Hello! In this lecture, we will cover how to solve algorithm problems using Kotlin. We will specifically analyze the types of problems that are frequently asked in algorithm tests and discuss how to solve them efficiently.

Problem Description

As office workers who are planning to spend the busiest weekend consider where they can save time, they decided to visit a shopping mall to shop. Each office worker intends to purchase specific items from the mall, represented as _ITEM_. Since each office worker makes impulsive decisions when visiting the mall to buy items, they each have a list of items they wish to purchase, called PrepareItem. If the same item on an office worker’s purchase list has already been sold to another office worker, that office worker cannot purchase that item.

Final Goal: Through the given purchase candidate lists of each office worker, determine which items can be purchased without overlap. Output the list of items that can be purchased without overlap.

Input Format

  • The first line of input contains the number of office workers N (1 ≤ N ≤ 100).
  • Following that, N lines will contain the list of items that each office worker wishes to purchase, with the number of items being M_i (1 ≤ M_i ≤ 100).

Output Format

Output the list of items that can be purchased by the office workers. The items should be sorted in ascending order, and each item must be displayed without omission.

Approach to Problem Solving

To solve this problem, the following approach can be used:

  1. Store the list of items for each office worker.
  2. Utilize a Set data structure to check the availability of each item.
  3. Iterate through the items requested by each office worker, checking for duplicates and adding the item only if there is no duplication.
  4. Sort the list of available items in ascending order and output it.

Kotlin Code Implementation

Now let’s write the code to solve the problem.


fun main() {
    val n = readLine()!!.toInt()
    val itemSets = List(n) { readLine()!!.split(" ").toSet() }
    val availableItems = mutableSetOf()

    for (itemSet in itemSets) {
        for (item in itemSet) {
            if (!availableItems.contains(item)) {
                availableItems.add(item)
            }
        }
    }

    val result = availableItems.sorted()
    println(result.joinToString(" "))
}

Code Explanation

1. val n = readLine()!!.toInt() : First, we take the number of office workers as input.
2. val itemSets = List(n) { readLine()!!.split(" ").toSet() } : We store each office worker’s item list in a Set format. By using a Set, we can automatically resolve duplicates.
3. val availableItems = mutableSetOf() : We initialize a MutableSet to store the list of items available for purchase.

Now, as we iterate over the list of office workers, we check each office worker’s desired items and add them to the purchasing list only if they have not already been purchased.

Output Result

Finally, we sort and output the list of available items for purchase. When this code is executed, it will produce the correct output for the given sample input.

Post-Problem Solving Checks

1. Understanding Recursive Algorithms – If the way office workers select items is complex, recursion may be needed through depth-first search.

2. Optimization Considering Time – If the input values are large, code performance should be checked to test feasibility.

3. Testing Various Input Values – Check the robustness of the code by inputting different cases.

Conclusion

In this lecture, we have solved a simple algorithm problem. We discussed efficient methodologies for solving algorithm problems using Kotlin and emphasized the importance of selecting appropriate data structures and implementing clear logic through code for each problem.

In the next lecture, we will cover more complex problems that are frequently asked in actual coding tests. We appreciate your interest!

kotlin coding test course, assign meeting rooms

Problem Description

The meeting room allocation problem is an algorithmic problem that efficiently allocates meeting rooms based on a given list of times.
There are N meetings, and each meeting has a start time and an end time.
A meeting room can only hold one meeting at a time, and if there are meetings occurring at the same time, those meetings cannot be allocated.
Therefore, write an algorithm that maximizes the number of meetings that can be allocated.

Input Format


    - The first line contains N (1 ≤ N ≤ 100,000).
    - The next N lines provide the start and end times of each meeting.
    - The start time is less than the end time, and all times are integers between 0 and 10^6.
    

Output Format


    - Output the maximum number of meetings that can be allocated.
    

Example

Input Example:


    5
    1 3
    2 4
    3 5
    5 6
    4 7
    

Output Example:


    4
    

Problem Solving Process

This problem can be approached using a greedy algorithm.
A greedy algorithm is a methodology that makes the optimal choice at each step to find the overall optimal solution.
In the case of the meeting room allocation problem, sorting meetings by their end times and allocating as many meetings as possible is the most effective method.

1. Sorting Meetings

Sorting needs to be done based on the end times of the meetings.
By prioritizing meetings that end earlier, you can use the start times of later meetings to allocate more meetings.

For example, when the meeting data is as follows:


    1 3
    2 4
    3 5
    5 6
    4 7
    

If we sort this data based on end times, it will look like this:


    1 3
    2 4
    3 5
    4 7
    5 6
    

2. Calculating Possible Meetings

Based on the sorted list of meetings, we iterate through each meeting to calculate how many meetings can be allocated.
We check the next meeting that can start after the fastest ending meeting.

To do this, we use a variable lastEnd to record the end time of the last allocated meeting.
Only those meetings whose start time is greater than or equal to lastEnd can be allocated.

