Category: Swift Coding Test

Coding tests are an important factor in assessing abilities as a software engineer. This article will address an algorithm problem that can be implemented in Swift with the topic “Making an Integer 1” and provide a detailed explanation of the solution process.

Problem Description

We aim to make a given integer x equal to 1 by applying the following rules:

• x = x - 1 (subtract 1)
• x = x / 2 (divide by 2, only if x is even)
• x = x / 3 (divide by 3, only if x is a multiple of 3)

Write a program to calculate the minimum number of operations needed to make x equal to 1 using the above three operations.

Input Format

The first line contains the integer x (1 ≤ x ≤ 10⁶).

Output Format

Output the minimum number of operations required to make the integer 1.

Example

Input:
10

Output:
3

Explanation:
10 → 9 → 3 → 1 (3 operations)

Approach to Solve the Problem

This problem can be attempted using Dynamic Programming. The state of the problem is the current integer x, and we can solve how to make it 1 by using previous states.

By using memoization, we can avoid duplicate calculations, and we can efficiently solve the problem by storing the minimum number of operations needed for each state.

Algorithm Plan

1. Create an array dp to store the minimum number of operations for all numbers from 1 to x.
2. Set the initial state dp[1] to 0 (1 is already 1).
3. Use a loop to iterate from 2 to x:
• dp[i] = dp[i-1] + 1 (the case of subtracting 1)
• If i is divisible by 2, then dp[i] = min(dp[i], dp[i/2] + 1) (the case of division by 2)
• Similarly, if i is divisible by 3, then dp[i] = min(dp[i], dp[i/3] + 1) (the case of division by 3)
4. Finally, output dp[x].

Swift Implementation Code

import Foundation

func minOperationsToOne(_ x: Int) -> Int {
    var dp = Array(repeating: Int.max, count: x + 1)
    dp[1] = 0
    
    for i in 2...x {
        dp[i] = dp[i - 1] + 1 // -1 operation
        
        if i % 2 == 0 {
            dp[i] = min(dp[i], dp[i / 2] + 1) // /2 operation
        }
        
        if i % 3 == 0 {
            dp[i] = min(dp[i], dp[i / 3] + 1) // /3 operation
        }
    }
    
    return dp[x]
}

let input = Int(readLine()!)!
print(minOperationsToOne(input))

Performance Evaluation

The time complexity of this algorithm is O(n). n is the given integer x. Since it considers a maximum of 3 operations for each integer, it is efficient.

The space complexity is O(n), as it uses the dp array to store all states. This can require a considerable amount of memory, so further optimization may be necessary depending on the needs.

Conclusion

The algorithm problem of making an integer 1 demonstrates the basic principles of dynamic programming well. By solving this problem, one can enhance their understanding of Swift coding tests and learn how to apply dynamic programming techniques.

Through the process of solving the problem, you can develop an understanding of each operation and cultivate a thought process to find the optimal solution. It is advisable to test the algorithm’s performance with various input values and also consider error handling and various optimization methods.

The problem of “Making an Integer 1” discussed in this article is a type that may frequently appear in real coding tests, so it is recommended to practice sufficiently to master it.

In this article, we will explore how to implement an absolute value heap using Swift. We will explain in detail the implementation specifics, test cases, and efficiency analysis while solving an algorithm problem. We will look at how to efficiently solve problems using Swift’s capabilities.

Problem Description

An absolute value heap is a data structure that follows these rules:

• The first element inserted is the one with the smallest absolute value.
• If the absolute values are the same, the smaller element value comes first.

Additionally, the operations that must be supported in the absolute value heap are as follows:

1. Insert a number into the heap.
2. Remove and return the number with the smallest absolute value from the heap.

The input consists of several commands, with each command being a form of inserting or removing a number. If the heap is empty, it outputs 0.

Input Format

The input is provided over multiple lines, and each line has the following format:

            1
            -1
            0
            1
            2 Ivy
            

Output Format

For each remove command, output the value at the top of the heap.

Problem Solving Process

To solve the problem, we will first define the structure of the absolute value heap and choose a data structure for it. A primary method to implement as a boilerplate is using an array-based heap structure. Here, we will utilize Swift’s basic data structure, Array.

Step 1: Designing the Heap Structure

To implement an absolute value heap, we need to define the structure and priority of the heap. We will prioritize the absolute value, and for elements with the same absolute value, we will base the priority on the actual number size. To do this, we will use a tuple that adopts the Comparable protocol.

Step 2: Implementing Insert Operation

When inserting into the heap, we add the new element to the end of the array and then move it up until the heap property is satisfied. This process is called ‘bubbling up’ or ‘up-heap’.

Step 3: Implementing Delete Operation

To delete the element with the smallest absolute value, we remove the root element, place the last element in the root position, and then move it down until the heap property is satisfied. This process is called ‘bubbling down’ or ‘down-heap’.

