Author: root

Problem Description

Implement the bubble sort algorithm to sort the given integer array in ascending order.
Bubble sort works by comparing two adjacent elements and sorting them.
Repeat through all elements in the array, comparing each element with the next one and
repeatedly moving larger values to the back. This process continues until the array is sorted.

Example Input and Output

Input

            [64, 34, 25, 12, 22, 11, 90]
            

Output

            [11, 12, 22, 25, 34, 64, 90]
            

Bubble Sort Algorithm

Steps of the Algorithm

1. Let the length of the given array be n, and repeat n-1 times.
2. In each iteration, compare two adjacent elements.
3. If the current element is greater than the next element, swap their positions.
4. After comparing all elements, the largest number will be positioned at the end of the array.
5. Repeat this process until the array is sorted.

Swift Code Implementation

            
    func bubbleSort(array: inout [Int]) {
        let n = array.count
        for i in 0.. array[j+1] {
                    // Swap
                    let temp = array[j]
                    array[j] = array[j+1]
                    array[j+1] = temp
                }
            }
        }
    }

    var arr = [64, 34, 25, 12, 22, 11, 90]
    bubbleSort(array: &arr)
    print("Sorted array: \(arr)")
            
            

Code Explanation

In the code above, the bubbleSort function sorts the array passed as an argument.
The inout keyword allows the function to modify the array directly,
and the for loop iterates through all elements of the array. In each inner loop,
adjacent numbers are compared to change their order.

Efficiency Analysis

The time complexity of the bubble sort algorithm is O(n^2) in the worst case.
This arises from the nested loops, while the space complexity is O(1), using only constant space.
Therefore, it can be inefficient for large datasets.
However, bubble sort is often used for educational purposes due to its simplicity and ease of understanding.

Expected Benefits

Through bubble sort, students can learn the basic concepts of arrays, loops, and conditionals.
This algorithm serves as a good starting point for a fundamental understanding of other sorting algorithms
and can be used as foundational knowledge when learning more complex data structures and algorithms.

Conclusion

In this lecture, we learned how to implement the bubble sort algorithm in Swift.
It is a simple algorithm, but it plays an important role in learning the basic concepts of programming.
In the next lecture, we will explore more efficient sorting algorithms.

Algorithm problems in coding tests often evaluate theoretical knowledge and practical problem-solving skills that are frequently used in actual work. In this article, we will discuss the problem of finding the Kth number in an array. This problem requires techniques for sorting an array and extracting values from specific indices.

Problem Description

Given an integer array array and an integer k, write a function that returns the Kth number after sorting array in ascending order. The Kth number uses 1-based indexing.

Input

• array: An array containing integers, e.g., [3, 1, 2, 4, 5]
• k: A positive integer to find the Kth number in the array

Output

Output the Kth number after sorting array. If K is greater than the length of the array, return -1.

Example

Input:
    array = [3, 5, 2, 1, 4]
    k = 3

    Output:
    3
    

Solution

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

1. Sort the given input array.
2. Reference the Kth number from the sorted array and return it.

Step-by-step Solution

Step 1: Sorting the Array

There are several ways to sort an array. Common sorting algorithms include Quick Sort, Merge Sort, and Insertion Sort. However, using the built-in sorting function provided by Swift allows for convenient sorting.

Step 2: Finding the Kth Number

To find the Kth number in the sorted array, we return the value at the K-1 index. If K exceeds the length of the array, we should return -1.

Swift Code Implementation

func findKthNumber(array: [Int], k: Int) -> Int {
        let sortedArray = array.sorted()
        guard k > 0 && k <= sortedArray.count else {
            return -1
        }
        return sortedArray[k - 1]
    }

// Test
let array = [3, 5, 2, 1, 4]
let k = 3
print(findKthNumber(array: array, k: k)) // Output: 3
    

Complexity Analysis

Time Complexity: The time complexity for sorting is O(n log n), where n is the length of the array.

Space Complexity: The additional space complexity for sorting is O(n).

Conclusion

Through this problem, we learned about sorting arrays and accessing specific indices. This foundational thinking can be applied to other algorithm problems as well.

If you want more problem-solving and algorithm tutorials, please continue to visit the blog. Thank you!

In this course, we will address algorithm problems using arrays and lists with the Swift programming language. An array is a data structure that can hold multiple pieces of data of the same type, while a list is a structure that stores data in a contiguous block of memory. Understanding the properties of arrays and lists can greatly assist in solving various algorithm problems.

Problem Description

Problem 1: Intersection of Two Arrays

Write a function to find the common elements and compute the intersection of the two given arrays. The arrays may have duplicate elements, and the resulting array should not include duplicate elements.

For example, let’s assume we are given the following two arrays.

let array1 = [1, 2, 3, 4, 5]
let array2 = [4, 5, 6, 7, 8]

In this case, the output of the function should be [4, 5].

