Category: Swift Coding Test

In this course, we will solve one problem to prepare for coding tests using the Swift language. The topic is to divide line segments into groups. We will understand the problem and explain the solution through the following procedures.

Problem Description

The problem is to determine the minimum number of groups into which the given line segments can be divided when there are overlaps. Line segments are expressed by their starting and ending coordinates, and overlapping segments should be grouped together. Each line segment is given in the form [start, end]. For example, it can be given as [1, 3], [2, 5], [6, 8], etc.

Input Format

Multiple line segments are received in the form of an array. For example:

let segments = [[1, 3], [2, 5], [6, 8], [7, 9]]

Output Format

The minimum number of groups needed to divide the segments will be printed. In the above example, it will be 2. ([1, 3] and [2, 5] overlap and belong to the same group, while [6, 8] and [7, 9] are in different groups.)

Solution

To solve the problem, we first need to sort the segments and then group together the overlapping segments. The strategy to solve this problem is as follows:

1. Sort the segments in ascending order based on their starting points.
2. Iterate through the sorted segments and check if they overlap with the last segment of the current group.
3. If they overlap, include them in the same group; if not, create a new group.

Implementation

Now let’s write code in Swift according to the above strategy:

func countGroups(segments: [[Int]]) -> Int {
    guard !segments.isEmpty else { return 0 }
    
    // 1. Sort segments
    let sortedSegments = segments.sorted { $0[0] < $1[0] }
    
    var groups = 1
    var lastEnd = sortedSegments[0][1]

    // 2. Explore segments
    for i in 1..
Code Explanation
The countGroups function takes an array of line segments as input and returns the minimum number of groups. Each step is as follows:

If the array is empty, return 0 for the number of groups.
Sort the input segments based on their starting points.
Set the initial value for the last endpoint as the endpoint of the first segment. Initialize the number of groups to 1.
Iterate through the remaining segments and check if the starting point of the current segment is less than or equal to the endpoint of the last segment to check for overlaps.
If they overlap, update the last endpoint; otherwise, start a new group and increase the number of groups.

Time Complexity Analysis
The time complexity of this algorithm is O(n log n). Sorting the segments takes O(n log n), and iterating through the segments to calculate the groups takes an additional O(n). Therefore, the final time complexity is O(n log n).
Key Points

It is important to accurately verify whether the segments overlap.
You need to understand how to effectively handle the segments after sorting.

Through this course, I hope you can improve your algorithm problem-solving skills using Swift. The problem and approach above will be useful for preparing for coding tests. I encourage you to practice with additional problems and have opportunities to learn various algorithms!

	

Problem Definition

The problem is to determine the direction of line segment AB given two points A(x1, y1) and B(x2, y2). The direction of the line segment reflects how the x-axis and y-axis change as it goes from A to B. We need to determine whether the direction of AB is upward, downward, leftward, or rightward.

The problem is as follows:

Given four integers x1, y1, x2, y2, determine the direction of line segment AB.

• 0: Horizontal line (y1 == y2)
• 1: Upward (y1 < y2)
• -1: Downward (y1 > y2)
• 2: Rightward (x1 < x2)
• -2: Leftward (x1 > x2)

Problem Analysis

To find the direction from point A to point B, we need to compare the x-coordinates and y-coordinates of the two points. The following cases are considered.

• If the y-coordinates are the same, the line segment is considered horizontal (y1 == y2).
• If A’s y-coordinate is less than B’s y-coordinate, the line segment is directed upward (y1 < y2).
• If A’s y-coordinate is greater than B’s y-coordinate, the line segment is directed downward (y1 > y2).
• If A’s x-coordinate is less than B’s x-coordinate, the line segment is directed rightward (x1 < x2).
• If A’s x-coordinate is greater than B’s x-coordinate, the line segment is directed leftward (x1 > x2).

These conditions can be used to resolve the problem.

Algorithm Design

To solve the problem, we design the following algorithm:

1. Input the coordinates of two points A(x1, y1) and B(x2, y2).
2. Compare y1 and y2 to determine the result in three cases (horizontal, upward, downward).
3. Compare x1 and x2 to define the cases for rightward or leftward directions.
4. Output the result.

Implementation

Below is the code implemented in Swift based on the above algorithm:

            
                import Foundation
                
                func determineDirection(x1: Int, y1: Int, x2: Int, y2: Int) -> String {
                    if y1 == y2 {
                        return "Horizontal line"
                    } else if y1 < y2 {
                        return "Upward"
                    } else {
                        return "Downward"
                    }
                }
                
                func determineHorizontalDirection(x1: Int, x2: Int) -> String {
                    if x1 < x2 {
                        return "Rightward"
                    } else if x1 > x2 {
                        return "Leftward"
                    } else {
                        return "Vertical line"
                    }
                }
                
                let x1 = 1, y1 = 2, x2 = 3, y2 = 4
                print(determineDirection(x1: x1, y1: y1, x2: x2, y2: y2))
                print(determineHorizontalDirection(x1: x1, x2: x2))
            
            

In the above Swift code, two functions are used to determine the direction of the line segment. The first function determineDirection assesses the direction based on the y-coordinates, while the second function determineHorizontalDirection assesses it based on the x-coordinates. Each case returns the appropriate string.

Test Cases

Now let’s look at a few test cases:

• Case 1: Line segment from A(1, 2) to B(3, 4)
• Case 2: Horizontal line from A(1, 3) to B(1, 3)
• Case 3: Line segment from A(5, 6) to B(2, 5)

We validate the algorithm through the results of each test case:

            
                // Case 1: A(1, 2), B(3, 4)
                let x1_case1 = 1, y1_case1 = 2, x2_case1 = 3, y2_case1 = 4
                print(determineDirection(x1: x1_case1, y1: y1_case1, x2: x2_case1, y2: y2_case1)) // Upward
                print(determineHorizontalDirection(x1: x1_case1, x2: x2_case1)) // Rightward
                
                // Case 2: A(1, 3), B(1, 3)
                let x1_case2 = 1, y1_case2 = 3, x2_case2 = 1, y2_case2 = 3
                print(determineDirection(x1: x1_case2, y1: y1_case2, x2: x2_case2, y2: y2_case2)) // Horizontal line
                print(determineHorizontalDirection(x1: x1_case2, x2: x2_case2)) // Vertical line
                
                // Case 3: A(5, 6), B(2, 5)
                let x1_case3 = 5, y1_case3 = 6, x2_case3 = 2, y2_case3 = 5
                print(determineDirection(x1: x1_case3, y1: y1_case3, x2: x2_case3, y2: y2_case3)) // Downward
                print(determineHorizontalDirection(x1: x1_case3, x2: x2_case3)) // Leftward
            
            

Conclusion

In this post, we examined the process of designing and implementing an algorithm to determine the direction of a line segment given two points. This algorithm allows for a straightforward determination of directionality through the comparison of the given two coordinates.

The problem of determining line segment direction requires a basic geometric approach, and it is important to clearly understand each condition and write the corresponding logic. Through such problems, one can develop algorithmic thinking and improve their skills.

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!