Category: Swift Coding Test

Swift Coding Test Course, Finding the Building Order

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!

This article was written as part of the Swift coding test course. I hope it helps you greatly in your job preparation!

Swift Coding Test Course, Creating Blu-ray

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

Swift Coding Test Course, I Will Become the President of the Women’s Association

Hello, everyone! In this post, we will address one of the algorithm problems called “I Will Be the Apartment Leader”.
This problem is one of the frequently asked topics in Swift coding tests.
I will explain the basic problem description, the process of solving it, and how to implement the code in detail.

Problem Description

The “I Will Be the Apartment Leader” problem is about calculating the number of residents in an apartment according to specific rules.
The given apartment consists of N floors and K units.
The number of residents in each unit on the K-th floor follows these rules:

  • On the 0-th floor, there is 1 resident in every K unit.
  • The number of residents in unit n on the k-th floor is the sum of the number of residents from unit 1 to n on the (k-1)-th floor.

In other words, the number of residents on the 0-th floor is 1 for everyone, and the number of residents in unit n on the 1st floor is the total from unit 1 to n on the 0-th floor.
We can use these rules to calculate the number of residents in unit n on the k-th floor.

Input and Output Format

Input

  • The first line contains the number of test cases T. (1 ≤ T ≤ 100)
  • Each test case consists of two integers K and N. (0 ≤ K ≤ 14, 1 ≤ N ≤ 14)

Output

  • For each test case, print the number of residents in unit N on the k-th floor.

Process of Solving the Problem

To solve this problem, we first declare a 2-dimensional array to calculate the number of residents in the apartment.
The size of the array is set to (15 x 15) to store the number of residents from the 0-th floor to the 14-th floor.
We will use this to solve the problem using Dynamic Programming.

Step 1: Initialize the Array


let maxK = 15
let maxN = 15
var dp = Array(repeating: Array(repeating: 0, count: maxN), count: maxK)

for i in 0..

Here, dp[k][n] represents the number of residents in unit n on the k-th floor.
We initialized all values on the 0-th floor to 1. For k > 0, we can calculate dp[k][n] as the sum of residents from the previous floor.

Step 2: Calculate the Number of Residents


for k in 1..

This is the process of filling the dp array with values.
The number of residents in unit n on the k-th floor is the sum from units 1 to n on the (k-1)-th floor.
We calculate the values for each floor and store them in the dp array through a loop.

Step 3: Print the Results


for _ in 0..

Finally, we print the number of residents in unit N on the k-th floor for each test case.
Through this process, we can solve the problem.

Implementation Code

Now, the full code that implements the above process in Swift is below. 


import Foundation

let maxK = 15
let maxN = 15
var dp = Array(repeating: Array(repeating: 0, count: maxN), count: maxK)

for i in 0..

Conclusion

In this article, we examined the process of solving the "I Will Be the Apartment Leader" problem.
We learned how to understand the rules of the problem and efficiently calculate the solution using Dynamic Programming.
I hope this helps you solidify your understanding of algorithms and prepare for coding tests.
Thank you!

Swift Coding Test Course, Merge Sort

Hello. Today, we will learn about the Merge Sort algorithm implemented in Swift. This tutorial will start from the basics of Merge Sort and explain its application in actual coding tests in detail.

What is Merge Sort?

Merge Sort is a sorting algorithm based on the divide and conquer strategy. It divides the array into two halves, sorts each sub-array, and then merges the two sorted sub-arrays into one sorted array. Merge Sort has an average and worst-case time complexity of O(n log n) and is a stable sort.

Principle of Merge Sort

The process of Merge Sort is as follows:

  1. Given the original array, it is split into two sub-arrays at the midpoint.
  2. Merge Sort is recursively performed on each sub-array.
  3. The two sorted sub-arrays are merged to create one sorted array.

The important point here is that the sub-arrays are sorted at each step. The entire array is sorted by merging these sorted arrays back together.

Problem: Sort Integers

Write a function in Swift that sorts a given list of integers in ascending order. An example input is as follows:


            let nums = [38, 27, 43, 3, 9, 82, 10]

To solve this problem, you need to apply the principles of Merge Sort explained above.

Implementing Merge Sort

Now, let’s implement Merge Sort in Swift. Below is the complete code:


            func mergeSort(_ array: [Int]) -> [Int] {
                // 1. If the size of the array is less than 1, return the original array
                if array.count < 2 {
                    return array
                }
                
                // 2. Split the array at the midpoint
                let midIndex = array.count / 2
                let leftArray = mergeSort(Array(array[0.. [Int] {
                var mergedArray: [Int] = []
                var leftIndex = 0
                var rightIndex = 0
                
                // 4. Merge by comparing the two arrays
                while leftIndex < left.count && rightIndex < right.count {
                    if left[leftIndex] < right[rightIndex] {
                        mergedArray.append(left[leftIndex])
                        leftIndex += 1
                    } else {
                        mergedArray.append(right[rightIndex])
                        rightIndex += 1
                    }
                }
                
                // 5. Add remaining elements
                while leftIndex < left.count {
                    mergedArray.append(left[leftIndex])
                    leftIndex += 1
                }
                
                while rightIndex < right.count {
                    mergedArray.append(right[rightIndex])
                    rightIndex += 1
                }
                
                return mergedArray
            }

            // Main
            let nums = [38, 27, 43, 3, 9, 82, 10]
            let sortedNums = mergeSort(nums)
            print(sortedNums) // Result: [3, 9, 10, 27, 38, 43, 82]

The above code sorts the input list of numbers using Merge Sort and prints the result. Comments are added at each step to aid understanding.

