSortExample {
        public static void main(String[] args) {
            int[] arr = {3, 6, 1, 8, 4, 7, 9, 2, 5};
            RadixSort.radixSort(arr);
            System.out.println("Sorted array: ");
            for (int num : arr) {
                System.out.print(num + " ");
            }
        }
    }
    

6. Conclusion

In this post, we covered the concept of Radix Sort, how it operates, its Java implementation, and problem-solving processes. Radix Sort is a useful algorithm that can be applied in various ways depending on submission and criteria. I hope you frequently utilize Radix Sort in algorithmic problem solving, and I will return with more valuable content next time. Thank you!

Java Coding Test Course – Greedy Algorithm

The Greedy Algorithm is a method of solving optimization problems by making the most optimal choice step by step. While the optimal choice at each step does not guarantee the optimal solution for the entire problem, it is often useful in many cases. In this article, we will solve a problem using the Greedy Algorithm.

Problem: Coin Change

Problem Description: Write a program to find the minimum number of coins needed to make a given amount. The types of coins are [500 won, 100 won, 50 won, 10 won].
For example, find the minimum number of coins needed to make 770 won.

Input

• Integer N (0 < N ≤ 10,000,000): Amount to be made

Output

• Integer K: Minimum number of coins needed

Input Example

770

Output Example

6

Problem Approach

To solve the problem, we apply the Greedy Algorithm by considering the types of coins given. The essence of the Greedy Algorithm is to use the coin with the highest value at each step. When the types of coins are sorted, we can continually reduce the remaining amount by using the highest coin first.

Code Implementation

Now, let’s implement the code to solve the problem using Java. Below is the Java code that solves the Coin Change problem.

import java.util.Scanner;

public class CoinChange {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        // Input the given amount
        int N = scanner.nextInt();
        
        // Types of coins to be used
        int[] coins = {500, 100, 50, 10};
        
        int count = 0; // Variable to count the number of coins
        
        for (int coin : coins) {
            count += N / coin; // Add the number of coins that can be made with the current coin
            N %= coin;         // Update the remaining amount
        }
        
        System.out.println(count); // Output the final number of coins
        scanner.close();
    }
}

Code Explanation

The code above finds the optimal solution through the following process:

1. User Input: The Scanner class is used to receive the amount that the user wants to create.
2. Pattern Setting: Define the array of coins to be used. The values of the coins are sorted in descending order.
3. Coin Count Calculation: Iterate over each coin starting from the largest. Calculate how many can be made with the current coin and update the remaining amount.
4. Result Output: Finally, the number of coins needed is outputted.

Time Complexity

The time complexity of this problem is O(1). The types of coins are fixed, and the result can be calculated in constant time regardless of the input amount.

Conclusion

The Greedy Algorithm is a method of solving problems by making the most optimal choice at each step. Through the Coin Change problem, we have learned the basic principles of the Greedy Algorithm and how to implement it in Java. This algorithm can be applied to various optimization problems and helps develop algorithmic thinking.

Further Reading

To gain a deeper understanding of the Greedy Algorithm, it is recommended to try other types of problems. For example, you can apply the Greedy Algorithm in problems such as Minimum Spanning Tree (MST) or Activity Selection Problems.

Appendix

The Greedy Algorithm can be effectively used to solve various problems, and eventually, you will challenge yourself with more complex optimization problems. In actual coding tests, it is important to read the problem carefully and determine whether the Greedy Algorithm is suitable. It is also beneficial to develop mathematical thinking and an understanding of data structures needed to solve problems together.