Problem Solving Process

Problem Analysis

To approach this problem, we need to think of an algorithm to compare the elements of the two arrays to find the common ones. A basic approach would be to iterate through both arrays using a loop to find common elements. However, this method could lead to a time complexity of O(n^2), which may reduce performance. Therefore, we should use a more efficient method.

Efficient Approach

To enhance efficiency, we can utilize sets. A set is a data structure that allows each element to be stored uniquely, providing fast search performance. Here is a summary of the solution process.

1. Convert the first array into a set. This has a time complexity of O(n).
2. Iterate through the second array while checking for existence in the set. This also has a time complexity of O(m).
3. Create a new array to store the common elements and return the result.

Code Implementation

Below is the Swift code written using the above approach.

func intersection(array1: [Int], array2: [Int]) -> [Int] {
    var set = Set(array1) // Convert the first array into a set
    var result = [Int]() // An array to hold the result
    
    for element in array2 {
        if set.contains(element) { // Check if the element from the second array is in the set
            result.append(element) // If it is, add it to the result array
            set.remove(element) // Remove from the set to avoid duplicates
        }
    }
    
    return result
}

// Example execution
let array1 = [1, 2, 3, 4, 5]
let array2 = [4, 5, 6, 7, 8]
let result = intersection(array1: array1, array2: array2)
print(result) // Output: [4, 5]

Time Complexity Analysis

Analyzing the time complexity of the above solution, converting the first array into a set takes O(n), and iterating through the second array takes O(m). Therefore, the total time complexity is O(n + m). This method is much more efficient than the existing O(n^2) method.

Problem Variations

By modifying the original problem, if the elements of the arrays are sorted, we can further improve the performance of the solution by utilizing binary search. Searching for elements in a sorted array can enhance performance to O(log n). Additionally, if you are familiar with the concept of binary search, you can utilize it to improve search speed.

Conclusion

In this course, we have solved the problem of finding the intersection of two arrays using arrays and lists. Understanding and utilizing arrays and lists, which are fundamental data structures in programming, is crucial. Continue to tackle various algorithm problems to enhance your skills.

The coding test is an important process for applicants to validate their programming skills. In this course, we will learn how to solve algorithm problems in Swift through the topic of Calculating the Amount of Water.

Problem Description

The problem is to calculate the amount of water that can be stored at each unit height when it rains, given an array of heights. For example, consider the following height array:

    heights = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
    

Calculate the amount of water that can be stored in this height array when it rains. The amount of water stored at each position is determined by the higher boundaries on both sides.

Approach to the Problem

The basic approach to solve this problem is as follows:

1. Calculate the maximum heights on the left and right for each position.
2. Determine if water can be stored based on the smaller of the two maximum heights that are lower than the current height.
3. Accumulate the amount of water at each position to derive the result.

Algorithm Explanation

To implement the above algorithm, we use two arrays to store the maximum heights from the left and right at each position. We can then calculate the amount of water at each position.

Step 1: Analyzing the Height Array

This is the process of obtaining the maximum heights on both sides for each position in the given height array.

Creating the Left Maximum Height Array

First, we create an array to store the maximum height from the left. Each element of the array stores the maximum value of all building heights to the left of the current position.

    var leftMax = [Int](repeating: 0, count: heights.count)
    leftMax[0] = heights[0]
    for i in 1..
Creating the Right Maximum Height Array
Next, we create an array to store the maximum height from the right. This array is also created in the same manner.
    var rightMax = [Int](repeating: 0, count: heights.count)
    rightMax[heights.count - 1] = heights[heights.count - 1]
    for i in (0..<(heights.count - 1)).reversed() {
        rightMax[i] = max(rightMax[i + 1], heights[i])
    }
    
Step 2: Calculating the Amount of Water
Now, we calculate the amount of water that can be stored at each position. The amount of water stored can be calculated as the minimum of the left maximum height and the right maximum height minus the current height.
    var waterTrapped = 0
    for i in 0..
Complete Swift Code
    func trap(_ heights: [Int]) -> Int {
        guard heights.count > 2 else { return 0 }

        var leftMax = [Int](repeating: 0, count: heights.count)
        var rightMax = [Int](repeating: 0, count: heights.count)

        leftMax[0] = heights[0]
        for i in 1..
Result Analysis
When the above code is executed, the amount of water stored in the given height array is output as 6. This can be visualized as follows:

Index 2 can store 1 unit of water.
Index 4 can store 1 unit of water.
Index 5 can store 3 units of water.
Index 6 can store 1 unit of water.
Index 8 can store 1 unit of water.

Conclusion
In this course, we covered an algorithm problem of calculating the amount of water using Swift. This problem can be solved using arrays, loops, and conditional statements, and requires comparing the maximum heights at each position. It is important to understand the structure of the problem well and organize the approach systematically when solving algorithm problems.
If you want to learn more algorithm problems and solutions, I recommend referring to various related materials for practice.