Code Explanation

Let's take a look at the code line by line:

  1. if array.count < 2 { return array }: If the size of the array is less than 1, it is already sorted, so it is returned as is.
  2. let midIndex = array.count / 2: Calculates the midpoint index of the array.
  3. let leftArray = mergeSort(Array(array[0..: Recursively sorts the left sub-array.
  4. let rightArray = mergeSort(Array(array[midIndex..: Recursively sorts the right sub-array.
  5. return merge(leftArray, rightArray): Merges the sorted left and right arrays and returns the result.

In the merge function, indices are tracked as the two arrays are compared and merged. If one array is exhausted, the remaining elements are added to produce the final result.

Advantages and Disadvantages of Merge Sort

Merge Sort has the following advantages and disadvantages:

Advantages:

  1. Stable sorting: The relative order of elements with equal values is preserved.
  2. Time complexity of O(n log n): It provides a relatively fast sorting performance proportional to the amount of data.
  3. Can be applied to linked lists as well as arrays of fixed size: It can be used in structures that utilize pointers.

Disadvantages:

  1. Requires additional memory: Because a new array is created during the merging process, the space complexity is O(n).
  2. In simpler cases (e.g., nearly sorted cases), simpler algorithms like insertion sort may perform better.

Additional Problems for Skill Improvement

Now that you understand Merge Sort, practice further with these additional problems:

  1. Write a function that takes an array of integers and returns a sorted array with duplicates removed.
  2. Implement an algorithm that finds the index of an element with a specific value in a sorted array using binary search.
  3. Write a function in Swift that merges multiple input arrays into one sorted array.

This concludes our discussion on implementing Merge Sort in Swift. It is important to solidify your understanding of basic data structures and algorithm theories when solving algorithm problems. We will meet again in the next tutorial with more in-depth content. Thank you!

Swift Coding Test Course, Bellman-Ford

Hello! In this course, we will discuss the Bellman-Ford algorithm using Swift. The Bellman-Ford algorithm is designed to find the shortest path from a single source to all vertices in a graph, and it has the advantage of being usable in graphs with negative edge weights. In this lecture, we will explain the concept of the Bellman-Ford algorithm and examine the specific implementation process of the algorithm.

1. Overview of the Bellman-Ford Algorithm

The Bellman-Ford algorithm was developed by Richard Bellman and Lester Ford in 1958. The algorithm operates in the following manner:

  • Updates the initialized distance array for all edges in the graph.
  • Performs relaxation of edges up to the maximum number of vertices – 1 to optimize the distance information.
  • Adds a negative cycle check to verify if a negative cycle exists.

1.1. Algorithm Procedure

  1. Set the initial distance to infinity and initialize the starting vertex’s distance to 0.
  2. Perform relaxation for all edges up to a maximum of V-1 times.
  3. Finally, check for negative cycles on all edges.

2. Defining the Problem

Now let’s define a problem where the Bellman-Ford algorithm can be used.

Problem: Finding the Shortest Path

Given the number of vertices in the graph V, the number of edges E, and the information of the edges, find the shortest path from the starting vertex to all other vertices. If the graph contains a negative cycle, output “Negative Cycle”.

Input Format

V (number of vertices)
E (number of edges)
Edge information (starting vertex, ending vertex, weight)

Output Format

Shortest distance to each vertex or "Negative Cycle"

3. Implementation of the Bellman-Ford Algorithm

Now that we have defined the problem, let’s implement the algorithm using Swift.

import Foundation

struct Edge {
    let from: Int
    let to: Int
    let weight: Int
}

func bellmanFord(vertices: Int, edges: [Edge], start: Int) -> [Int] {
    var distances = Array(repeating: Int.max, count: vertices)
    distances[start] = 0

    // Relaxation phase
    for _ in 0..

3.1. Example Usage

Now let's test the above algorithm using a graph with multiple edges and vertices.

let edges = [
    Edge(from: 0, to: 1, weight: 4),
    Edge(from: 0, to: 2, weight: 1),
    Edge(from: 2, to: 1, weight: 2),
    Edge(from: 1, to: 3, weight: 1),
    Edge(from: 2, to: 3, weight: 5)
]

let result = bellmanFord(vertices: 4, edges: edges, start: 0)
if !result.isEmpty {
    for (index, distance) in result.enumerated() {
        print("Vertex \(index): \(distance)")
    }
}

4. Time Complexity of the Algorithm

The Bellman-Ford algorithm has a time complexity of O(VE) in the worst case, where there are V vertices and E edges. As a result, it can be efficiently used on relatively small graphs, but it may take a long time when the number of edges increases.

5. Conclusion

In this lecture, we covered the basic concept of the Bellman-Ford algorithm, defined the problem using it, implemented the algorithm, and explored an example. We learned that the Bellman-Ford algorithm is a valuable algorithm that can find the shortest path even in graphs with negative weight. I encourage you to use this algorithm to solve various problems.

In the next session, we will cover Dijkstra's algorithm. Thank you!