Author: root

Hello, today we will discuss one of the algorithm problems that frequently appears in Swift coding tests, “Gift Delivery.” This problem often comes up in interviews and coding tests, and helps develop algorithmic thinking.

Problem Description

You and your friends have decided to hold a birthday party. All friends want to give and receive gifts from each other. However, it has been decided that a friend cannot give a gift to themselves. Each friend has promised to give a gift to a specific friend. Given that there are a total of N friends and the list of gifts they promised to give, write a program to output the gifts each friend will receive.

Input Format

• The first line contains the number of friends N (1 ≤ N ≤ 100).
• The second line contains the indices of the friends to whom each friend promised to give a gift. (Friends are numbered from 1 to N)

Output Format

Output the friend number of the gift each friend will receive, one per line.

Example Input

5
2 3 4 5 1
    

Example Output

1
2
3
4
5
    

Problem Solving Process

Step 1: Understand the Problem

First, we need to understand the problem. From the given input, we can know who each friend will give a gift to. Each friend needs to know the index of the friend from whom they will receive a gift. Hence, the output can be thought of as an array based on the indices each friend has promised to give.

Step 2: Design Data Structure

We will prepare an array to store the input values and an array for each friend’s desired gifts. Since friend numbers start from 1, it is convenient to declare the array with size N + 1.

Step 3: Design Problem Solving Algorithm

The algorithm is as follows:

1. Receive the input and declare a gift array of size N + 1.
2. Store the input gift boxes in the gift array.
3. Print the gifts corresponding to each friend’s index from the app.

Step 4: Swift Implementation

Now let’s implement the algorithm using Swift:

import Foundation

// Input the number of friends
let N = Int(readLine()!)!

// Gift delivery array
var gift = Array(repeating: 0, count: N + 1)

// Input gift distribution
let gifts = readLine()!.split(separator: " ").map { Int($0)! }

// Packing into the array
for i in 1...N {
    gift[gifts[i - 1]] = i
}

for i in 1...N {
    print(gift[i])
}

Step 5: Code Explanation

To briefly explain the code:

1. First, we receive the number of friends, N.
2. We declare an array called gift with a size of N + 1. (Since array indices start from 1)
3. Then, through the second input, we receive the indices of friends to whom gifts will be given as an array called gifts.
4. Using a loop, we store the gifts each friend will receive in the gift array, saving them at the respective friend’s index using index i.
5. After completing all processes, we print the gifts that friends will receive.

Conclusion

This problem is a basic array problem given in most coding tests. By understanding the problem and applying appropriate data structures and algorithms, it can be easily solved. Approaching algorithm problems in various ways greatly helps improve your skills. I hope everyone becomes more familiar with Swift through this process!

Coding tests have become an essential process for software developers today. The ability to solve various algorithmic problems plays a crucial role in real software development tasks. In this course, we will take a deep dive into the ‘Insertion Sort’ algorithm and learn how to implement it in the Swift language. Insertion sort is a simple yet very useful sorting algorithm, which we will cover in detail starting from the basic concepts to actual problem-solving processes.

1. What is Insertion Sort?

Insertion sort is one of the algorithms used to sort a given array, where each element determines its position through iteration and sorts itself accordingly. This algorithm can be likened to sorting cards in a card game. Each card goes to the position that requires sorting, and through this, a sorted array is formed.

2. Basic Principle of Insertion Sort

Insertion sort operates through the following steps:

1. Starting from the second element, each element is compared to the elements before it at its current position.
2. If the current element is smaller than the previous element, the current element is moved backward.
3. This process is repeated until the current element reaches its appropriate position.
4. The process continues until all elements have been checked.

3. Time Complexity of Insertion Sort

The average and worst-case time complexity of insertion sort is O(n²). This means that as the amount of data increases, performance can degrade. However, it operates quickly at O(n) for sorted or nearly sorted data. In terms of memory usage, insertion sort is not inefficient, with a space complexity of O(1).

4. Implementing Insertion Sort in Swift

Now, let’s actually implement insertion sort in Swift. Below is the code that implements the insertion sort algorithm:

func insertionSort(array: inout [Int]) {
    for i in 1..= 0 && array[j] > key {
            array[j + 1] = array[j]
            j -= 1
        }
        array[j + 1] = key
    }
}

var numbers = [5, 3, 1, 4, 2]
insertionSort(array: &numbers)
print(numbers) // Output: [1, 2, 3, 4, 5]

5. Algorithm Problem: Finding Maximum Profit Using Insertion Sort

Problem: Sort a given integer array using insertion sort and find the maximum difference between two adjacent elements based on the sorted array. This problem can be solved by sorting the array first and then finding the maximum value among the differences of each element and its adjacent elements.

Problem Solving Process

1. First, sort the array using insertion sort.
2. Traverse the sorted array and calculate the differences between adjacent elements.
3. Find the maximum value among the calculated differences.