Swift Code Implementation

Now, let’s look at the actual Swift code based on the algorithm we have organized.

            import Foundation

            struct AbsoluteHeap {
                private var heap: [(value: Int, absolute: Int)] = []

                mutating func insert(_ value: Int) {
                    let absoluteValue = abs(value)
                    heap.append((value, absoluteValue))
                    upHeap(index: heap.count - 1)
                }

                mutating func pop() -> Int {
                    guard !heap.isEmpty else { return 0 }
                    let root = heap[0].value
                    heap[0] = heap.removeLast()
                    downHeap(index: 0)
                    return root
                }

                private mutating func upHeap(index: Int) {
                    var childIndex = index
                    let child = heap[childIndex]
                    var parentIndex = (childIndex - 1) / 2

                    while childIndex > 0 && (heap[parentIndex].absolute > child.absolute ||
                                              (heap[parentIndex].absolute == child.absolute && heap[parentIndex].value > child.value)) {
                        heap[childIndex] = heap[parentIndex]
                        childIndex = parentIndex
                        parentIndex = (childIndex - 1) / 2
                    }

                    heap[childIndex] = child
                }

                private mutating func downHeap(index: Int) {
                    var parentIndex = index
                    let count = heap.count

                    while true {
                        let leftChildIndex = 2 * parentIndex + 1
                        let rightChildIndex = 2 * parentIndex + 2
                        var candidateIndex = parentIndex

                        if leftChildIndex < count && (heap[leftChildIndex].absolute < heap[candidateIndex].absolute ||
                                                       (heap[leftChildIndex].absolute == heap[candidateIndex].absolute && heap[leftChildIndex].value < heap[candidateIndex].value)) {
                            candidateIndex = leftChildIndex
                        }
                        
                        if rightChildIndex < count && (heap[rightChildIndex].absolute < heap[candidateIndex].absolute ||
                                                        (heap[rightChildIndex].absolute == heap[candidateIndex].absolute && heap[rightChildIndex].value < heap[candidateIndex].value)) {
                            candidateIndex = rightChildIndex
                        }

                        if candidateIndex == parentIndex { break }
                        heap.swapAt(parentIndex, candidateIndex)
                        parentIndex = candidateIndex
                    }
                }
            }

            var absoluteHeap = AbsoluteHeap()
            let operations = [1, -1, 0, 1, 2]
            for op in operations {
                if op != 0 {
                    absoluteHeap.insert(op)
                } else {
                    print(absoluteHeap.pop())
                }
            }
            

Test Cases

Test Case 1

Input values used in the script: [1, -1, 0, 1, 2]

Expected output: 1 (when pop is performed)

Test Case 2

Additional input values used in the script: [2, -4, 0]

Expected output: -4 (when pop is performed)

Conclusion

We have now successfully implemented an absolute value heap using Swift. In this tutorial, we utilized the essence of algorithms, carefully chose data structures, and applied basic Swift syntax and structure to implement the heap's insertion and deletion processes. Through this implementation, I hope to solidify your prior knowledge of solving algorithm problems.

I look forward to solving various algorithm problems together in the future!

1. Problem Definition

The critical path problem is an important problem in graph theory, which calculates the minimum time required to complete a given set of tasks. Each task has a structure of dependencies on other tasks, which determines the execution order of the tasks. Through this problem, we need to create an optimal schedule that considers the dependencies between tasks.

2. Problem Description

Based on the given list of tasks and the execution time of each task, calculate the minimum time required to complete all tasks. The tasks are given in the following format:

    tasks = [
        (1, 3, [2]),        // Task 1 can start after Task 2 is completed
        (2, 2, []),         // Task 2 can be performed independently
        (3, 4, [1, 2])      // Task 3 can start after Tasks 1 and 2 are completed
    ]
    

Input Format

Each task is composed of a tuple, where the first element is the task ID, the second element is the execution time of the task, and the third element is a list of task IDs that it depends on.

Output Format

Output an integer representing the minimum time required to complete all tasks.

3. Solution Process

To solve this problem, we will follow these procedures.

1. Creating the Dependency Graph: Represent the dependencies between tasks in the form of a graph.
2. Finding the Longest Path: Use Depth-First Search (DFS) to find the longest path in order to calculate the start time of each task.
3. Calculating the Final Time: Calculate the time required to complete all tasks.

3-1. Creating the Dependency Graph

First, we transform the input tasks into a graph structure. Each task is represented as a node, and the dependencies are represented as edges.

3-2. Finding the Longest Path

Using depth-first search, we determine the timing for when each task can start. We record the start time of the task considering when its dependency tasks are finished. Then, we must visit all tasks to find the longest path.

