Category: Python Coding Test

Problem Description

The company is trying to develop a new Blu-ray disc production system. Each Blu-ray disc has a specific size, and it must be optimized to store the maximum amount of data. Based on the given capacity of the Blu-ray and the sizes of each file, write a program that can fit as many files as possible onto the Blu-ray.

Problem Definition: Given N files, each with a positive integer size, and a given capacity C of the Blu-ray, write a program to calculate the maximum number of files that can fit without exceeding the Blu-ray’s capacity.

Input Format:
The first line contains the capacity C of the Blu-ray (1 ≤ C ≤ 10000) and the number of files N (1 ≤ N ≤ 100).
The second line contains the sizes of N files (1 ≤ file size ≤ 1000).

Output Format:
Print the maximum number of files that can fit on the Blu-ray.

Problem Analysis

This problem involves finding the maximum number of files that can be combined without exceeding the capacity C of the Blu-ray. To maximize the number of files that can fit on the Blu-ray, the files must be sorted appropriately in advance, and a suitable search method must be used to solve the problem.

The basic strategy to solve the problem is as follows:

• Sort the file sizes in ascending order.
• Sequentially add the sorted files until the total size exceeds the capacity C of the Blu-ray.
• If it exceeds the capacity, stop adding files and return the count of files added so far.

Problem Solving Process

Let’s look at the process of solving the problem step by step.

Step 1: Input

First, we get the capacity of the Blu-ray, the number of files, and the sizes of the files. Input takes place via standard input. We can use Python’s input() function to receive the data.

Step 2: Data Sorting

Sort the input file sizes in ascending order. This is necessary to ensure we add the smallest files first. Python’s sorted() function can easily sort the data.

Step 3: Add Files and Sum Sizes

While iterating through the sorted file list, check if adding the current file would exceed the total capacity C of the Blu-ray. If it does not exceed, add the current file to the Blu-ray and count the number of files.

Step 4: Output Result

After iterating through all the files, output the final count of files that can fit on the Blu-ray.

Step 5: Complete Code

def maximum_files_in_blu_ray(capacity, files):
    # Sort file sizes
    files.sort()
    count = 0
    total_size = 0

    for file in files:
        if total_size + file <= capacity:
            total_size += file
            count += 1
        else:
            break

    return count

# Get input
capacity, n = map(int, input().split())
files = list(map(int, input().split()))

# Call function and output result
result = maximum_files_in_blu_ray(capacity, files)
print(result)

            

Example Input and Output

Example 1

Input:

10 5
1 2 3 4 5
            

Output:

4
            

Example 2

Input:

7 4
1 2 3 4
            

Output:

3
            

Result Analysis

By using the above code, we can fit the optimal number of files according to the size of the given Blu-ray. This problem serves as a fundamental example of a greedy algorithm, presenting one method that satisfies the problem's conditions.

Conclusion

In this lecture, we have learned about basic list manipulation, sorting, and searching in Python through the Blu-ray making problem. These fundamental problem-solving techniques can be very useful in actual coding tests or algorithm problems. Keep improving your skills through a variety of challenges.

Coding tests are one of the essential skills required in the recent IT industry. It is necessary to deeply understand the core of the problem, rather than mechanically solving it, and to utilize the correct algorithms and data structures. In this course, we will select an algorithmic problem with the theme of helping the underprivileged and explain the process of solving that problem in detail.

Problem Description

The Helping the Underprivileged program simulates the process of gathering donations from donors to meet the amount needed by welfare organizations. Donors can contribute different amounts, and when the total amount of donations reaches a specific amount, the program must output the total amount of donations and the number of donors.

Problem Definition

Implement a program that satisfies the following conditions.

• The number of donors is N. (1 ≤ N ≤ 100)
• Each donor can contribute an amount of 1,000 or more.
• Set a target amount M. (M is a natural number of 1,000 or more)

Input

The first line contains the number of donors N and the target amount M, separated by a space.

The second line contains the amounts each donor will contribute, separated by spaces.

Output

Print the total amount of donations and the number of donors. If the total amount is greater than or equal to M, also print the message “Target Achieved”.

Problem Solving Process

Now, let’s take a step-by-step look at the algorithm to solve the above problem.

Step 1: Problem Analysis

To solve the problem, we need to calculate the total amount of donations using the number of donors and the contribution amount from each donor provided as input. Then we check this total amount against the target amount M.

