KOTLIN CODING TEST COURSE, IMPLEMENTING ABSOLUTE VALUE HEAP

Kotlin is a programming language that has recently gained popularity among many developers due to its modern and concise syntax as well as its powerful features. In coding tests, specifically algorithm problem-solving, Kotlin demonstrates its usefulness as well. In this post, I would like to explain in detail how to implement an Absolute Value Heap using Kotlin.

Problem Description

The definition of an absolute value heap is as follows:

An absolute value heap is a priority queue that always places the smallest absolute value at the top.
If there are multiple values with the same absolute value, the actual smaller value is prioritized.

The operations of the heap are as follows:

  • INSERT X: Adds integer X (to the priority queue)
  • EXTRACT: Deletes and outputs the integer with the smallest absolute value

Input Format

The program’s input is provided over multiple lines, each containing a command.
Each line is given in the form of ‘INSERT X’ or ‘EXTRACT’.

Output Format

Each time an ‘EXTRACT’ command is called, the outputted value is printed on a new line.

Example


    INPUT:
    INSERT -5
    INSERT 3
    INSERT -1
    EXTRACT
    EXTRACT
    INSERT -2
    EXTRACT
    EXTRACT
    EXTRACT
    

OUTPUT:
-1
-2
3

Problem Analysis

To solve the given problem, we need to consider the following:

  • How should we structure the priority queue?
  • Should we utilize Java’s heap structure or Kotlin’s collections?
  • How do we sort based on absolute values?

To create an absolute value heap, we can use a min-heap structure.
In Java, we can easily implement this using PriorityQueue.
We can also utilize this in Kotlin, and we can implement the desired functionality using a basic heap structure.

Implementation Process

1. Define Data Structure

The most crucial aspect of the absolute value heap is how we store the data.
Since we need to sort based on absolute values, we will use PriorityQueue to add and remove elements.
When adding elements, it is advisable to set both the absolute value and the original value.

2. Define Comparator

To define priorities in the PriorityQueue, we define a custom sorting rule (Comparator).
Here, values with smaller absolute values should come first, and in the case of equal absolute values, the actual smaller value should be prioritized.


    val comparator = Comparator> { a, b ->
        if (Math.abs(a.first) == Math.abs(b.first)) {
            a.first.compareTo(b.first)
        } else {
            Math.abs(a.first).compareTo(Math.abs(b.first))
        }
    }
    

3. Process Commands

To handle the input commands, we run a loop.
If the command is INSERT X, we add the value to the heap, and if the command is EXTRACT, we remove the value from the heap and output it.


    val priorityQueue = PriorityQueue>(comparator)

    val reader = BufferedReader(InputStreamReader(System.`in`))
    val n = reader.readLine().toInt()
    
    repeat(n) {
        val command = reader.readLine().split(" ")
        when (command[0]) {
            "INSERT" -> {
                val value = command[1].toInt()
                priorityQueue.add(Pair(value, value))
            }
            "EXTRACT" -> {
                if (priorityQueue.isNotEmpty()) {
                    println(priorityQueue.poll().first)
                } else {
                    println(0)
                }
            }
        }
    }
    

Full Code


    import java.io.BufferedReader
    import java.io.InputStreamReader
    import java.util.PriorityQueue

    fun main() {
        val comparator = Comparator> { a, b ->
            if (Math.abs(a.first) == Math.abs(b.first)) {
                a.first.compareTo(b.first)
            } else {
                Math.abs(a.first).compareTo(Math.abs(b.first))
            }
        }

        val priorityQueue = PriorityQueue>(comparator)

        val reader = BufferedReader(InputStreamReader(System.`in`))
        val n = reader.readLine().toInt()

        repeat(n) {
            val command = reader.readLine().split(" ")
            when (command[0]) {
                "INSERT" -> {
                    val value = command[1].toInt()
                    priorityQueue.add(Pair(value, value))
                }
                "EXTRACT" -> {
                    if (priorityQueue.isNotEmpty()) {
                        println(priorityQueue.poll().first)
                    } else {
                        println(0)
                    }
                }
            }
        }
    }
    

Conclusion

Implementing an absolute value heap is an essential part of dynamic data structures, and it can assist in solving various problems.
Through this tutorial, we learned how to implement a heap simply and effectively in Kotlin.
I hope you were able to realize how important it is to understand the essence of a problem and choose an efficient data structure while solving algorithm problems.
I look forward to improving our skills by tackling various algorithm problems together in the future!

kotlin coding test course, finding the critical path

Written on: October 1, 2023

Author: [Your Name]

Problem Definition

This problem involves finding the critical path from a start node to a finish node in a given connected graph. The critical path indicates the minimum time or distance required to perform the tasks in the graph and plays a key role in project management and optimal resource allocation.