Swift Implementation Code

func maxAdjacentDifference(array: inout [Int]) -> Int? {
    // Sort the array
    insertionSort(array: &array)
    
    var maxDifference = 0
    for i in 0.. maxDifference {
            maxDifference = difference
        }
    }
    return maxDifference
}

// Test the function with an example
var testArray = [3, 5, 1, 9, 2]
if let maxDiff = maxAdjacentDifference(array: &testArray) {
    print("The maximum difference is: \(maxDiff)")
} else {
    print("The array is empty.")
}

6. Practical Applications

Insertion sort can be applied to various problems due to its simplicity. For example, it is useful for finding data that meets specific conditions or for real-time data processing. It performs well when the rate of data updates is fast and when most of the inserted data is already sorted. Therefore, when adding data in real-time, insertion sort may be suitable without the need for complicated sorting algorithms.

7. Conclusion

In this article, we learned about the insertion sort algorithm and solved the problem of finding the maximum adjacent difference in an array through it. Insertion sort is easy to understand and has concise code, making it suitable for foundational algorithm learning. It will greatly help in laying the groundwork for data structures and algorithms that frequently appear in actual coding tests. In the next course, we will cover more complex sorting algorithms.

If you have any questions or need further information, please leave a comment. Thank you!

Author: [Your Name]

Written on: [Date]

1. Problem Definition

Recently, the importance of coding tests in IT companies has increased, requiring the ability to solve various algorithm problems.
In this course, we will explore algorithm problem-solving using the Swift programming language through the “Word Finder” problem.
This problem involves finding the position of a specific word in a given list of words, requiring an understanding of basic data structures and algorithms.

Problem Description

This problem is to find a specific word in a given list and output its index number.
For example, given the following list of words:

        Word list: ["apple", "banana", "cherry", "date", "fig", "grape"]
        Word to find: "cherry"
        

In this case, the index of “cherry” is 2. If the word to find is not in the list, it should return -1.

2. Problem Analysis

To solve this problem, it is common to iterate through the given list of words and compare it with the target word.
However, to handle it more efficiently, we can utilize the binary search algorithm.
Binary search is a method used to effectively find a specific value in sorted data, reducing time complexity to O(log n).

Sorting the List

First, the list of words must be sorted to use binary search.
Therefore, we will sort the given list of words before applying binary search.

3. Algorithm Design

The algorithm consists of the following steps:

1. Sort the word list.
2. Use binary search to find the specific word.
3. Return the found index or -1 if it does not exist.

4. Code Implementation

Now, let’s implement the above algorithm using Swift.

        import Foundation

        func binarySearch(words: [String], target: String) -> Int {
            var low = 0
            var high = words.count - 1

            while low <= high {
                let mid = (low + high) / 2
                if words[mid] == target {
                    return mid
                } else if words[mid] < target {
                    low = mid + 1
                } else {
                    high = mid - 1
                }
            }
            return -1
        }

        func findWordIndex(words: [String], target: String) -> Int {
            let sortedWords = words.sorted()
            return binarySearch(words: sortedWords, target: target)
        }

        // Example
        let words = ["apple", "banana", "cherry", "date", "fig", "grape"]
        let target = "cherry"
        let index = findWordIndex(words: words, target: target)
        print("Index number: \(index)") // Result: Index number: 2
        

5. Code Explanation

The above code consists of the following steps:

• binarySearch function:

  Searches for the target word in the provided list using binary search.
  It adjusts the current search interval with the low and high variables and calculates the mid index for comparison.
  If the searching word matches, it returns the corresponding index; otherwise, it narrows the search interval.

• findWordIndex function:

  This function sorts the list of words and then calls binary search.
  It returns the index of the word to find.

6. Time Complexity

The time complexity of this algorithm can be divided into two parts.
First, the sorting of the word list is O(n log n).
Secondly, the binary search part is O(log n).
Therefore, the overall time complexity can be evaluated as O(n log n).

7. Various Test Cases

To enhance the accuracy of the algorithm, various test cases can be created.
For example:

Test Case 1

        Input: ["apple", "banana", "cherry", "date"], "banana"
        Output: 1
        

Test Case 2

        Input: ["apple", "banana", "cherry", "date"], "kiwi"
        Output: -1
        

Test Case 3

        Input: [], "apple"
        Output: -1
        

You can run several test cases to check if the algorithm works correctly.

8. Conclusion and Further Learning

In this course, we explored the algorithmic problem-solving method using Swift through the “Word Finder” problem.
Coding tests are an important process to assess understanding of algorithms and data structures, so practicing various problems is beneficial.
Also, consider using the BLS (Best Learning Strategy) principle to repetitively learn algorithms and improve your skills.

Additionally, solving problems on platforms like LeetCode and HackerRank is also a good way to improve algorithm problem-solving skills.
This will help you acquire the skills required in actual coding tests and will be beneficial for job preparation as well.

Thank you!

Hello! Today we will learn about one of the problems frequently encountered in Swift coding tests, ‘Finding the Order of Building’. This problem will provide a good opportunity to lay the foundation of algorithms through the process of finding the order to construct buildings based on specific conditions.

Problem Definition

This problem involves returning a possible order of construction for given building information. The buildings must satisfy specific conditions, which are as follows:

• Each building has a unique height.
• If a taller building is constructed first, any building taller than that must be constructed afterwards.
• The order of the buildings must be sorted in descending order based on height.