3-3. Calculating the Final Time

Based on the time required to complete all tasks, we determine the completion time of the final task to calculate the minimum time.

4. Swift Code Implementation

Below is the Swift code based on the algorithm above.

    import Foundation

    struct Task {
        let id: Int
        let time: Int
        let dependencies: [Int]
    }

    func minimumCompletionTime(tasks: [Task]) -> Int {
        var graph = [Int: [Int]]() // Dependency graph
        var inDegree = [Int: Int]() // In-degree
        var completionTime = [Int: Int]() // Store completion time

        // Initialize the graph and in-degrees
        for task in tasks {
            graph[task.id] = []
            inDegree[task.id] = task.dependencies.count
            completionTime[task.id] = 0
        }

        // Create the graph
        for task in tasks {
            for dependency in task.dependencies {
                graph[dependency]?.append(task.id)
            }
        }

        var queue = [Int]() // Tasks that can start
        for task in tasks {
            if inDegree[task.id] == 0 {
                queue.append(task.id)
                completionTime[task.id] = task.time // Initialize the time for independent tasks
            }
        }

        var maxCompletionTime = 0

        // Queue for topological sorting
        while !queue.isEmpty {
            let currentTaskId = queue.removeFirst()
            let currentTaskTime = completionTime[currentTaskId]!

            // Visit child tasks of the current task
            for dependentTaskId in graph[currentTaskId]! {
                completionTime[dependentTaskId] = max(completionTime[dependentTaskId]!, currentTaskTime + tasks.first { $0.id == dependentTaskId }!.time)

                // Decrease the in-degree
                inDegree[dependentTaskId]! -= 1
                // Add to the queue if the in-degree is 0
                if inDegree[dependentTaskId] == 0 {
                    queue.append(dependentTaskId)
                }
            }
            maxCompletionTime = max(maxCompletionTime, currentTaskTime)
        }

        return maxCompletionTime
    }

    // Test case
    let tasks = [
        Task(id: 1, time: 3, dependencies: [2]),
        Task(id: 2, time: 2, dependencies: []),
        Task(id: 3, time: 4, dependencies: [1, 2])
    ]

    let result = minimumCompletionTime(tasks: tasks)
    print("Minimum time required to complete all tasks: \(result) seconds")
    

5. Code Explanation

The code above constructs a graph to solve the critical path problem and calculates the completion time for each task through depth-first search. It checks the dependencies for each task, updating completion times, and ultimately outputs the minimum time needed to complete all tasks.

6. Conclusion

The critical path problem is an important algorithmic problem for managing dependencies between tasks. By solving this problem, one can enhance their understanding of graph theory, and it is also a frequently posed type in coding tests. Let’s practice various forms of problems based on the methods described above!

This course covers the problem of finding binomial coefficients. The binomial coefficient is an important concept in combinatorics, representing the number of ways to choose k items from n items. This type of problem is frequently asked in algorithm coding tests.

Problem Description

Write a program to calculate the binomial coefficient C(n, k) for the given two integers n (0 ≤ n ≤ 30) and k (0 ≤ k ≤ n). The binomial coefficient C(n, k) is defined as follows:

C(n, k) = n! / (k! * (n - k)!)

An example of the problem is as follows:

Example

• Input: 5 2
• Output: 10

Problem Solving Process

Recursive Property of Binomial Coefficient

The binomial coefficient has the following recursive property:

C(n, k) = C(n - 1, k - 1) + C(n - 1, k)

Here, C(n, 0) = 1 and C(n, n) = 1. By utilizing this property, we can use a recursive function to calculate the binomial coefficient. However, this method can be inefficient due to deep recursive calls.

Solution via Dynamic Programming

This problem can be solved using dynamic programming. Dynamic programming improves algorithm performance by avoiding redundant calculations. The following table can be used to derive the value of C(n, k).

Dynamic Programming Approach

To calculate the binomial coefficient using dynamic programming, we declare the following 2D array. dp[i][j] will store the value for C(i, j).

var dp = [[Int]](repeating: [Int](repeating: 0, count: n + 1), count: n + 1)

for i in 0...n {
    for j in 0...i {
        if j == 0 || j == i {
            dp[i][j] = 1
        } else {
            dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
        }
    }
}

Swift Code Implementation

Based on the above dynamic programming approach, let’s write a Swift program to calculate the binomial coefficient.

import Foundation

func binomialCoefficient(n: Int, k: Int) -> Int {
    var dp = [[Int]](repeating: [Int](repeating: 0, count: k + 1), count: n + 1)

    for i in 0...n {
        for j in 0...min(i, k) {
            if j == 0 || j == i {
                dp[i][j] = 1
            } else {
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
            }
        }
    }
    return dp[n][k]
}

// Example Input
let n = 5
let k = 2

// Print Result
let result = binomialCoefficient(n: n, k: k)
print("C(\(n), \(k)) = \(result)")

Result Verification

Running the above code will output 10 for C(5, 2). This accurately calculates the number of ways to choose 2 items from 5 items.