For example, assume a diagram is given that represents the time required to complete each task. There may be conditions that other tasks must be completed first for each task to be completed. The problem can be solved using a Directed Acyclic Graph (DAG) that reflects these conditions.

Input Format

  • The first line contains a natural number N (the number of tasks).
  • The next N lines contain the task number, required time, and dependent task number separated by spaces. If there are no dependent tasks, it is indicated by -1.

Output Format

The length of the critical path (total time taken) is printed.

Example

Input

            5
            1 3 -1
            2 2 1
            3 1 1
            4 4 2
            5 2 3
            

Output

            9
            

Solution Process

To solve this problem, the following steps are followed:

  1. Graph Modeling: Each task is represented as a node, and the dependencies are represented as edges to create a Directed Acyclic Graph (DAG).
  2. Topological Sorting: To perform tasks along the edges, the dependencies must be considered. Therefore, topological sorting is performed to order the nodes.
  3. Longest Path Calculation: Following the results of the topological sorting, the required times of each task are accumulated to calculate the longest path. The maximum value of the required times for the nodes is stored to derive the final result.

Kotlin Code Implementation

            fun main() {
                val n = readLine()!!.toInt()
                val time = IntArray(n + 1)
                val graph = Array(n + 1) { mutableListOf() }
                val indegree = IntArray(n + 1)
                
                for (i in 1..n) {
                    val input = readLine()!!.split(" ").map { it.toInt() }
                    time[i] = input[1]
                    if (input[2] != -1) {
                        graph[input[2]].add(i)
                        indegree[i]++
                    }
                }

                val dp = IntArray(n + 1)
                val queue: Queue = LinkedList()

                for (i in 1..n) {
                    if (indegree[i] == 0) {
                        queue.offer(i)
                        dp[i] = time[i]
                    }
                }

                while (queue.isNotEmpty()) {
                    val current = queue.poll()
                    for (next in graph[current]) {
                        dp[next] = maxOf(dp[next], dp[current] + time[next])
                        indegree[next]--
                        if (indegree[next] == 0) {
                            queue.offer(next)
                        }
                    }
                }

                println(dp.maxOrNull())
            }
            

Code Explanation

The above code consists of the following steps:

  1. Read the number of tasks N.
  2. Initialize an array time to store the execution times of tasks and an array graph to store the graph edges.
  3. Read the dependency relationships of each task to set up the graph and the indegree.
  4. Use a Queue to perform topological sorting and calculate the longest path. Accumulate the required times of each task in the dp array.
  5. Print the result of the longest path.

Problem-Solving Strategy

The problem of finding the critical path is an important factor in time management in project management. Therefore, to solve such problems:

  • Code Optimization: It is important to optimize the algorithm with performance considerations.
  • Analysis and Planning: Systematically analyze the problem requirements and plan solutions.
  • Practice Similar Problems: Gain experience and practice considering various situations through similar problems.

This article is part of a lecture series on algorithm problem solving using Kotlin. Additional questions or feedback are always welcome!

Kotlin coding test course, calculating the binomial coefficient

Introduction

Kotlin is a modern programming language widely used for coding tests and algorithm problem-solving. In this tutorial, we will explore how to calculate the binomial coefficient. The binomial coefficient is a very important concept used to compute combinations. This post will cover two methods for calculating the binomial coefficient: a recursive method and a dynamic programming method.

Problem Description

Let’s solve the problem of calculating nCk (n choose k) for given natural numbers n and k. The binomial coefficient is defined as n! / (k! * (n-k)!). The values of n and k must satisfy the following conditions:

  • 0 ≤ k ≤ n
  • n is provided as a natural number.

