Category: Kotlin coding test

1. Problem Definition

This is a problem to write a program in Kotlin to calculate the average from a given set of N integers.
The number of integers provided as input ranges from 1 to 100,000, and each integer is between -50,000 and 50,000.
The algorithm should have a time complexity of O(N) and must perform the summation of the numbers and calculate the average.

2. Problem Approach and Solution

To find the average, we need to calculate the sum of the given integers and then divide that sum by the number of integers.
Below are the steps to solve the problem.

• Step 1: Store the given integers in an array or list.
• Step 2: Traverse all the elements of the array to calculate the sum.
• Step 3: Divide the calculated sum by the number of input integers to compute the average.
• Step 4: Output the average value rounded to two decimal places.

3. Time Complexity Analysis

The solution approach for the given problem involves traversing the list once to calculate the sum.
Since each element is visited only once, the time complexity is O(N), making it very efficient.
The space complexity is also O(1), as there is minimal additional memory usage.

4. Kotlin Implementation

Below is an example of Kotlin code for the problem. This code is written following the steps described above.

fun main() {
    // 1. Read the number of inputs.
    val n = readLine()!!.toInt()
    // 2. Declare a list to store the integers.
    val numbers = readLine()!!.split(" ").map { it.toInt() }

    // 3. Calculate the sum of the integers.
    val sum = numbers.sum()

    // 4. Calculate the average.
    val average = sum.toDouble() / n

    // 5. Print the average rounded to two decimal places.
    println(String.format("%.2f", average))
}
    

5. Input and Output Example

Below are examples of the input and output of the program execution.

Input Example:
5
10 20 30 40 50

Output Example:
30.00
    

6. Example Explanation

Looking at the input example above, it is input to calculate the average of the integers 10, 20, 30, 40, and 50.
The total number of integers is 5, and their sum is 150. Therefore, the average, which is 150 divided by 5, equals 30.00.

7. Additional Considerations

While it is unrealistic for the number of integers to be 0 when calculating the average,
it is advisable to include additional checks in the code to handle exceptional cases.
For example, if the input count is 0, the code could be modified to print a message like “No input numbers.”
This emphasizes the importance of exception handling in programming.

8. Summary

We learned how to calculate the average from N integers using Kotlin.
The process involved handling input data, calculating the sum, and outputting the average in a step-by-step manner.
We also analyzed the time complexity and space complexity to explain the algorithm’s efficiency.
Additionally, we introduced how to write more stable code by including exception handling.

9. Conclusion

When solving algorithm problems, it is essential to understand the problem, design an approach,
and practice a lot in the area of implementation in code. By using Kotlin, you will enhance your problem-solving skills
which will be beneficial for employment. I hope you continue to tackle various problems and build your skills.

Written on: November 1, 2023

Introduction

Hello, in this Kotlin coding test tutorial, we will cover the algorithmic problem of ‘Finding Cities at a Specific Distance’.
This problem involves graph traversal and requires an algorithmic approach to find nodes in cities at a specific distance based on given conditions.
We will look at how to solve the problem step by step using Kotlin.

Problem Description

The given graph consists of N cities and M bidirectional roads.
Each city is identified by numbers from 1 to N, and the roads connect two cities.
The goal is to find all cities that are exactly at the given distance K.

Input Format
The first line contains the number of cities N (1 ≤ N ≤ 30000) and the number of roads M (1 ≤ M ≤ 200000), followed by the distance K (0 ≤ K ≤ 30000) we want to find.
From the second line, M lines follow, each containing information about two connected cities A and B (1 ≤ A, B ≤ N).

Output Format
Output the numbers of cities that are exactly at distance K in ascending order; if there are no such cities, output -1.

Problem Solving Approach

This problem involves exploring vertices in a graph that match the given conditions.
By using the BFS (Breadth-First Search) algorithm, we can find cities at a specific distance by searching to the appropriate depth.
BFS can be implemented using an empty queue and an adjacency list.

Implementation Method

First, we construct an adjacency list based on the given city and road information.
Then we use the BFS algorithm to find the cities that reach the given distance K.

The implementation steps are as follows:

1. Process input values: Read the number of cities, number of roads, and distance K.
2. Construct adjacency list: Store other cities connected to each city in list form.
3. Initialize BFS: Explore from the starting city to the distance K.
4. Store results: Save city numbers that have reached the distance K in a list.
5. Output results: Sort and print the cities reached at distance K.

