Category: Kotlin coding test

Today, we will solve an interesting algorithm problem called ‘Counting Stair Numbers’ using Kotlin. In this tutorial, we will explain the definition of the problem, the rules, the solution strategy, and the process of implementing the code in detail. The topics covered are as follows.

1. Problem Definition

Stair numbers are defined based on the number of digits, and each digit must follow the following rules:

• The difference between the last digit and the one before it must be 1.
• For example, 123, 321, and 456 are stair numbers. However, 122, 200, and 300 are not stair numbers.

Given an integer N, the problem is to find the number of stair numbers consisting of N digits. I will explain how to calculate the output value considering the input value N.

2. Problem Rules

– The maximum number of digits in input is 100, and when N is 1, there are 10 stair numbers including 0 to 9.
– Each digit can be considered a number from 0 to 9, and the first digit cannot be 0.
– As the number of digits increases, the combinations of stair numbers increase.

3. Solution Strategy

This problem can be solved using dynamic programming. The basic idea is to store the number of possible stair numbers for each digit and use it to calculate the number of stair numbers for the next digit.

• For each digit, consider the possible last digit and calculate a new value based on the value from the previous digit.
• For example, dp[n][digit] is an array that stores the number of stair numbers when the last digit of an n-digit number is digit.
• Therefore, dp[n][digit] can be calculated as the sum of dp[n-1][digit-1] and dp[n-1][digit+1].

3.1. Recurrence Relation

The dp array can be constructed using the following recurrence relations.

    dp[n][0] = dp[n-1][1]
    dp[n][9] = dp[n-1][8]
    dp[n][digit] = dp[n-1][digit-1] + dp[n-1][digit+1] (1 <= digit <= 8)
    

4. Code Implementation

Now, let’s implement the code to solve the problem. Below is the code using Kotlin.

    fun countStairNumbers(n: Int): Int {
        if (n == 1) return 10 // When N is 1, the stair numbers are from 0 to 9.

        val mod = 1000000000 // Modular operation used when outputting the result
        val dp = Array(n + 1) { IntArray(10) }

        // Initial values
        for (j in 1..9) {
            dp[1][j] = 1 // The first digit can be from 1 to 9
        }

        // Constructing the DP array
        for (i in 2..n) {
            for (j in 0..9) {
                if (j == 0) {
                    dp[i][j] = dp[i - 1][1] // 0 is only possible from 1 in the previous digit
                } else if (j == 9) {
                    dp[i][j] = dp[i - 1][8] // 9 is only possible from 8 in the previous digit
                } else {
                    dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j + 1]) % mod
                }
            }
        }

        // Calculating total stair numbers
        var result = 0
        for (j in 0..9) {
            result = (result + dp[n][j]) % mod
        }

        return result
    }

    fun main() {
        val n = readLine()!!.toInt() // Receiving input value
        println(countStairNumbers(n)) // Outputting the stair numbers
    }
    

5. Execution and Results

Running the above code will output the number of stair numbers based on the input N value. Below are the results for some test cases.

• Input: 1 → Output: 10
• Input: 2 → Output: 17
• Input: 3 → Output: 32

It can be confirmed that the number of stair numbers is correctly calculated for each input value.

6. Conclusion

In this tutorial, we used dynamic programming to solve the ‘Counting Stair Numbers’ problem. We examined how to construct the recurrence relations and the DP array during the problem-solving process, and wrote the actual code. Such problems are frequently presented in coding tests, so it is important to practice adequately.

We will continue to tackle various algorithm problems in the future, and hope to help you improve your problem-solving skills through Kotlin. Thank you!

1. Introduction to the Problem

The pathfinding problem is about finding a route between specific nodes (vertices) in a given graph.
This problem is one of the fundamental and important problems for evaluating coding tests and algorithm problem-solving abilities.
In this course, we will explain how to solve such problems using Kotlin.

2. Problem Definition

Determine whether a path exists between the start node and the goal node in the given graph,
and if a path exists, print the corresponding path.

Input Format

• The first line: the number of vertices N (1 <= N <= 1000)
• The second line: the number of edges M (0 <= M <= 10000)
• From the third line to M lines: edge information (vertex A, vertex B)
• The last line: start node and goal node (start, goal)

Output Format

• If a path exists, print that path; if not, print “No path exists.”

3. Problem-Solving Strategy

To solve the pathfinding problem, the following algorithms can be used:

• Breadth-First Search (BFS)
• Depth-First Search (DFS)

Here, we will look at how to search for the shortest path using BFS. Since BFS explores in levels,
it is advantageous for finding the shortest path.