For example, if n is 5 and k is 2, then 5C2 is 10. This is because 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 × 4) / (2 × 1) = 10.

Solution Method

1. Recursive Method

The binomial coefficient can be calculated recursively. The key point to note is the underlying formula. The binomial coefficient has the following property and can be implemented recursively:

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

Based on the above formula, let’s implement the binomial coefficient recursively.


fun binomialCoefficient(n: Int, k: Int): Int {
    // Base case
    if (k == 0 || k == n) return 1
    // Recursive case
    return binomialCoefficient(n - 1, k - 1) + binomialCoefficient(n - 1, k)
}
            

2. Dynamic Programming Method

The recursive method can be inefficient in terms of execution time and memory. To improve this, we can use dynamic programming. This method saves previous computation results to avoid unnecessary calculations. Let’s create a DP table to compute the binomial coefficient.


fun binomialCoefficientDP(n: Int, k: Int): Int {
    // Create a 2D array to store results
    val dp = Array(n + 1) { IntArray(k + 1) }
    
    // Fill the table according to base cases
    for (i in 0..n) {
        for (j in 0..minOf(i, k)) {
            if (j == 0 || j == i) {
                dp[i][j] = 1
            } else {
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
            }
        }
    }
    
    // Result is stored at dp[n][k]
    return dp[n][k]
}
            

Code Execution Example

We can calculate the binomial coefficient using the two methods above. Here’s how to execute it in Kotlin.


fun main() {
    val n = 5 // value of n
    val k = 2 // value of k

    // Using the recursive method
    val resultRecursive = binomialCoefficient(n, k)
    println("Recursive method: $n C $k = $resultRecursive")

    // Using the dynamic programming method
    val resultDP = binomialCoefficientDP(n, k)
    println("Dynamic programming method: $n C $k = $resultDP")
}
            

Executing the above code will yield the following results:


Recursive method: 5 C 2 = 10
Dynamic programming method: 5 C 2 = 10
            

Conclusion

In this post, we explored two methods for calculating the binomial coefficient using Kotlin. The recursive method is intuitive for understanding, but it can be inefficient with larger input values. On the other hand, dynamic programming uses more memory but significantly reduces time complexity. We hope this helps you understand ways to efficiently solve algorithm problems.

We hope this tutorial is helpful in your preparation for coding tests. Thank you!

Kotlin coding test course, finding the binomial coefficient 1

Hello! Today we will learn how to calculate binomial coefficients using Kotlin. The binomial coefficient is an important concept in combinations, representing the number of ways to choose k objects from a given n objects. This topic frequently appears in coding tests, so it is essential to understand it.

Problem Description

Given two integers n and k, write a program to calculate nCk (n choose k). The binomial coefficient is defined by the following formula:

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

Here, n! signifies the factorial of n. A factorial is the product of all natural numbers from 1 to n.

Input Format

The first line contains integers n (0 ≤ n ≤ 30) and k (0 ≤ k ≤ n).

Output Format

Print the value of nCk.

Example Input

5 2

Example Output

10

Problem Solving Methods

To solve this problem, you can use two methods:

  1. Recursive factorial calculation
  2. Dynamic programming for binomial coefficient calculation

1. Recursive Factorial Calculation

The most basic method is to calculate the factorial recursively. This method is easy to understand but can become inefficient as n grows larger. Here is a Kotlin example of calculating factorial recursively:

fun factorial(n: Int): Long {
    return if (n == 0 || n == 1) 1 else n * factorial(n - 1)
}

2. Dynamic Programming for Binomial Coefficient Calculation

To efficiently calculate the binomial coefficient, dynamic programming can be utilized. The binomial coefficient can be computed using the properties of Pascal’s triangle, defined by the following recurrence relation:

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

The base cases are as follows:

  • C(n, 0) = 1 (n >= 0)
  • C(n, n) = 1 (n >= 0)

Let’s calculate the binomial coefficient using a dynamic programming array based on Pascal’s triangle:

fun binomialCoefficient(n: Int, k: Int): Long {
    val dp = Array(n + 1) { LongArray(k + 1) }
    for (i in 0..n) {
        for (j in 0..minOf(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]
}

Code Integration Example

Based on the above methods, let’s write the final code:

fun main() {
    val input = readLine()!!.split(" ").map { it.toInt() }
    val n = input[0]
    val k = input[1]
    
    println(binomialCoefficient(n, k))
}

fun binomialCoefficient(n: Int, k: Int): Long {
    val dp = Array(n + 1) { LongArray(k + 1) }
    for (i in 0..n) {
        for (j in 0..minOf(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]
}

Conclusion

We have learned how to calculate binomial coefficients. We can solve this problem using recursion and dynamic programming, and it is important to understand the advantages and disadvantages of each method. The binomial coefficient can be applied in many fields beyond combinatorics, so it is beneficial to have a solid understanding of this concept.