Step 2: Algorithm Design

The algorithm can be designed as follows:

1. Input the number of donors N and the target amount M.
2. Input the amounts donated by N donors as a list.
3. Calculate the total amount of donations by summing all the elements in the list.
4. If the total amount of donations is greater than or equal to M, output the total amount and the number of donors, including the message “Target Achieved”.
5. If not, output only the total amount and the number of donors.

Step 3: Code Implementation

Now, let’s implement the code in Python based on the algorithm designed above.

    
def main():
    # Input the number of donors N and the target amount M
    N, M = map(int, input().split())
    
    # Input the list of amounts donated by donors
    donations = list(map(int, input().split()))
    
    # Calculate the total amount of donations
    total_donations = sum(donations)
    
    # Output the result
    print(f"Total Donations: {total_donations}, Number of Donors: {N}")
    
    # Compare with the target amount
    if total_donations >= M:
        print("Target Achieved")

if __name__ == "__main__":
    main()
    
    

Step 4: Code Explanation

The above code performs the following functions:

• On the first line, it inputs the number of donors and the target amount, converting them to integers using the map function.
• On the second line, it inputs and stores the donation amounts from each donor in a list.
• It uses the sum() function to calculate the total amount of donations and outputs this amount.
• It checks whether the total amount of donations meets or exceeds the target amount and outputs a message based on that check.

Step 5: Performance Review

The time complexity of this problem is O(N). Since the maximum number of donors is 100, this level of time complexity can be considered very efficient for coding tests. The space complexity is O(N) as a list is used to store donation amounts.

Step 6: Example and Test Cases

To increase the reliability of the code, various test cases have been prepared:

Test Case 1

Input:

    5 10000
    3000 4000 2000 5000 1500
    

Output:

    Total Donations: 15500, Number of Donors: 5
    Target Achieved
    

Test Case 2

Input:

    3 20000
    5000 6000 7000
    

Output:

    Total Donations: 18000, Number of Donors: 3
    

Through these various test cases, we can verify that the code works without issues.

Conclusion

In this course, we systematically examined the process of solving a Python algorithm problem with the theme of helping the underprivileged. I would like to emphasize the importance of accurately understanding the problem and progressively building the algorithm to solve it. This process is useful not only for coding tests but also in real development environments.

I hope you will continue to build your skills through various algorithm problems and develop your problem-solving abilities. Thank you.

I Will Become the Resident Association President – Python Coding Test Course

In this article, we will cover a famous problem called “I Will Become the Resident Association President” through an algorithm problem-solving course using Python.
This problem requires a solution using dynamic programming based on the given conditions.
We will define the problem, demonstrate the method through examples, and finally write the code to solve the problem.

Problem Description

Problem:
The resident association president is a resident living in apartment number B on the A-th floor, among N residents.
The apartment has K floors, and each floor has apartments numbered from 1 to K.
The problem is to calculate the total number of residents living in apartment B on the A-th floor.
The number of residents living in apartment B on the A-th floor varies depending on the apartment number on each floor, and the rule is as follows:

• The number of residents in apartment B on the A-th floor = The number of residents in apartment 1 on the A-th floor + The number of residents in apartment 2 on the A-th floor + … + The number of residents in apartment B on the A-th floor
• The number of residents in apartment B on the 1st floor is B.

For example, if we want to know the number of residents in apartment 4 on the 3rd floor, we need to add the number of residents from apartment 1 to apartment 4 on the 3rd floor.
The problem is to find the number of residents in apartment B on the A-th floor for the given A and B.

Input and Output Format

Input:
The first line contains the number of test cases T.
Following this, A and B will be given for each test case over T lines.

Output:
For each test case, print the number of residents in apartment B on the A-th floor.

Example

Input Example:
2
1 3
2 3

Output Example:
3
6

Problem-Solving Approach

To solve this problem, we can use a dynamic programming approach.
We can create a two-dimensional table to store the number of residents for each apartment number on each floor to solve the problem.

Step 1: Table Initialization

The number of residents on the 1st floor is always the same for each apartment number, so we initialize the table based on this.

Step 2: Set Dynamic Programming Relation

The number of residents in apartment B on each floor can be expressed as the sum of the number of residents from all apartments on the previous floor.
Therefore, the recurrence relation is as follows:

    dp[A][B] = dp[A-1][1] + dp[A-1][2] + ... + dp[A-1][B]

Step 3: Repetitive Process

Using the above recurrence relation, we calculate the values for A-th floor and apartment B through a loop.
This way, we will eventually obtain the number of residents in apartment B on the A-th floor.