Code Implementation

Below is the solution code implemented using the BFS algorithm in Kotlin:

import java.util.*

fun main() {
    val scanner = Scanner(System.`in`)
    val n = scanner.nextInt() // Number of cities
    val m = scanner.nextInt() // Number of roads
    val k = scanner.nextInt() // Desired distance
    val x = scanner.nextInt() // Starting city

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

    // Input road information
    for (i in 0 until m) {
        val a = scanner.nextInt()
        val b = scanner.nextInt()
        graph[a].add(b)
        graph[b].add(a)
    }

    // BFS implementation
    val distance = IntArray(n + 1) { -1 }
    distance[x] = 0
    val queue: Queue = LinkedList()
    queue.add(x)

    while (queue.isNotEmpty()) {
        val current = queue.poll()
        for (neighbor in graph[current]) {
            if (distance[neighbor] == -1) {
                distance[neighbor] = distance[current] + 1
                queue.add(neighbor)
            }
        }
    }

    // Find result cities
    val result = mutableListOf()
    for (i in 1..n) {
        if (distance[i] == k) {
            result.add(i)
        }
    }

    // Output results
    if (result.isEmpty()) {
        println(-1)
    } else {
        result.sort()
        for (city in result) {
            println(city)
        }
    }
}
        

Solution Analysis

The above code is a typical BFS implementation designed to solve the given problem.
In particular, by representing the road connectivity as an adjacency list, it optimizes memory usage and allows for fast searching.
BFS operates by exploring all nodes and placing adjacent nodes into a queue.
This way, it updates the distance to each node and ultimately collects nodes that stop at distance K.

If no city is found at distance K, returning -1 is also a good programming practice.
This code structure can be easily modified to suit different types of graph traversal problems.

Conclusion

Thus, we have learned about the process of implementing Kotlin code using the BFS algorithm to solve the ‘Finding Cities at a Specific Distance’ problem.
This problem helps in understanding the basics of graph problems and can develop problem-solving skills for various problem types.
It will also be beneficial for effectively applying algorithms in work or personal projects.
We plan to cover more diverse algorithm problems in the future, so please stay tuned.

Hello. Today we will learn how to find the diameter of a tree using Kotlin.
In this tutorial, we will explain the basic concepts of trees, the definition of diameter, and the algorithmic approach in detail,
and we will practice through example code.

Tree and Tree Diameter

A tree is a non-linear data structure in which each node can have zero or more children.
A tree has the following characteristics:

• A tree has a root node, and the root is connected to other nodes.
• A child node must have exactly one parent node.
• The path from the root to a leaf node defines the height of the tree.

The diameter of a tree refers to the length of the longest path within the tree.
The diameter can always be defined as the distance between two leaf nodes.
For example, consider the tree below:

        1
       / \
      2   3
     / \
    4   5
    

The diameter of this tree is the longest path from node 4 to node 5, which is 4-2-5.
Therefore, the length of the diameter is 2.

Algorithm for Finding the Diameter of a Tree

To find the diameter, you can use DFS (Depth-First Search) or BFS (Breadth-First Search).
Generally, DFS is the more commonly used method.
The basic idea of the algorithm is as follows:

1. Select an arbitrary node (root node) and perform the first DFS. During this, find the farthest node.
2. Starting from the node found in the first DFS, perform a second DFS.
3. Return the maximum distance (diameter) calculated in the second DFS.

Kotlin Code Implementation

Now, we will implement this algorithm in Kotlin.
We will use data classes to represent nodes and edges, and
we will create a DFS method to calculate the diameter of the tree.

Data Class and Graph Initialization

Kotlin
data class Edge(val to: Int, val weight: Int)

class Tree(val size: Int) {
    private val graph = Array(size) { mutableListOf() }

    fun addEdge(from: Int, to: Int, weight: Int) {
        graph[from].add(Edge(to, weight))
        graph[to].add(Edge(from, weight))
    }

    fun diameter(start: Int): Int {
        // Perform the first DFS
        val (farthestNode, _) = dfs(start, BooleanArray(size), 0)
        // Perform the second DFS
        val (_, diameter) = dfs(farthestNode, BooleanArray(size), 0)
        return diameter
    }

    private fun dfs(node: Int, visited: BooleanArray, distance: Int): Pair {
        visited[node] = true
        var maxDistance = distance
        var farthestNode = node

        for (edge in graph[node]) {
            if (!visited[edge.to]) {
                val (newFarthestNode, newDistance) = dfs(edge.to, visited, distance + edge.weight)
                if (newDistance > maxDistance) {
                    maxDistance = newDistance
                    farthestNode = newFarthestNode
                }
            }
        }

        return Pair(farthestNode, maxDistance)
    }
}

Code Explanation

– Edge data class: A class that defines an edge.
It includes the node and weight.

