Java Coding Test Course, Queueing

In this lecture, we will cover useful algorithm problems for job preparation using Java. The topic is “Queue Formation,” and through this problem, you can enhance your basic understanding of data structures, sorting, and algorithms and develop skills that can be applied in actual coding tests.

Problem Description

This is a problem of lining up students based on their heights. We want to arrange students with different heights in a line. In other words, when there are N students and their height information is provided, please write a program that sorts them and outputs their heights in order.

Input

  • The first line contains the number of students N. (1 ≤ N ≤ 100,000)
  • The second line contains each student’s height, separated by spaces. (1 ≤ height ≤ 200)

Output

Output the students’ heights in ascending order.

Example Input

5
170 165 180 175 160
    

Example Output

160 165 170 175 180
    

Problem Solving Process

1. Understanding the Problem

To solve the problem, we first need to clearly understand the requirements. We need to sort the students’ heights provided as input and output them. Considering the possible large input size, it is necessary to choose an efficient sorting algorithm.

2. Algorithm Selection

The number of students can be up to 100,000, and each height value ranges from 1 to 200. In this case, we can use comparison-based sorting algorithms such as Quick Sort and Merge Sort, but given the limited range of heights, it is more efficient to use the Counting Sort algorithm.

3. Explanation of Counting Sort Algorithm

Counting Sort can perform sorting with a time complexity of O(N + k), where k is the range of values of the data being sorted. In this problem, since each student’s height is an integer between 1 and 200, k becomes 200. The Counting Sort process is as follows:

  1. Create an array to store the number of times each height appears.
  2. Read each input height one by one and increase the corresponding index.
  3. Output the sorted values using the count array.

4. Java Code Implementation

Below is the Java code based on the above algorithm:

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        // Input number of students N
        int N = scanner.nextInt();
        // Declare array to hold the number of heights (1-200)
        int[] heightCount = new int[201];

        // Input heights of students and count them
        for (int i = 0; i < N; i++) {
            int height = scanner.nextInt();
            heightCount[height]++;
        }

        // Output sorted heights
        StringBuilder result = new StringBuilder();
        for (int i = 1; i <= 200; i++) {
            while (heightCount[i] > 0) {
                result.append(i).append(" ");
                heightCount[i]--;
            }
        }
        System.out.println(result.toString().trim());

        scanner.close();
    }
}
    

5. Code Explanation

In the code above, we create a count switch array based on the number of students, counting each student’s height by index. Then, we traverse the count array to output the indices where values exist in the result string.

6. Time Complexity

The time complexity of this algorithm is O(N + k). N is the number of students, and k is the range of heights. Therefore, this algorithm is very efficient and suitable for meeting the constraints of the problem.

Conclusion

In this lecture, we learned how to solve the “Queue Formation” problem using the Counting Sort algorithm. This technique is particularly useful in situations where the range of values is limited or easily predictable. Since you are likely to encounter similar problems in actual coding tests, try to further develop your own algorithm skills!

Java Coding Test Course, Jumong’s Command

Hello, everyone! Today, I would like to talk about preparing for coding tests using Java. In this lecture, we will solve an algorithm problem with the theme “Jumong’s Command,” which will help enhance our skills in handling arrays and strings, as well as basic algorithm design.

Problem Description

Jumong wants to give commands to 10 warriors and record their performances. Each warrior can deliver different performances each day, and these performances are given in the form of an array. However, Jumong needs to effectively compare and evaluate the warriors’ performances, so he wants to know the best performance among them. Write a program that finds the highest score in the given performance array and records that score.

Problem Specification

Input: An array scores consisting of integers is given. The length of this array is between 1 and 100,000. Each integer is between 0 and 10,000.

Output: Print the highest score in the scores array.

Example

Input: [85, 92, 75, 88, 95, 100, 60]
Output: 100

Problem Solving Process

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

Step 1: Problem Analysis

To find the highest score, we can iterate through the scores array, comparing each score. At this time, we should consider various approaches based on the length of the array.

Step 2: Algorithm Design

We will use the linear search method. This method checks each element once to find the maximum value. The time complexity of this algorithm is O(n), which increases with the size of the input array.

Step 3: Implementing Java Code

Now, let’s implement the algorithm in Java code. Below is an example of how to write the code to solve this problem.

public class JumongCommand {
    public static int findMaxScore(int[] scores) {
        int maxScore = scores[0]; // Initialize the first score as the max score
        for (int i = 1; i < scores.length; i++) {
            if (scores[i] > maxScore) {
                maxScore = scores[i]; // Update if the current score is greater than the max score
            }
        }
        return maxScore; // Return the final maximum score
    }

    public static void main(String[] args) {
        int[] scores = {85, 92, 75, 88, 95, 100, 60};
        int maxScore = findMaxScore(scores);
        System.out.println("The highest score is: " + maxScore); // Print the maximum score
    }
}

Step 4: Testing and Validation

After writing the code, it is necessary to conduct tests with various input values. Below are some test cases.

public static void main(String[] args) {
    // Test case 1
    int[] scores1 = {85, 92, 75, 88, 95, 100, 60};
    System.out.println(findMaxScore(scores1)); // Output: 100