4. Algorithm Implementation (Kotlin Example)

            
                import java.util.*

                fun bfs(graph: Array>, start: Int, goal: Int): List? {
                    val queue: Queue = LinkedList()
                    val visited = BooleanArray(graph.size)
                    val parent = IntArray(graph.size) { -1 }

                    queue.add(start)
                    visited[start] = true

                    while (queue.isNotEmpty()) {
                        val current = queue.poll()

                        if (current == goal) {
                            return constructPath(parent, start, goal)
                        }

                        for (neighbor in graph[current]) {
                            if (!visited[neighbor]) {
                                visited[neighbor] = true
                                parent[neighbor] = current
                                queue.add(neighbor)
                            }
                        }
                    }
                    return null
                }

                fun constructPath(parent: IntArray, start: Int, goal: Int): List {
                    val path = mutableListOf()
                    var at = goal
                    while (at != -1) {
                        path.add(at)
                        at = parent[at]
                    }
                    path.reverse()
                    return path
                }

                fun main() {
                    val scanner = Scanner(System.`in`)
                    val N = scanner.nextInt() // Number of vertices
                    val M = scanner.nextInt() // Number of edges
                    
                    val graph = Array(N) { mutableListOf() }
                    
                    for (i in 0 until M) {
                        val u = scanner.nextInt()
                        val v = scanner.nextInt()
                        graph[u].add(v)
                        graph[v].add(u) // Undirected graph
                    }

                    val start = scanner.nextInt()
                    val goal = scanner.nextInt()

                    val path = bfs(graph, start, goal)
                    if (path != null) {
                        println("Path: ${path.joinToString(" -> ")}")
                    } else {
                        println("No path exists.")
                    }
                }
            
        

5. Code Explanation

The above code is an implementation of pathfinding based on BFS.

• bfs(graph, start, goal): Searches for a path from the start node to the goal node in the given graph.
  It uses a queue to proceed with the search and records visited nodes in the visited array.
  When it reaches the goal node, it returns the path based on parent information.
• constructPath(parent, start, goal): Reconstructs the path from the goal node to the start node.
  It traces the path using the parent information of each node and returns the array.

6. Time Complexity Analysis

The time complexity of the BFS algorithm is O(V + E), where V is the number of vertices and E is the number of edges.
In this problem, there can be a maximum of 1000 vertices and 10000 edges, making it possible to solve efficiently.

7. Conclusion

In this course, we explored how to solve the pathfinding problem using Kotlin.
We presented the most efficient way to search for paths in graphs using the BFS algorithm,
and verified the process of solving the problem with actual code.
Additionally, understanding the implementation of DFS and the differences from BFS can provide further insights.

I hope this course will be of great help in preparing for coding tests and understanding algorithms. Thank you!

Problem Description

Problems that frequently appear in coding tests require the use of various algorithms. This problem is based on an algorithmic challenge regarding a game where a player defeats monsters.
Given the health of the monsters and the player’s attack power, the task is to calculate the minimum number of attacks needed for the player to defeat the monsters.
Each monster has unique health, and the player can deal a certain amount of damage with each attack.

Problem

        
        Given N monsters and the player's attack power K, for each monster's health provided,
        write a program to calculate the minimum number of attacks required for the player to defeat all monsters.

        Input format:
        - The first line contains the number of monsters N (1 ≤ N ≤ 10^5) and the attack power K (1 ≤ K ≤ 10^4).
        - The second line provides the health of each monster H[i] (1 ≤ H[i] ≤ 10^5) as N integers.

        Output format:
        - Output the minimum number of attacks needed for the player to defeat the monsters.
        
        

Approach

To solve this problem, we will use the following approach:

1. First, we calculate the number of attacks required to defeat each monster based on the player’s attack power and the monster’s health.
2. After calculating the required attacks for each monster, we sum them up to get the total number of attacks needed to defeat all monsters.
3. Finally, we output the calculated total number of attacks.

Step-by-Step Solution

Step 1: Input Handling

We will input the number of monsters N and the player’s attack power K.
Then, we need to input the health of N monsters as an array. This will be handled using standard input in Kotlin.

Step 2: Calculate Attacks for Each Monster

For each monster’s health H[i], the number of attacks required to defeat the monster can be calculated as (H[i] + K – 1) / K.
This is obtained by dividing the monster’s health by the player’s attack power and rounding up.
For example, if H[i] is 10 and K is 3, the minimum attack count is (10 + 3 – 1) / 3 = 4.

Step 3: Calculate Total Attacks for All Monsters

After obtaining the attack counts for each monster, we can sum them all to get the final total number of attacks needed.

