Kotlin Coding Test Course, Quick Sort

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.

Kotlin Coding Test Course, Kevin Bacon’s Six Degrees of Separation

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.

kotlin coding test course, cocktail making

Today, we will explore how to solve a simple algorithm problem using Kotlin.
The topic is “Making Cocktails.” This problem helps us learn efficient algorithm design and implementation
by combining various ingredients to create the desired cocktail.
After understanding the basic concept of the problem, we can write the code and conduct tests to
familiarize ourselves with several important elements in actual coding tests.

Problem Description

We are now bartenders at a cocktail bar. The customer has requested a specific color, and we need to
create a cocktail that matches that color using various ingredients.

The following input is given:

  • The number of ingredients N (1 <= N <= 100)
  • An array of colors representing each ingredient (colors are provided as strings)
  • The color requested by the customer targetColor (string)

We need to determine whether we can make a cocktail of the requested color.
If it can be made, we should output “Possible”; otherwise, we output “Impossible.”

Input Example

        5
        "red", "blue", "green", "yellow", "red"
        "purple"
    

Output Example

Possible

Algorithm Design

To solve this problem, we need to perform a few simple operations.
First, we must check all the given colors to see if any match the customer’s request.
We can approach it as follows:

  1. Search the array of colors to see if it contains a color that matches the customer’s request.
  2. If a matching color is found, print “Possible” and terminate the iteration.
  3. If no matching color is found after checking all colors, print “Impossible.”

Code Writing

Now, let’s write Kotlin code based on the algorithm designed above.
Using Kotlin’s concise syntax, we can implement the solution to the problem as follows.

        fun canMakeCocktail(colors: Array, targetColor: String): String {
            for (color in colors) {
                if (color == targetColor) {
                    return "Possible"
                }
            }
            return "Impossible"
        }

        fun main() {
            val colors = arrayOf("red", "blue", "green", "yellow", "red")
            val targetColor = "purple"
            println(canMakeCocktail(colors, targetColor))
        }
    

Code Explanation

In the code above, the canMakeCocktail function takes two parameters.
The first parameter is an array of colors, and the second parameter is the color requested by the customer.
It uses a for loop to iterate through the array, checking for a match with the requested color.

If a matching color is found, it returns “Possible” and terminates.
If no matches are found after checking all colors, it returns “Impossible.”
The main function sets the given array of colors and the requested color,
and calls the canMakeCocktail function to output the result.

Test Cases

Now, let’s test the code. We will check our code’s correctness through various input scenarios.

1. Typical case

        Input: ["red", "blue", "green", "yellow", "red"], "green"
        Output: Possible
    

2. Case where requested color does not exist

        Input: ["red", "blue", "green", "yellow", "red"], "orange"
        Output: Impossible
    

3. Case with duplicate colors

        Input: ["red", "blue", "blue", "yellow", "red"], "blue"
        Output: Possible
    

4. Case where all colors are the same

        Input: ["red", "red", "red", "red", "red"], "red"
        Output: Possible
    

5. Case where color array is empty

        Input: [], "red"
        Output: Impossible
    

Result Analysis

All test cases produced the expected results.
This confirms that the written code meets the requirements and operates
as an algorithm that considers various scenarios.

Conclusion

In this lesson, we learned how to solve a simple algorithm problem using Kotlin.
Through the cocktail-making problem, we explored array searching, conditional statements, and function definitions.
When solving algorithm problems, it is essential to clearly understand the problem requirements and to select
appropriate data structures and algorithms accordingly.

I hope you all continue to enhance your algorithm skills through various problems.
Next time, I will return with a more complex problem.
Thank you!

Kotlin coding test course, Card Sorting

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!

Kotlin Coding Test Course, Understanding Friend Relationships

Problem Description

There are N students who are friends with each other. The friendships are given in pairs,
and if student A is friends with student B, then A is a friend of B and B is a friend of A.
Now, you need to implement a function to identify all friends of friends for a specific student and
return the count of those friends.

For example, if the friendships are given as follows:

  • (1, 2)
  • (1, 3)
  • (2, 4)
  • (3, 5)

Student 1’s friends are 2 and 3, and their friends are 4 and 5.
Therefore, the total number of friends of friends for student 1 is 4.

Input Format

friends: List>, target: Int

friends: A list of tuples representing friendships,
target: The student whose friends of friends you want to find

Output Format

– An integer representing the number of friends of friends

Problem Solving Process

To understand friendships, we can apply graph theory.
Each student becomes a node, and friendships are represented as edges.
This allows us to explore the given student and their friends’ friends.

1. Constructing the Graph

First, we need to convert the friendships into a graph format.
Let’s create a map with each student as keys and their friends as values.


fun buildGraph(friends: List>): Map> {
    val graph = mutableMapOf>()
    for ((a, b) in friends) {
        graph.computeIfAbsent(a) { mutableListOf() }.add(b)
        graph.computeIfAbsent(b) { mutableListOf() }.add(a)
    }
    return graph
}
            

2. Finding Friends of Friends

Once the graph is constructed, we need to find the friends of a specific student and
obtain the list of their friends’ friends.
Care must be taken to ensure that the friends list does not contain duplicates.


fun findFriendsOfFriends(friends: List>, target: Int): Int {
    val graph = buildGraph(friends)
    val directFriends = graph[target] ?: return 0
    val friendsOfFriends = mutableSetOf()
    
    for (friend in directFriends) {
        val friendsOfCurrentFriend = graph[friend]
        if (friendsOfCurrentFriend != null) {
            for (fof in friendsOfCurrentFriend) {
                if (fof != target && !directFriends.contains(fof)) {
                    friendsOfFriends.add(fof)
                }
            }
        }
    }
    return friendsOfFriends.size
}
            

3. Complete Code

Now, let’s organize the complete code.


fun main() {
    val friendsList = listOf(Pair(1, 2), Pair(1, 3), Pair(2, 4), Pair(3, 5))
    val targetStudent = 1
    val result = findFriendsOfFriends(friendsList, targetStudent)
    println("Number of friends of friends for student $targetStudent: $result")
}
            

Complexity Analysis

The time complexity of this algorithm is O(N + M).
Here, N is the number of students, and M is the number of friendships.
The space complexity is also O(N + M).
This is because each student and friendship is stored in graph form.

Example and Test Cases

Let’s verify that the function works correctly with various examples.

  • Example 1:

    Input:

    
                        friendsList = listOf(Pair(1, 2), Pair(1, 3), Pair(2, 4), Pair(3, 5))
                        targetStudent = 1
                        

    Output: “Number of friends of friends for student 1: 2”

  • Example 2:

    Input:

    
                        friendsList = listOf(Pair(1, 2), Pair(1, 3), Pair(2, 3))
                        targetStudent = 1
                        

    Output: “Number of friends of friends for student 1: 0”

  • Example 3:

    Input:

    
                        friendsList = listOf(Pair(1, 2), Pair(3, 4), Pair(4, 5))
                        targetStudent = 1
                        

    Output: “Number of friends of friends for student 1: 0”

Conclusion

Understanding friendship relationships is a common type of algorithm problem.
It can be effectively solved through graph structures,
and care must be taken at each step.
I hope this tutorial helped you learn how to solve problems using Kotlin’s features.
As you tackle more algorithm problems, may you continue to build your coding skills.