Time Complexity Analysis

The time complexity of this algorithm is O(n*k). In this case, n is 30, and k is proportional to n, so it will be at most 30. The number of binomial coefficients calculated in this manner is efficient, making it highly suitable for solving the problem.

Conclusion

In this course, we addressed the problem of calculating binomial coefficients and learned how to implement efficient algorithms using dynamic programming. By using a 2D array to store each binomial coefficient, we avoided recursive calls and improved performance. This problem helps establish foundational concepts in combinatorics and dynamic programming.

In the next course, we will cover another algorithm problem. I hope you continue to enhance your algorithm-solving skills through ongoing learning!

Hello everyone! Today we will tackle the problem of calculating binomial coefficients using Swift. This course is designed to be easy to understand for those preparing for employment and will explain the process of solving the problem step by step.

Problem Description

Binomial coefficient refers to the number of combinations that represent ‘the number of ways to choose k from n’. Mathematically, the binomial coefficient is defined as follows:

C(n, k) = n! / (k! * (n – k)!)

Here, n is the total number of elements, k is the number of elements to choose, and ! signifies factorial.

Problem

    Write a program to calculate the binomial coefficient C(n, k) for given n and k. 
    n and k are integers between 0 and 30, inclusive.

Problem Solving Process

Step 1: Understanding the Problem

Understanding the problem is the most important step in solving algorithmic problems. In this problem, we need to calculate the binomial coefficient for the given two integers n and k. The binomial coefficient is a form of combination, governed by a mathematical formula. Let’s organize the necessary information to solve this problem.

Conditions to calculate the binomial coefficient:

• 0 ≤ k ≤ n
• 0 ≤ n ≤ 30

Step 2: Mathematical Approach

To easily calculate the binomial coefficient, we can use either a recursive function or the dynamic programming approach. Here, we will consider both the recursive method and dynamic programming, focusing on the latter.

Recursive Method

The binomial coefficient has the following property:

C(n, k) = C(n-1, k-1) + C(n-1, k)

This shows that we can divide it into two cases: one where we select the nth element and one where we do not. However, this method can be inefficient due to repeated calculations.

Dynamic Programming

Using dynamic programming allows us to avoid redundant calculations. First, we create an array to store the binomial coefficients and calculate the needed values while storing results in the array.

Step 3: Algorithm Design

Now it’s time to design the algorithm using dynamic programming. We will follow this process to design the algorithm:

1. Define a two-dimensional array to create space for storing binomial coefficient values. Set the size of the array to (n+1) x (k+1).
2. Fill in the base conditions. C(n, 0) = 1, C(n, n) = 1.
3. Calculate the binomial coefficient for the given n and k and store values in the array.
4. Output the calculated binomial coefficient.

Step 4: Code Implementation

Now let’s write the actual Swift code. We can implement a program that calculates the binomial coefficient using dynamic programming as follows.

import Foundation

func binomialCoefficient(n: Int, k: Int) -> Int {
    // Initialize the dynamic array
    var dp = Array(repeating: Array(repeating: 0, count: k + 1), count: n + 1)
    
    // Set base conditions
    for i in 0...n {
        dp[i][0] = 1 // C(n, 0) = 1
        dp[i][i] = 1 // C(n, n) = 1
    }
    
    // Calculate the binomial coefficients
    for i in 1...n {
        for j in 1...min(i, k) {
            dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
        }
    }
    
    return dp[n][k]
}

// Input values
let n = 5
let k = 2
let result = binomialCoefficient(n: n, k: k)
print("C(\(n), \(k)) = \(result)")

Step 5: Code Explanation

The above code defines a function that calculates the binomial coefficient. The key concepts used here are dynamic programming and storing results in an array.

• First, the dp array is used to store each binomial coefficient.
• Secondly, the base conditions are set by initializing dp[i][0] and dp[i][i] to 1. This capitalizes on the fact that there is only one way to choose 0 or n elements from n elements.
• Then, using the properties we created above, we calculate the remaining binomial coefficients via a loop.

Step 6: Time Complexity

The time complexity of this algorithm is O(n * k), which increases proportionately with n and k. This is a significant advantage as it allows us to calculate the binomial coefficient efficiently using dynamic programming.

Conclusion

Today we learned how to calculate the binomial coefficient using Swift. We explained the process step by step, starting from basic concepts to the implementation of algorithms. This can be applied not only to actual algorithm problems but also to similar problems, so be sure to practice!

Now it is your turn to find your own way to calculate binomial coefficients. Do you understand the code and the algorithm? In the next lesson, we will tackle another problem related to binomial coefficients!

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