Author: root

Problem Description

Imagine! You, from the future, decide to travel to a specific time using a time machine. However, this time machine can only operate through a complex algorithm. The problem is to calculate the minimum number of jumps needed to reach the given time.

Problem Definition

Two integers S and E are given. S is the starting time and E is the arrival time. The time machine operates according to the following rules:

• From the current time, you can jump to the next time by performing +1, +2, or *2 operations.
• Write a program to calculate and output the minimum number of jumps taken by the time machine to operate.

Input

In the first line, two integers S (1 ≤ S ≤ 10⁵) and E (1 ≤ E ≤ 10⁵) are given.

Output

Output the minimum number of jumps required for the time machine to reach E.

Example

    Input:
    4 7

    Output:
    2
    

Problem Solving Process

This problem can be solved using the BFS (Breadth-First Search) algorithm. BFS starts from a specific node and explores all nodes connected to it before exploring the nodes of the next layer. In this problem, each time zone can be represented as a node in a graph, while each jump can be represented as an edge.

Step-by-Step Solution

1. Initialize the Queue: To perform BFS, first initialize the queue. Add the starting time S and the current jump count to the queue.
2. Visit Array: Create a visit array to record the times that have already been visited. This prevents visiting the same time multiple times.
3. Perform BFS: Remove a node from the queue and explore the next times by performing all possible jumps. If the next jump reaches the arrival time E, return the current jump count.
4. Termination Condition: If the queue becomes empty without reaching the arrival time, terminate the jump process.

Code Implementation

fun minJumps(S: Int, E: Int): Int {
    val queue: Queue> = LinkedList()
    val visited = BooleanArray(100001) // maximum time range 100000
    queue.add(Pair(S, 0))
    visited[S] = true

    while (queue.isNotEmpty()) {
        val (currentTime, jumps) = queue.poll()

        // Reached the destination
        if (currentTime == E) {
            return jumps
        }

        // Next possible times
        val nextTimes = listOf(currentTime + 1, currentTime + 2, currentTime * 2)

        for (nextTime in nextTimes) {
            if (nextTime in 1..100000 && !visited[nextTime]) {
                visited[nextTime] = true
                queue.add(Pair(nextTime, jumps + 1))
            }
        }
    }
    return -1 // unreachable case
}

fun main() {
    val input = readLine()!!.split(" ")
    val S = input[0].toInt()
    val E = input[1].toInt()
    println(minJumps(S, E))
}
    

Time Complexity Analysis

The time complexity of this algorithm is O(n). Since all time zones will be visited once, performing BFS for up to 100,000 nodes is efficient.

Conclusion

In this lecture, we explored how to apply the BFS algorithm to solve the time machine problem and calculate the shortest distance. Understanding and applying the advantages of implementing it in Kotlin as well as the basic principles of BFS will help you in coding tests. In the next session, we will prepare a lecture to explore other problems and more diverse algorithms!

Quick Sort is a very efficient sorting algorithm, especially with an average time complexity of O(n log n) and a space complexity of O(log n). In this article, we will take a detailed look at the concept of Quick Sort, how to implement it in Kotlin, algorithm problems, and the problem-solving process.

1. Concept of Quick Sort

Quick Sort is a type of divide and conquer algorithm that operates in the following steps:

1. Select one element from the given array as a pivot. Usually, the first element, the last element, or the middle element of the array is chosen.
2. Split the array into two sub-arrays based on the pivot. Elements smaller than the pivot are positioned on the left, and larger elements on the right.
3. Recursively apply Quick Sort on the left and right sub-arrays.

This process is repeated until the array cannot be divided any further. The greatest advantage of Quick Sort is that it is very fast on average, and it is an in-place sorting algorithm. Therefore, it uses very little additional memory.

2. Example of Quick Sort in Action

Assume the array to be sorted is [3, 6, 8, 10, 1, 2, 1]. Below is the step-by-step process of Quick Sort.

Step 1: Choose a Pivot

Pivot: 1

Step 2: Split the Array

[1, 2, 1, 10, 6, 8, 3]

Step 3: Recursive Call

• The left array [1, 1] is already sorted.
• Apply Quick Sort again on the right array [10, 6, 8, 3].

Step 4: Final Sort

[1, 1, 2, 3, 6, 8, 10]

3. Algorithm Problem

Now, let’s solve a problem that returns a sorted array using Quick Sort.

Problem: Sort the given integer array using Quick Sort.

