Java Coding Test Course, Creating a Blu-ray

Coding tests are becoming an increasingly important element in the field of software engineering. In this article, we will discuss how to acquire the skills necessary to solve complex algorithm problems using Java, focusing on the problem titled ‘Making Blu-rays.’

Problem Description

Requirements: Based on the given movie list and the playback times of each movie, find a way to use the minimum number of Blu-rays to play all movies, considering the capacity of a Blu-ray. The maximum capacity that can be held in one Blu-ray is specified.

Input:

  • maxSize: The maximum capacity of each Blu-ray (integer)
  • movies: A list of movie playback times (integer array)

Output:

  • The minimum number of Blu-rays required to play all movies (integer)

Example


maxSize: 10
movies: [1, 2, 3, 4, 5, 6]
Output: 3

Problem Approach

To solve this problem, we can approach it with the following steps:

  1. Consider adding as many movies as possible without exceeding the capacity of one Blu-ray.
  2. Sort the list of movies so that shorter movies are played first.
  3. Calculate the playback time of each Blu-ray and use a new Blu-ray if it exceeds the maximum capacity.
  4. Count the number of Blu-rays needed.

Java Code Implementation


import java.util.Arrays;

public class BluRayMaker {
    
    public static int minBluRays(int maxSize, int[] movies) {
        Arrays.sort(movies); // Sort the movies.
        int bluRayCount = 0;
        int currentBluRaySize = 0;

        for (int i = movies.length - 1; i >= 0; i--) {
            // Add the movie to the current Blu-ray.
            if (currentBluRaySize + movies[i] <= maxSize) {
                currentBluRaySize += movies[i];
            } else {
                // Use a new Blu-ray if the capacity is exceeded.
                bluRayCount++;
                currentBluRaySize = movies[i]; // Start the Blu-ray with the current movie.
            }
        }

        // Add to the count if there is a remaining Blu-ray.
        if (currentBluRaySize > 0) {
            bluRayCount++;
        }

        return bluRayCount;
    }

    public static void main(String[] args) {
        int maxSize = 10;
        int[] movies = {1, 2, 3, 4, 5, 6};
        System.out.println("Minimum number of Blu-rays needed: " + minBluRays(maxSize, movies));
    }
}

Code Explanation

The code above defines a class for creating Blu-rays and implements an optimal movie selection method. The code works as follows:

  1. Sort the movie list and process it from the longest movie onward.
  2. Add movies to the current Blu-ray without exceeding the maximum capacity.
  3. If the capacity is exceeded, finish the current Blu-ray and start a new one.
  4. After processing all movies, count the last Blu-ray if it exists.

Analysis and Complexity

The time complexity of this problem is O(N log N). This is due to the time required to sort the given movie list. An additional O(N) time is spent traversing the movie list once. The space complexity is O(1), as no additional data structures are required.

Conclusion

Learning how to approach and solve algorithm problems like this is very useful for preparing for coding tests. As these types of problems can frequently appear in practice, it is important to try solving a variety of challenges. “Making Blu-rays” is a good problem to practice, as it requires both a grasp of basic Java syntax and algorithm design skills.

In the next lesson, we will tackle more complex algorithm problems and provide answers to various questions you may encounter in real coding tests. Thank you!