    // Test case 2
    int[] scores2 = {50, 55, 60, 45, 40};
    System.out.println(findMaxScore(scores2)); // Output: 60

    // Test case 3
    int[] scores3 = {100, 100, 100, 100, 100};
    System.out.println(findMaxScore(scores3)); // Output: 100

    // Test case 4 (single value)
    int[] scores4 = {42};
    System.out.println(findMaxScore(scores4)); // Output: 42

    // Test case 5 (empty case)
    int[] scores5 = {};
    try {
        System.out.println(findMaxScore(scores5)); // Exception occurs
    } catch (Exception e) {
        System.out.println("There are no scores."); // Error handling
    }
}

Step 5: Time Complexity Analysis

The time complexity of this algorithm is O(n). This is because it checks each element in the array once, resulting in a time expenditure proportional to the size n of the input array. The space complexity is O(1), as it does not use any additional data structures, thereby maintaining fixed space.

Conclusion

In this lecture, we implemented an algorithm to find the maximum value in an array using Java. I hope this problem has helped you build a foundation in array processing and algorithm design. Next time, we will tackle more complex problems. Thank you!

Java Coding Test Course, Pebble Picking

Written on: October 4, 2023

Problem Description

You are assigned the task of collecting pebbles on the beach. You must extract pebbles according to specific rules.
The pebbles come in various colors, and there is a list of N pebbles given at the location.
Each pebble in this list has a unique color.
You can perform two operations repeatedly:
1. Take out a pebble
2. Put a pebble back.

What is the minimum number of operations (taking out or putting back) required to retrieve all the pebbles?
If it is impossible to retrieve all the pebbles, output ‘Impossible’.

Input Format

  • The first line contains the number of pebbles N (1 ≤ N ≤ 100).
  • The second line contains the colors of the N pebbles. The colors are given as integers from 1 to 100.

Output Format

Output the minimum number of operations required to retrieve all the pebbles. If it is impossible, print ‘Impossible’.

Problem Analysis

This problem involves managing the color palette of the pebbles and processing with minimal operations.
Each pebble can be grouped by color, and this grouping plays an important role in retrieving the desired pebbles.
It is necessary to have an approach that strategically combines taking out and putting back pebbles based on the rules.

Approach

To effectively solve the problem, we can approach it with the following steps.

  1. Count the frequency of pebble colors: Count how many pebbles there are of each color.
  2. Calculate the minimum number of operations: If the number of pebbles is odd, the operation of taking out should always take precedence, and final remaining operations should be considered additionally.
  3. Validation: Check if it is possible to retrieve all the pebbles.

Sample Code

            
            import java.util.HashMap;
            import java.util.Scanner;

            public class PebbleRemoval {
                public static void main(String[] args) {
                    Scanner scanner = new Scanner(System.in);
                    int N = scanner.nextInt();
                    HashMap colorFrequency = new HashMap<>();
                    
                    for (int i = 0; i < N; i++) {
                        int color = scanner.nextInt();
                        colorFrequency.put(color, colorFrequency.getOrDefault(color, 0) + 1);
                    }
                    
                    int operations = 0;
                    int oddCount = 0;

                    for (int freq : colorFrequency.values()) {
                        if (freq % 2 != 0) {
                            oddCount++;
                        }
                        operations += freq;
                    }
                    
                    if (oddCount > 1) {
                        System.out.println("Impossible");
                    } else {
                        System.out.println(operations);
                    }

                    scanner.close();
                }
            }
            
        

Code Explanation

The above Java code demonstrates a basic structure for counting the frequency of pebble colors.

  • It uses a HashMap to count the frequency of each color.
  • It checks all frequencies to see if they are odd and tracks the take-out operations.
  • Finally, it prints ‘Impossible’ if take-out operations are not feasible.

Author: AI Algorithm Tutor

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Java Coding Test Course, Understanding Combinations

1. Introduction

Programming algorithms and data structures play a very important role in modern software development.
One of the major parts of coding tests is combination problems.
Understanding this provides the foundation needed to algorithmically solve problems.
In this article, we will explain the concept of combinations, how to solve this problem using Java, and detail the process of solving example problems.

2. What are Combinations?

A combination refers to the number of ways to choose R objects from N objects.
Generally, a combination represents a set of selected objects without regard to the order.
For example, when there are three objects {A, B, C}, the combinations of selecting two out of them are as follows:

  • {A, B}
  • {A, C}
  • {B, C}

The number of combinations is calculated using the following mathematical formula:

C(N, R) = N! / (R! * (N – R)!)

In this formula, N is the total number of objects, R is the number of objects to be selected, and “!” denotes factorial.

3. Problem Definition

The problem we will solve in this article is as follows:

Given N integers, output all combinations that can be made by selecting K of them.
The order of input does not matter.

4. Problem-Solving Approach

This problem can be solved using recursion.
It is a method of generating combinations by dividing into cases of selection and non-selection.
Here is the basic mechanism of the algorithm to solve this problem:

  1. Declare an array for generating combinations.
  2. Define a recursive function to create combinations.
  3. Set base conditions to terminate the recursive calls.
  4. Once a combination is complete, print the result.