Code Solution

def calculate_people(A, B):
    # Function to calculate the number of residents in apartment B on the A-th floor
    dp = [[0] * (B + 1) for _ in range(A + 1)]
    
    # Initialize the 1st floor
    for j in range(1, B + 1):
        dp[1][j] = j
    
    # Calculate number of residents using dynamic programming
    for i in range(2, A + 1): # A-th floor
        for j in range(1, B + 1): # B-th apartment
            dp[i][j] = sum(dp[i-1][k] for k in range(1, j + 1))
    
    return dp[A][B]

# Processing test cases
T = int(input())
for _ in range(T):
    A, B = map(int, input().split())
    print(calculate_people(A, B))

Conclusion

Through this problem, we explored the application of dynamic programming. We could understand how important it is to analyze the problem and design an algorithm to solve the given problem.
This approach can also help in finding more efficient solutions when solving other coding test problems.

More practice is needed to solve problems effectively. I hope you can develop an algorithmic mindset by solving a variety of problems.

Wishing you a successful coding test!

Hello! Today, we will take an in-depth look at the Merge Sort algorithm, which is frequently asked in Python coding tests. Merge Sort is one of the sorting algorithms that uses the divide and conquer method to sort data. Since sorting is a crucial process in data processing, understanding and implementing this algorithm will greatly help in coding tests as well as in actual development.

What is Merge Sort?

Merge Sort works by recursively dividing the list, sorting the divided lists, and then merging them back together. The process for Merge Sort is as follows:

1. Divide the list into halves.
2. Recursively perform merge sort on each sublist.
3. Merge the two sorted sublists into one sorted list.

The characteristic of Merge Sort is that it is a stable sort algorithm, and its time complexity is O(n log n), making it very efficient. Furthermore, it guarantees a performance of O(n log n) even in the worst case, making it useful for sorting large datasets.

Time Complexity of Merge Sort

Analyzing the time complexity of Merge Sort gives us the following results:

• When the size of the input array is n, the time required to divide the array into two parts is O(1).
• Since merge sort is called recursively on each part, the depth will be log n.
• The merging stage requires O(n) time to combine the two sublists into one.

Thus, the overall time complexity is O(n log n). Additionally, Merge Sort requires O(n) of memory space.

Implementing Merge Sort

Now let’s implement Merge Sort in Python. The code below implements Merge Sort:

def merge_sort(array):
    if len(array) <= 1:
        return array

    mid = len(array) // 2
    left_half = merge_sort(array[:mid])
    right_half = merge_sort(array[mid:])

    return merge(left_half, right_half)

def merge(left, right):
    result = []
    left_index, right_index = 0, 0

    while left_index < len(left) and right_index < len(right):
        if left[left_index] <= right[right_index]:
            result.append(left[left_index])
            left_index += 1
        else:
            result.append(right[right_index])
            right_index += 1

    result.extend(left[left_index:])
    result.extend(right[right_index:])
    
    return result

# Example usage
array = [38, 27, 43, 3, 9, 82, 10]
sorted_array = merge_sort(array)
print(sorted_array)  # [3, 9, 10, 27, 38, 43, 82]

The code above consists of two functions. The merge_sort function recursively divides the array, and the merge function merges two sorted lists. To briefly explain this code, it first returns the array if its length is 1 or less. Then, it divides the array at the midpoint and calls merge_sort on each sublist again. Finally, it merges the two sublists into one sorted list using the merge function.

Checking the Result

You can sort an input array using the merge_sort function defined above, as shown below. The output will be the sorted list.

array = [38, 27, 43, 3, 9, 82, 10]
sorted_array = merge_sort(array)
print(sorted_array)  # [3, 9, 10, 27, 38, 43, 82]

Applications of Merge Sort

Merge Sort is used in many applications that require large data processing or stability. For instance, the usefulness of Merge Sort can be found in database sorting, large-scale data analysis, and data sorting in distributed systems.

Comparison of Sorting Algorithms

Merge Sort has several pros and cons compared to Quick Sort and Insertion Sort:

• Quick Sort tends to perform faster on average, but it could have a performance of O(n2) in the worst case.
• Insertion Sort performs well on small datasets, but it is inefficient for large data processing.
• Merge Sort guarantees a performance of O(n log n) at all times, making it suitable for specific problems where stability is required.

Conclusion

In this lesson, we have delved deeply into Merge Sort. Understanding Merge Sort plays an important role in grasping the basic concepts of data sorting and serves as a foundation for learning other sorting algorithms. Knowing these algorithms will greatly assist not only in preparing for coding tests but also in actual development.

I hope this lesson helps you effectively understand and utilize Merge Sort, and next time we will explore other sorting algorithms. If you have any questions, feel free to leave a comment!