Swift Coding Test Course, Finding the Critical Path

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!

Swift Coding Test Course, Finding Binomial Coefficient 2

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!

Swift Coding Test Course, Finding Binomial Coefficient 1

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!

Please subscribe and like! Thank you!

Swift Coding Test Course, Find the Number of Colloquial Expressions

Author: [Author Name]

Date: [Date]

1. What is a Binary Friend Number?

A Binary Friend Number refers to a binary string that satisfies the following conditions:

  • It is composed only of 0s and 1s.
  • There cannot be consecutive digits ‘1’. In other words, the substring ’11’ must not exist.
  • The string must start and end with 0.

For example, ‘010’, ‘0010’, and ‘1000’ are Binary Friend Numbers. In contrast, ’11’, ‘110’, ‘0110’, and ‘0001’ are not Binary Friend Numbers.

2. Problem Definition of Binary Friend Numbers

Given a natural number n, we define the problem of finding the count of Binary Friend Numbers of length n. This problem can be efficiently solved using Dynamic Programming.

3. Problem Examples

For example, the Binary Friend Numbers of length 1 are ‘0’ and ‘1’, totaling 2. However, the Binary Friend Numbers of length 2 are ’00’, ’01’, and ’10’, totaling 3. The Binary Friend Numbers of length 3 are ‘000’, ‘001’, ‘010’, ‘100’, and ‘101’, totaling 5, while for length 4, they are ‘0000’, ‘0001’, ‘0010’, ‘0100’, ‘0101’, ‘1000’, ‘1001’, and ‘1010’, totaling 8. One can find patterns in this manner.

4. Approach to the Problem

This problem has the following recursive properties:

  • A Binary Friend Number of length n can be derived from Binary Friend Numbers of length n-1 and has two cases: one that ends with 0 and one that ends with 1.
  • Thus, it can be expressed as dp[n] = dp[n-1] + dp[n-2].

5. Implementation Using Dynamic Programming

Now, based on the above relationship, let’s write a code to compute Binary Friend Numbers in Swift. Below is an example of Swift code:

            
                func countBinaryFriends(n: Int) -> Int {
                    guard n > 1 else { return n }
                    
                    var dp = [Int](repeating: 0, count: n + 1)
                    dp[1] = 2 // 0, 1
                    dp[2] = 3 // 00, 01, 10
                    
                    for i in 3...n {
                        dp[i] = dp[i - 1] + dp[i - 2]
                    }
                    
                    return dp[n]
                }

                let n = 4 // Example input
                print(countBinaryFriends(n: n)) // Output the count of Binary Friend Numbers
            
        

6. Time Complexity and Space Complexity

The above algorithm has a time complexity of O(n) concerning n. Additionally, since it utilizes a dp array, it has a space complexity of O(n). If optimized, the space complexity can be reduced to O(1) by only remembering the two previous values:

            
                func optimizedCountBinaryFriends(n: Int) -> Int {
                    guard n > 1 else { return n }
                    
                    var prev1 = 2 // dp[1]
                    var prev2 = 3 // dp[2]
                    var current = 0

                    for i in 3...n {
                        current = prev1 + prev2
                        prev1 = prev2
                        prev2 = current
                    }
                    
                    return current
                }

                let n = 4 // Example input
                print(optimizedCountBinaryFriends(n: n)) // Output the count of optimized Binary Friend Numbers
            
        

7. Conclusion

Through the above process, we were able to solve the problem of finding Binary Friend Numbers. This problem serves as a good example for understanding the basics of Dynamic Programming. I hope you understand and remember the patterns that occur during the process of finding Binary Friend Numbers, as this will help in solving similar problems.

8. Additional Learning Materials

Additionally, it is important to review and practice problems related to algorithms and data structures. Below are recommended resources:

Swift Coding Test Course, Binary Tree

Hello! Today, we will solve a coding test problem related to binary trees. Binary trees are a data structure that appears frequently in many problems. A binary tree is a tree structure where each node can have a maximum of two children, allowing for various traversal methods. Our goal is to understand binary trees and utilize them to solve specific problems.

Problem Description

Given the following binary tree, write a function that traverses all nodes using DFS (Depth-First Search) and returns the values of the nodes as a list.

Problem Definition

func depthFirstTraversal(root: TreeNode?) -> [Int] {}

Input: The root node of the binary tree, root

Output: An integer array containing the values of the nodes from the DFS traversal

Example

Input:


        1
       / \
      2   3
     / \
    4   5
    

Output:

[1, 2, 4, 5, 3]

Definition of Binary Tree

A binary tree is a tree structure that can have two child nodes. Each node has a value, and binary trees are usually defined recursively. Since a node may be empty, the root node should be handled as nil.

Problem Solving Process

To solve this problem, we will explore the tree using the DFS method. DFS means Depth-First Search, which involves completely traversing one branch before moving on to the next. Below, we explain the process of tree traversal using DFS.

Step 1: Define Tree Node

First, we need to define a tree node. We will implement the TreeNode class to define a tree node.


    class TreeNode {
        var value: Int
        var left: TreeNode?
        var right: TreeNode?
        
        init(value: Int) {
            self.value = value
        }
    }
    

Step 2: Implement DFS Function

Now we will implement a function to traverse the binary tree using the DFS method. The priority is current node -> left child -> right child. We will perform this recursively.


    func depthFirstTraversal(root: TreeNode?) -> [Int] {
        guard let node = root else { return [] }
        
        // Add the current node's value to the array
        var result = [node.value]
        
        // Explore the left subtree
        result += depthFirstTraversal(root: node.left)
        
        // Explore the right subtree
        result += depthFirstTraversal(root: node.right)
        
        return result
    }
    

Step 3: Test and Validate

Now, we will create a binary tree to test the function we have implemented. We will use the example above to create the tree.


    let root = TreeNode(value: 1)
    let leftChild = TreeNode(value: 2)
    let rightChild = TreeNode(value: 3)
    let leftLeftChild = TreeNode(value: 4)
    let leftRightChild = TreeNode(value: 5)
    
    root.left = leftChild
    root.right = rightChild
    leftChild.left = leftLeftChild
    leftChild.right = leftRightChild
    
    let result = depthFirstTraversal(root: root)
    print(result)  // [1, 2, 4, 5, 3]
    

Conclusion

We have implemented a DFS algorithm to traverse binary trees. Through this problem, we were able to understand the structure of binary trees and the concept of DFS traversal. In Swift, trees can be easily traversed using recursion, and problems like this are common in coding tests, making it important to practice. In the next session, we will explore another algorithmic problem. Thank you!