We hope you continue to explore various algorithm problem-solving processes using Kotlin. Thank you!

course on Kotlin Coding Test, Finding Cichlid Numbers

Hello, everyone! In this lecture, we will solve the problem of finding “binary friends” using Kotlin. A binary friend is a sequence made up of 0s and 1s that satisfies the condition of alternating ‘0’s and ‘1’s. This problem can be efficiently solved using dynamic programming techniques.

1. Problem Definition

The problem of finding binary friends can be defined as follows:

Given an integer N, find the number of binary friends with length N.

Examples

  • N = 1: The binary friends are “0”, “1” → Count: 2
  • N = 2: The binary friends are “00”, “01”, “10”, “11” (but ’00’ is not a binary friend, so it is excluded) → Count: 1 (only 01 and 10 are possible)
  • N = 3: The binary friends are “001”, “010”, “100”, “101”, “110” → Count: 5

2. Understanding the Problem

To solve this problem, it is essential to understand how binary friends are structured. Binary friends have the following conditions:

  1. It cannot start with 0.
  2. 0 cannot appear consecutively more than 2 times.
  3. 1 cannot appear consecutively more than 2 times.

Thus, when the length N of the binary friends is given, if the last digit is 0, the preceding digit must be 1. Conversely, if the last digit is 1, the preceding digit can be either 0 or 1, but the conditions for it to be a binary friend must still be maintained.

3. Dynamic Programming Approach

This problem can be efficiently solved through dynamic programming. We will construct an array divided into two states. Here:

  • dp[i] represents the number of binary friends of length i.
  • dp0[i] represents the number of binary friends of length i that end with 0.
  • dp1[i] represents the number of binary friends of length i that end with 1.

Recurrence Relation

Therefore, we can establish the following recurrence relations:

  • dp0[i] = dp1[i-1]
  • dp1[i] = dp0[i-1] + dp1[i-1]
  • dp[i] = dp0[i] + dp1[i]

Initial Values

The initial values are as follows:

  • dp[1] = 2 (binary friends are 0 and 1)
  • dp0[1] = 1 (case that ends with 1)
  • dp1[1] = 1 (case that ends with 0)

4. Code Implementation

Now, let’s implement the code to find binary friends using Kotlin.


fun countBinaryFriends(n: Int): Int {
    if (n == 1) {
        return 2
    }
    
    val dp0 = IntArray(n + 1)
    val dp1 = IntArray(n + 1)
    
    dp0[1] = 1
    dp1[1] = 1
    
    for (i in 2..n) {
        dp0[i] = dp1[i - 1]
        dp1[i] = dp0[i - 1] + dp1[i - 1]
    }
    
    return dp0[n] + dp1[n]
}

fun main() {
    val n = readLine()!!.toInt()
    println(countBinaryFriends(n))
}
    

5. Code Explanation

The above code defines a function countBinaryFriends to find binary friends of length N. It takes an integer N as input and returns the count of binary friends related to it. It first handles the case for length 1, then updates the arrays dp0 and dp1 using dynamic programming. Finally, it sums dp0[n] and dp1[n] to return the result.

6. Time Complexity

The time complexity of this algorithm is O(N). It calculates values by traversing the array once, making it efficient. The space complexity is also O(N), allowing dynamic memory management for finding binary friends.

7. Conclusion

Through this lecture, we solved the problem of finding binary friends using Kotlin. This problem is a good example for understanding dynamic programming and is one of the frequently asked problems in practical coding tests. If you know the basic principles of dynamic programming and can apply them, it will enhance your ability to approach and solve various problems.

I hope to continue solving various algorithm problems together and improve coding skills. Thank you!