– Tree: A class that represents a tree. It initializes the size of the tree and the graph data structure.
Edges can be added using the addEdge method.

– diameter method: A method that calculates the diameter.
It performs the first and second DFS.

– dfs method: A recursive method that performs depth-first search.
It checks the visited nodes and calculates the maximum distance.
It returns the farthest node and the distance based on the maximum distance.

Test Cases

To test the diameter of the tree, we will add edges as shown below and print the results.

Kotlin
fun main() {
    val tree = Tree(5)
    tree.addEdge(0, 1, 1)
    tree.addEdge(0, 2, 2)
    tree.addEdge(1, 3, 1)
    tree.addEdge(1, 4, 1)

    val diameter = tree.diameter(0)
    println("The diameter of the tree is: $diameter")
}

Execution Result

    The diameter of the tree is: 4
    

Conclusion

In this tutorial, we explored how to find the diameter of a tree using Kotlin.
This approach using DFS is useful for dealing with tree data structures and can be applied in various applications.
I hope this tutorial helps in solving tree-related problems.

We plan to cover various algorithms and coding test problems in the future, so please look forward to it!
Thank you.

1. Tree (Introduction)

A tree is a data structure used to represent hierarchical structures, where each node can have multiple children. A tree consists of nodes and edges, where the top node is called the root node, and the nodes at the ends of the tree are called leaf nodes. Trees are essential concepts for solving various problems and implementing algorithms.

1.1. Properties of Trees

• A tree with n nodes always has n-1 edges.
• The root node is unique, and each node has only one parent.
• Not all leaf nodes are located at the same depth.

2. Algorithm Problem

Problem: Calculate the maximum depth of a binary tree

Write a function to calculate the maximum depth of a given binary tree. The depth of a binary tree is the number of nodes along the path from the root node to the deepest leaf node.

Input

• A binary tree is defined with node types, and each node contains an integer value.
• A node can have child nodes, and it can have up to two children.

Output

• Returns the maximum depth as an integer.

3. Problem-Solving Process

3.1. Understanding the Problem

To understand the problem, examine the characteristics of the given binary tree. The depth refers to the longest path from the root to a leaf node, which can be obtained through exploration. A common method is depth-first search (DFS).

3.2. Defining the Tree Node Class

First, we will define a class to represent a node of the tree. It can be implemented in Kotlin as follows:

data class TreeNode(val value: Int, var left: TreeNode? = null, var right: TreeNode? = null)

3.3. Implementing the Algorithm to Find Maximum Depth

The maximum depth can be calculated recursively. Below is an example implemented using DFS.

fun maxDepth(node: TreeNode?): Int {
        if (node == null) return 0
        val leftDepth = maxDepth(node.left)
        val rightDepth = maxDepth(node.right)
        return Math.max(leftDepth, rightDepth) + 1
    }

The above function works as follows:

• If the tree node is null, the depth is 0.
• Recursively call the depths of the left and right child nodes and choose the deeper of the two.
• Add 1 to return the maximum depth.

3.4. Complete Example Implementation

Let’s implement the entire code to find the maximum depth:

fun main() {
        val root = TreeNode(1)
        root.left = TreeNode(2)
        root.right = TreeNode(3)
        root.left!!.left = TreeNode(4)
        root.left!!.right = TreeNode(5)

        val depth = maxDepth(root)
        println("The maximum depth is: $depth")  // Output: The maximum depth is: 3
    }

4. Conclusion

The binary tree is an important topic for solving algorithm problems. By solving the problem, one can understand the structure of trees and practice recursive thinking, which is a useful skill in coding tests. It is recommended to try various tree-related problems in the future.

5. Additional Practice Problems

• Finding a specific value in a binary search tree
• Printing all paths in a binary tree
• Checking the balance of a binary tree

6. References

To enhance your understanding of trees, please refer to the following resources:

• Wikipedia: Tree (Computer Science)
• GeeksforGeeks: Tree Data Structure
• Programiz: Tree Data Structure