Input: [10, 7, 8, 9, 1, 5]

Output: [1, 5, 7, 8, 9, 10]

4. Kotlin Implementation

Below is the Kotlin code to solve the above problem.

fun quickSort(arr: IntArray, low: Int, high: Int) {
    if (low < high) {
        val pi = partition(arr, low, high)
        quickSort(arr, low, pi - 1)
        quickSort(arr, pi + 1, high)
    }
}

fun partition(arr: IntArray, low: Int, high: Int): Int {
    val pivot = arr[high]
    var i = (low - 1)

    for (j in low until high) {
        if (arr[j] <= pivot) {
            i++
            val temp = arr[i]
            arr[i] = arr[j]
            arr[j] = temp
        }
    }

    val temp = arr[i + 1]
    arr[i + 1] = arr[high]
    arr[high] = temp
    return i + 1
}

fun main() {
    val arr = intArrayOf(10, 7, 8, 9, 1, 5)
    quickSort(arr, 0, arr.size - 1)
    println("Sorted array: ${arr.joinToString(", ")}")
}

5. Code Explanation

5.1 quickSort Function

The quickSort function is called recursively, taking the starting and ending indices of the array as arguments. If the starting index is less than the ending index, the partition function is called to find the position of the pivot, based on which the array is rearranged.

5.2 partition Function

The partition function is responsible for dividing the array based on the pivot. It checks all elements and moves the elements smaller than the pivot to the left array. Finally, it places the pivot in the correct position and returns that index.

6. Complexity Analysis

The time complexity of Quick Sort is as follows:

• Best case: O(n log n)
• Average case: O(n log n)
• Worst case: O(n²) – This can occur if an already sorted array is provided as input.

The space complexity is O(log n). This is due to stack memory usage from recursive calls.

7. Conclusion

In this article, we explored how to implement the Quick Sort algorithm in Kotlin. Quick Sort is widely used due to its efficiency and simplicity. It often appears in situations like coding tests, so a fundamental understanding and practice are very important.

8. Additional Practice Problems

Below are some additional practice problems where Quick Sort can be applied:

• Implement a method to find the k-th smallest element in the array.
• Implement Quick Sort non-recursively.
• Solve the problem of merging two sorted arrays.

Quick Sort is a great way to learn various important concepts in algorithms and data structures. Continue practicing and expand your understanding into deep knowledge.

In this course, we will explore ‘Kevin Bacon’s Six Degrees of Separation’ and examine the process of solving related algorithm problems. Kevin Bacon’s Six Degrees of Separation is the theory that everyone is connected to Kevin Bacon within six degrees. Based on this, we will solve problems using graph theory.

Problem Description

The problem is to represent given people and their relationships in the form of a graph, and to identify how a specific individual relates to all other individuals. We define the problem as follows:

Problem: Kevin Bacon's Six Degrees of Separation

    
Given n people and m relationships.
    
People are represented by integers from 1 to n.
    
Each relationship is given by the IDs of two people, indicating that they are friends.
    
Please calculate each person's Kevin Bacon number. The Kevin Bacon number is the length of the shortest path from that person to another person.
    
Finally, print the ID of the person with the smallest Kevin Bacon number. If there are multiple individuals, select the one with the smallest ID.

Input Format

  
ID1 ID2
ID1 ID3
...

Output Format

The ID of the person with the smallest Kevin Bacon number

Problem Solving Strategy

To solve the problem, we will follow these steps:

1. Store people and relationships in the form of a graph based on the input.
2. Use Dijkstra’s algorithm or BFS (Breadth First Search) to calculate each person’s Kevin Bacon number.
3. Compare all people’s Kevin Bacon numbers and print the ID of the person with the smallest number.

Code Implementation

We will implement this problem in Kotlin.

import java.util.*

fun main() {
    val reader = Scanner(System.`in`)
    val n = reader.nextInt() // Number of people
    val m = reader.nextInt() // Number of relationships

    // Initialize adjacency list
    val graph = Array(n + 1) { mutableListOf() }

    // Receive relationships input
    for (i in 0 until m) {
        val u = reader.nextInt()
        val v = reader.nextInt()
        graph[u].add(v)
        graph[v].add(u)
    }

    // Array to store each person's Kevin Bacon number
    val kevinBaconScores = IntArray(n + 1)

    // Define BFS method
    fun bfs(start: Int) {
        val queue: Queue> = LinkedList()
        val visited = BooleanArray(n + 1)
        queue.add(Pair(start, 0)) // Starting person with distance 0
        visited[start] = true

        while (queue.isNotEmpty()) {
            val (current, distance) = queue.poll()
            for (neighbor in graph[current]) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true
                    queue.add(Pair(neighbor, distance + 1))
                    kevinBaconScores[start] += distance + 1 // Calculate Kevin Bacon score
                }
            }
        }
    }