This approach will repeatedly perform selections and non-selections for each number to generate all combinations.

5. Java Code Implementation


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

public class Combination {
    public static void main(String[] args) {
        int[] numbers = {1, 2, 3, 4, 5}; // Array of numbers to generate combinations
        int K = 3; // Number to select
        List> result = new ArrayList<>(); // List to store results
        generateCombinations(numbers, K, 0, new ArrayList(), result);
        
        // Print results
        for (List combination : result) {
            System.out.println(combination);
        }
    }

    // Recursive function to generate combinations
    private static void generateCombinations(int[] numbers, int K, int start, List combination, List> result) {
        // Base condition: when the number of selected items reaches K, add to results
        if (combination.size() == K) {
            result.add(new ArrayList<>(combination));
            return;
        }

        // Generate combinations
        for (int i = start; i < numbers.length; i++) {
            combination.add(numbers[i]); // Select
            generateCombinations(numbers, K, i + 1, combination, result); // Recursive call
            combination.remove(combination.size() - 1); // Deselect
        }
    }
}

        

The above code demonstrates the method of creating combinations and printing the results through recursion.
The `generateCombinations` function generates combinations starting from the index of the current number.
At this time, the selected number is added to the `combination` list, and once K numbers are selected, it is added to the results.

6. Time Complexity Analysis

The time complexity of the combination problem is O(C(N, K)).
Here, C(N, K) represents the number of combinations of selecting K from N objects.
The depth of the recursive calls is K, which is related to the number of calls made to select K items.
Therefore, as N increases, the number of combinations increases sharply.
Practically, the time taken to calculate combinations can become very large, especially when K is close to half of N.

7. Conclusion

Combination problems are frequently posed in algorithm questions and have various variations.
In this column, we learned about the concept of combinations and how to implement it using Java.
By understanding the structure of combinations algorithmically, it becomes easier to solve more complex problems.
I wish you good results in coding tests through practice with combination problems!

Java Coding Test Course, Finding Non-Square Numbers

Coding tests are an essential part of modern software development. Especially in object-oriented languages like Java, it is crucial to develop algorithm problem-solving skills. In this course, we will address the algorithm problem titled ‘Finding Non-Perfect Squares’. This problem involves finding the remaining numbers in an integer array excluding the perfect squares.

Problem Description

This problem requires identifying non-perfect square numbers from a given integer array. A perfect square is a number that results from squaring an integer, such as 0, 1, 4, 9, 16, ....

Input

  • An integer N is given (1 ≤ N ≤ 10000)
  • An array containing N integers is provided.

Output

  • A new array composed of integers from the original array that are not perfect squares is returned.

Problem-Solving Strategy

To solve this problem, we will follow these steps:

  1. Iterate through the array to check if each number is a perfect square.
  2. If it is not a perfect square, add it to a new array.
  3. Return the result array.

How to Check for Perfect Square

To determine if a number x is a perfect square, you can first calculate the square root of x and then convert that square root into an integer and square it again. If it matches x, then x is a perfect square.

boolean isPerfectSquare(int x) {
        if (x < 0) return false;
        int s = (int) Math.sqrt(x);
        return s * s == x;
}

Java Implementation

Now we will implement this algorithm in Java.

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

public class FindNonPerfectSquares {
    
    public static List<Integer> findNonPerfectSquares(int[] arr) {
        List<Integer> result = new ArrayList<>();

        for (int num : arr) {
            if (!isPerfectSquare(num)) {
                result.add(num);
            }
        }
        
        return result;
    }
    
    private static boolean isPerfectSquare(int x) {
        if (x < 0) return false;
        int s = (int) Math.sqrt(x);
        return s * s == x;
    }

    public static void main(String[] args) {
        int[] inputArray = {1, 2, 3, 4, 5, 9, 10, 15, 16, 20};
        List<Integer> nonPerfectSquares = findNonPerfectSquares(inputArray);
        System.out.println("Non-perfect squares: " + nonPerfectSquares);
    }
}

Code Explanation

The above code iterates through the provided array using a method called findNonPerfectSquares. For each number, it calls the isPerfectSquare method to skip perfect squares and adds only non-perfect squares to the result list.

Testing

Let’s run the code in practice. The main method defines a test input array, calls the function, and outputs the list of non-perfect squares.

Result Analysis

From the input array {1, 2, 3, 4, 5, 9, 10, 15, 16, 20}, the non-perfect squares are {2, 3, 5, 10, 15, 20}. This is an important step to verify the accuracy of the algorithm.

Complexity Analysis

The time complexity of this algorithm is O(N), as it traverses the array just once. The space complexity is also O(N) in the worst case due to the space required to store the result list.

Conclusion

In this lecture, we tackled the algorithm problem of finding non-perfect squares. Through this problem, we learned a basic method for solving algorithm problems by iterating through an array and adding values based on conditions. It is important to consistently practice solving such problems in preparation for various algorithm challenges that frequently appear.

In the next lecture, we will address more complex algorithm problems. Thank you!