For example, if there are buildings with heights [5, 3, 4, 1], the possible order of the buildings will be [5, 4, 3, 1], and building 5 will be constructed first.

Input and Output

Input

An integer array heights containing the heights of the buildings is given.

Output

Returns the order of buildings that can be constructed in the form of an array based on height order.

Problem Solving Approach

To solve this problem, the following steps are taken:

1. Sort the given array of building heights in descending order.
2. Traverse the sorted array and add values to the result array.
3. Return the result array.

Example

For example, if the input array is [5, 3, 4, 1], the result after sorting will be as follows:

Input: [5, 3, 4, 1]
Output: [5, 4, 3, 1]

Swift Code Implementation

Now, let’s implement the above approach in Swift.

func buildingOrder(heights: [Int]) -> [Int] {
        return heights.sorted(by: >)
    }

    // Example usage
    let heights = [5, 3, 4, 1]
    let orderedBuildings = buildingOrder(heights: heights)
    print(orderedBuildings) // Output: [5, 4, 3, 1]

The above code is a simple function that sorts the given heights array in descending order. It uses sorted(by: >) to sort the array and returns the sorted result.

Time Complexity Analysis

The time complexity of this algorithm is the same as that used for sorting the array, which is O(n log n). Here, n is the number of buildings. Since this problem is based on sorting, the time increases as the input size grows, but it can be considered an efficient method.

Conclusion

Today we explored the problem of finding the order to construct buildings using Swift. I hope this problem has helped to enhance your understanding of the fundamental concepts of algorithms and sorting. I encourage you to build your skills through more algorithmic challenges in the future!

Solving algorithm problems is a very important process in preparing for coding tests. In this course, we will learn how to solve the problem titled ‘Creating Blu-rays’ using the Swift programming language. This problem is often encountered in actual job exams and will provide a good opportunity to understand both data structures and algorithms at the same time.

Problem Description

The given ‘Creating Blu-rays’ problem involves N movies, and the lengths of each movie are provided in the array filmLengths. You need to fit these movies onto Blu-rays, with each Blu-ray having a maximum capacity of maxCapacity. The challenge is to find a way to include all movies while minimizing the number of Blu-rays used.

The function signature is as follows:

func minimumBluRays(filmLengths: [Int], maxCapacity: Int) -> Int

Input Example

let films = [90, 85, 75, 60, 120]
let maxCap = 180

Output Example

minimumBluRays(filmLengths: films, maxCapacity: maxCap) // Result: 3

Approach

To solve this problem, we can consider two approaches.

1. Greedy Algorithm: This involves sorting the movie lengths in ascending order and then placing the longest movie onto a Blu-ray with the remaining movies. This method allows us to maximize the number of movies added to the Blu-ray each time.
2. Binary Search: This involves determining the maximum number of Blu-rays that can be used through binary search. We define a range for the possible maximum number of Blu-rays, and count the number needed based on a mid value. This method can also provide an efficient solution.

Solution Process

Here, we will take a closer look at how to solve the problem using the greedy algorithm.

Step 1: Sort Movies

First, sort the given movie lengths in ascending order. By doing this, you can place the shortest movies in order onto the Blu-ray, balancing the longest movie with the others.

let sortedFilms = filmLengths.sorted()

Step 2: Implement Blu-ray Placement Logic

We implement the logic to determine if movies can be placed onto each Blu-ray. While managing the current capacity of the Blu-ray, if the movie to be added does not exceed the capacity, we add it to that Blu-ray. If it exceeds the capacity, a new Blu-ray is created.

var bluRayCount = 1
var currentCapacity = 0

for film in sortedFilms {
    if currentCapacity + film <= maxCapacity {
        currentCapacity += film
    } else {
        bluRayCount += 1
        currentCapacity = film
    }
}

Step 3: Final Code

Combine the above steps to complete the final code.

func minimumBluRays(filmLengths: [Int], maxCapacity: Int) -> Int {
    let sortedFilms = filmLengths.sorted()
    var bluRayCount = 1
    var currentCapacity = 0

    for film in sortedFilms {
        if currentCapacity + film <= maxCapacity {
            currentCapacity += film
        } else {
            bluRayCount += 1
            currentCapacity = film
        }
    }
    return bluRayCount
}

// Example usage
let films = [90, 85, 75, 60, 120]
let maxCap = 180
print(minimumBluRays(filmLengths: films, maxCapacity: maxCap)) // Result: 3

Time Complexity

The time complexity of this solution is O(N log N). Sorting the movie lengths takes O(N log N), followed by a loop that takes O(N), making it an overall efficient solution.

Conclusion

This problem frequently appears in actual coding tests and can be effectively solved using the greedy algorithm. By understanding the problem and setting up the approach correctly, you can achieve good results in various coding tests. Continually practice solving these problems while familiarizing yourself with the features of the Swift language.

References

• Introduction to Algorithms, Thomas H. Cormen
• Algorithms, Robert Sedgewick
• Swift Programming: The Big Nerd Ranch Guide, Matthew Mathias