    // Perform BFS for all people
    for (i in 1..n) {
        bfs(i)
    }

    // Find the person with the minimum Kevin Bacon number
    var minScore = Int.MAX_VALUE
    var resultId = -1
    
    for (i in 1..n) {
        if (kevinBaconScores[i] < minScore) {
            minScore = kevinBaconScores[i]
            resultId = i
        }
    }

    // Print the result
    println(resultId)
}

Process Description

First, we input the number of people (n) and relationships (m) to create the graph, storing the relationships in adjacency list form. Then, we use the BFS algorithm to calculate each person’s Kevin Bacon number. BFS explores all connected individuals starting from each person, increasing the distance and ultimately recording each person’s Kevin Bacon score.

Finally, we compare all individuals’ Kevin Bacon scores to find the smallest value. In case of a tie, the person with the smaller ID is chosen.

Conclusion

Through this course, we learned how to solve algorithm problems using Kevin Bacon’s Six Degrees of Separation. Understanding graph theory and the BFS algorithm is very useful for technical interviews. Practice these problems extensively to prepare effectively for coding tests.

Hello! In this Kotlin coding test course, we will cover an interesting problem called “Card Sorting.” This will be an opportunity to enhance logical thinking and understanding of various data structures while solving algorithmic problems.

Problem Description

The problem is to sort the given cards. Each card consists of the following information:

• Value: The number on the card (e.g., 1, 2, 3)
• Color: The color of the card (e.g., Clubs, Diamonds, Hearts, Spades)

Please write a program to sort the given cards by size and color. The size of the card should be sorted in ascending order based on the numbers, and if the numbers are the same, it should be sorted by color. The order of colors is Clubs < Diamonds < Hearts < Spades.

Input Format

The input is provided as an array consisting of several cards. Each card is represented as an element in the array in the Pair(value, color) format. For example, it can be in the form of [Pair(3, 'Hearts'), Pair(2, 'Spades'), Pair(3, 'Diamonds'), Pair(1, 'Clubs'), Pair(2, 'Hearts')].

Output Format

The output is the sorted array of cards. The information of each card should be expressed in the Pair(value, color) format.

Process of Solving the Problem

1. Understanding the Problem

The key point of the problem is to sort the cards. Two criteria are needed to perform the sorting:

• Value: The number on the card
• Color: Processed according to the defined order of card types

2. How to Sort the Data?

To define the sorting criteria, it is necessary to map the priorities of colors to numbers. For example, the mapping can be done as follows:

• Clubs: 1
• Diamonds: 2
• Hearts: 3
• Spades: 4

This allows for easy creation of sorting conditions based on color.

3. Implementing in Kotlin

Now, let’s write the code based on the explanation above. Below is an example implementation of a card sorting program using Kotlin:

data class Card(val value: Int, val color: String)

fun main() {
    val colors = mapOf(
        "Clubs" to 1,
        "Diamonds" to 2,
        "Hearts" to 3,
        "Spades" to 4
    )
    
    val cards = listOf(
        Card(3, "Hearts"),
        Card(2, "Spades"),
        Card(3, "Diamonds"),
        Card(1, "Clubs"),
        Card(2, "Hearts")
    )
    
    val sortedCards = cards.sortedWith(compareBy({ it.value }, { colors[it.color] }))
    
    println("Sorted card list:")
    for (card in sortedCards) {
        println("Value: ${card.value}, Color: ${card.color}")
    }
}

4. Explanation of the Code

Let me explain each component of the code above:

• data class Card: Defines a data class that includes the value and color of the card.
• colors map: Creates a map that maps the card colors to integers. This value is referenced during sorting.
• cards list: Creates a list of the given card list.
• sortedWith: Sorts the card list by value and color. The compareBy function allows for sorting by multiple criteria.

5. Checking the Results

When you run the program, the output will be as follows:

Sorted card list:
Value: 1, Color: Clubs
Value: 2, Color: Hearts
Value: 2, Color: Spades
Value: 3, Color: Diamonds
Value: 3, Color: Hearts

Conclusion

In this course, we dealt with the card sorting problem, exploring the problem-solving process using sorting algorithms and implementing it in Kotlin code. Understanding various data structures and algorithms is crucial in coding tests, so continued practice will be beneficial. We will cover more interesting problems in the next course. Thank you!