Code

        
        fun main() {
            // Input
            val (N, K) = readLine()!!.split(" ").map { it.toInt() }
            val healthList = readLine()!!.split(" ").map { it.toInt() }

            // Variable to calculate the total required attacks
            var totalAttacks = 0

            // Calculate the required attack counts for each monster
            for (health in healthList) {
                totalAttacks += (health + K - 1) / K  // Rounding up
            }

            // Output the result
            println(totalAttacks)
        }
        
    

Result

Running the code above will output the minimum number of attacks needed for the player to defeat all monsters.
Through this problem, one can understand the relationship between the monster’s health and the player’s attack power, and learn how to approach problems from a mathematical perspective.

Conclusion

In this Kotlin coding test tutorial, we addressed a simple algorithmic problem based on game development.
By combining the necessary mathematical approaches and coding techniques to solve the problem, we can see the importance of writing efficient code and obtaining optimized results.
If we continue to encounter and practice solving a variety of algorithmic problems, we can grow into more professional developers.

Problem Description

There are several residents in our village. Among these residents, some tell lies to express their opinions. Can we identify the liars among the information we have?

Problem Summary

The given N residents provide information about their neighbors. Each resident inputs who they consider their neighbors and what those neighbors said about them. We are trying to determine whether each resident is deceiving themselves. Ultimately, this problem is about figuring out whether the statements of the residents are consistent with each other.

Input Format

The first line contains the number of residents N. Then, for N lines, the index of each resident, the number of neighbors M that the resident thinks they have, and information about M neighbors is provided in a scattered manner. The neighbor’s information contains the index of that neighbor and whether the statement about that index is false or true.

Output Format

Based on whether each resident is a liar or not, the result is printed as YES or NO.

Example Input

    3
    0 2 1 1 0
    1 1 0
    2 1 1
    

Example Output

    NO
    NO
    YES
    

Problem Solving Process

To solve this problem, we follow the steps outlined below.

Step 1: Understand the Problem

To distinguish between truth and lies in the residents’ statements, we need to check the consistency between the statements of different residents based on what they said about their neighbors. For example, if a resident A claims that B is a neighbor, and B claims A is lying, we can conclude that A is indeed a liar.

Step 2: Design the Data Structure

To solve this problem, we can use an array or list to store and manage the information of the residents. A suitable method is to use the index of each resident (starting from 0) and a list to hold the information of their neighbors. Using a Pair class to compose resident information is also a good idea.

Step 3: Algorithm Design

After collecting information about the given residents, we will determine whether their statements are consistent with each other based on this information. We can design the following algorithm.

    1. Input the number of residents N.
    2. Initialize a list to store each resident's information.
    3. For each resident, input and store the information about their neighbors and statements.
    4. Check whether the neighbors' statements of each resident are consistent.
    5. If they are not consistent, determine that the resident is a liar.
    6. Print the results.
    

Step 4: Implement the Kotlin Code

Now we will implement the algorithm in Kotlin code. Below is the Kotlin code to solve the problem.

    data class Neighbor(val index: Int, val isLiar: Boolean)

    fun findLiar(N: Int, statements: List>): List {
        val result = MutableList(N) { "NO" }

        for (i in 0 until N) {
            val neighbors = statements[i]
            for (neighbor in neighbors) {
                val expectedTruth = !neighbor.isLiar
                if (expectedTruth && result[neighbor.index] == "YES") {
                    result[i] = "YES"
                    break
                } else if (!expectedTruth && result[neighbor.index] == "NO") {
                    result[i] = "YES"
                    break
                }
            }
        }
        return result
    }

    fun main() {
        val N = readLine()!!.toInt()
        val statements = mutableListOf>()

        for (i in 0 until N) {
            val input = readLine()!!.split(" ").map { it.toInt() }
            val M = input[1]
            val neighbors = mutableListOf()

            for (j in 0 until M) {
                val neighborIndex = input[2 + j * 2]
                val liar = input[3 + j * 2] == 1
                neighbors.add(Neighbor(neighborIndex, liar))
            }
            statements.add(neighbors)
        }

        val result = findLiar(N, statements)
        for (r in result) {
            println(r)
        }
    }
    

Step 5: Test and Validate

Once the code is completed, it should be validated through various test cases. It is necessary to verify the overall stability and reliability of the code through minimal input, maximal input, and various scenarios.

Conclusion

This problem taught us how to approach and solve algorithm problems in Kotlin. Analyzing the residents’ statements to distinguish between truth and lies greatly helps in developing logical thinking necessary for problem-solving. Additionally, learning how to approach complex problems through simple data structures and algorithm variations will be beneficial.

Now, you should carefully implement the algorithm to find the liars!