	

The programming language Swift is widely used in the Apple ecosystem and is often utilized for iOS and macOS application development.
It is important for developers to have problem-solving skills for algorithms. Especially for employment, it is necessary to demonstrate
the ability to solve various problems. Today, we will look into the maze exploration problem. To solve this problem, we will compare
the Depth-First Search (DFS) algorithm and the Breadth-First Search (BFS) algorithm.

Problem Definition

The problem is to find the shortest path from the starting point to the destination in a maze represented by a given 2D array.
The maze consists of 0s and 1s, where 0 represents a traversable space and 1 represents a wall.
The starting point is (0, 0) and the destination is (n-1, m-1).
Here is an example maze:

            0 0 1 0 0
            1 0 1 0 1
            0 0 0 0 1
            0 1 1 0 0
            0 0 0 0 0
        

Input Format

A 2D array of size n x m is inputted. The array consists of 0s and 1s.

Output Format

The length of the shortest path from the starting point to the destination is outputted. If it is not reachable, -1 is outputted.

Approach to Solve the Problem

Various search algorithms can be used to solve this problem. Among them,
Depth-First Search (DFS) and Breadth-First Search (BFS) are the most commonly used.
BFS is suitable for shortest path problems. I will provide a brief explanation of each algorithm.

BFS (Breadth-First Search)

BFS visits all vertices of a graph level by level. It visits all adjacent vertices from the starting vertex,
then explores adjacent vertices in the next step to explore all paths. BFS is implemented using a queue,
and it can find the shortest path by recording the depth of the path each time a vertex is visited. The time complexity of BFS is O(V + E).

DFS (Depth-First Search)

DFS starts at one vertex of the graph and explores as deeply as possible before backtracking to explore alternative paths.
DFS is implemented using a stack, and the order of visiting depends on the depth. Since DFS explores all paths, it does not guarantee
the shortest path. Therefore, BFS is more suitable for maze exploration problems. The time complexity of DFS is O(V + E),
and its space complexity is O(V).

Algorithm Design

Now, let’s design an algorithm to solve the maze exploration problem using BFS.
We need to define and initialize the necessary variables to proceed to the next steps.

            // maze size n, m
            let n = maze.count
            let m = maze[0].count
            // direction array (up, down, left, right)
            let directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
            var queue: [(Int, Int)] = []
            // add starting point (0, 0) to queue
            queue.append((0, 0))
            // visited array
            var visited = Array(repeating: Array(repeating: false, count: m), count: n)
            visited[0][0] = true
            // distance array
            var distance = Array(repeating: Array(repeating: -1, count: m), count: n)
            distance[0][0] = 0
        

Code Implementation

Let’s solve the problem with code. Below is the BFS algorithm implemented in Swift.

            func bfs(maze: [[Int]]) -> Int {
                let n = maze.count
                let m = maze[0].count
                let directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
                var queue: [(Int, Int)] = []
                var visited = Array(repeating: Array(repeating: false, count: m), count: n)
                var distance = Array(repeating: Array(repeating: -1, count: m), count: n)
                
                queue.append((0, 0))
                visited[0][0] = true
                distance[0][0] = 0
                
                while !queue.isEmpty {
                    let (x, y) = queue.removeFirst()
                    
                    for direction in directions {
                        let newX = x + direction.0
                        let newY = y + direction.1
                        
                        if newX >= 0 && newY >= 0 && newX < n && newY < m &&
                           maze[newX][newY] == 0 && !visited[newX][newY] {
                            visited[newX][newY] = true
                            distance[newX][newY] = distance[x][y] + 1
                            queue.append((newX, newY))
                        }
                    }
                }
                
                return distance[n-1][m-1] != -1 ? distance[n-1][m-1] : -1
            }

            let maze = [
                [0, 0, 1, 0, 0],
                [1, 0, 1, 0, 1],
                [0, 0, 0, 0, 1],
                [0, 1, 1, 0, 0],
                [0, 0, 0, 0, 0]
            ]

            let result = bfs(maze: maze)
            print(result) // Result: 8
        

Code Explanation

In the above code, we used the BFS algorithm to explore the maze and find the shortest path.
Initially, we start at the coordinate (0, 0), and set up the queue, visited array, and distance array.
When we dequeue one coordinate from the queue, we check all adjacent coordinates. If it is a traversable coordinate, we add it to the queue and update the
visited array and distance array. Finally, we return the distance value of the destination point.
If it is unreachable, we return -1.

Conclusion

In this tutorial, we learned how to solve the maze exploration problem with Swift.
Through the BFS algorithm, we effectively utilized the queue and dimensional arrays to find the shortest path.
Such search algorithms are often encountered in job interviews, so we need to practice sufficiently to solve a variety of problems.
These fundamental concepts are very useful when solving algorithm problems in Swift.