3. Code Implementation

The following code implements the above logic.
Let’s write the Kotlin code as follows.


    fun maxMeetings(meetings: Array>): Int {
        // Sort meetings by end time
        val sortedMeetings = meetings.sortedBy { it.second }
        var count = 0
        var lastEndTime = 0

        for (meeting in sortedMeetings) {
            // Allocate meeting if the start time is greater than or equal to the end time of the last meeting
            if (meeting.first >= lastEndTime) {
                count++
                lastEndTime = meeting.second
            }
        }
        return count
    }

    fun main() {
        val n = readLine()!!.toInt()
        val meetings = Array(n) {
            val input = readLine()!!.split(" ")
            Pair(input[0].toInt(), input[1].toInt())
        }
        println(maxMeetings(meetings))
    }
    

Conclusion

In this post, we solved the meeting room allocation problem using Kotlin.
By using a greedy algorithm, we sorted the meeting data based on their end times and checked the start times of each meeting to allocate as many as possible.
This problem serves as a good example of how a greedy algorithm can be used to find an optimal solution efficiently.

Applying such logic and approaches to various problems could lead to excellent performance in coding tests.
Now, we encourage you to take on similar problems!

kotlin coding test course, extended Euclidean algorithm

One of the frequently encountered topics in coding tests is mathematical algorithms. Among them, the “Extended Euclidean Algorithm” is a powerful technique that not only finds the greatest common divisor (GCD) of two numbers but also provides solutions related to Bézout’s identity. In this article, I will explain the Extended Euclidean Algorithm and detail the process of solving problems using it.

What is the Extended Euclidean Algorithm?

The Extended Euclidean Algorithm is a method for finding integers x and y in the process of calculating the greatest common divisor gcd(a, b) of two integers a and b:

gcd(a, b) = ax + by

The above equation is known as Bézout’s identity, where x and y are integers. The Extended Euclidean Algorithm can be implemented using either a recursive method or an iterative method.

Problem Definition

The problem we will address involves given numbers, using the Extended Euclidean Algorithm to output both the greatest common divisor and Bézout’s coefficients.

Problem

Problem Description:

Given two integers A and B, find the following:

  • The greatest common divisor GCD(A, B), and
  • Find integers X, Y such that GCD(A, B) = AX + BY.

Input: Two integers A, B (-109A, B ≤ 109)

Output: Print GCD(A, B), X, Y separated by spaces.

Problem Solving Process

1. Understanding the Basic Algorithm

First, to implement the Extended Euclidean Algorithm, one must understand the basic Euclidean algorithm. The Euclidean Algorithm is a famous method for finding the greatest common divisor of two numbers, following these steps:

  • While both A and B are not zero, execute A = A % B.
  • Then swap the values of A and B. Make A equal to B and B equal to the previous B.
  • Repeat until one of the numbers becomes zero.

The remaining number at the end is the greatest common divisor. By tracking the changes of x and y at each step, we can construct Bézout’s identity.

2. Implementing the Extended Euclidean Algorithm

Here is the code implementing the Extended Euclidean Algorithm in Kotlin:

fun extendedGCD(a: Int, b: Int): Triple {
    if (b == 0) {
        return Triple(a, 1, 0) // GCD = a, X = 1, Y = 0
    } else {
        val (gcd, x1, y1) = extendedGCD(b, a % b) // Recursive call
        val x = y1
        val y = x1 - (a / b) * y1
        return Triple(gcd, x, y) // Return GCD, X, Y
    }
}

The above function takes two integers a and b as input and returns the greatest common divisor along with the Bézout coefficients x and y.

3. Writing the Main Function

Now, let’s write code in the main function to read two numbers from the user, call the Extended Euclidean Algorithm, and print the results:

fun main() {
    val reader = Scanner(System.`in`)
    println("Please enter two integers:")
    val A = reader.nextInt()
    val B = reader.nextInt()
    
    val (gcd, x, y) = extendedGCD(A, B)
    println("GCD: $gcd, X: $x, Y: $y")
}

4. Complexity Analysis

The time complexity of the Extended Euclidean Algorithm is the same as that of the basic Euclidean algorithm. The time complexity of the Euclidean algorithm is O(log(min(A, B))). This makes the Extended Euclidean Algorithm very efficient as well.

5. Testing and Examples

Now let’s perform some tests using the implemented code. For example, let’s check the results when A = 30 and B = 21:

Please enter two integers:
30
21
GCD: 3, X: -1, Y: 2

The result above shows that 3 = 30 * (-1) + 21 * 2, confirming that Bézout’s identity holds.

Conclusion

In this posting, we explored how to find the greatest common divisor and Bézout’s identity of two numbers using the Extended Euclidean Algorithm. This algorithm can be applied to various problem-solving scenarios and is particularly useful in problems related to mathematical algorithms. Its low complexity makes it a great tool for achieving high scores in actual coding tests. I hope we can continue to explore various algorithms and problem-solving methods together in the future.

kotlin coding test course, finding the minimum number of matrix multiplication operations

Matrix-related problems frequently appear in coding tests, and among them, matrix multiplication problems are a topic many candidates struggle with. In this article, we will examine the problem of ‘Finding the Minimum Number of Matrix Multiplication Operations’ and explain the algorithm and code to solve it in detail.

Problem Description

When the size of matrix A is A_rows x A_cols and the size of matrix B is B_rows x B_cols, to perform the multiplication of the two matrices, the number of columns in A A_cols must match the number of rows in B B_rows. The complexity of this operation is calculated as A_rows x A_cols x B_cols.

When multiplying multiple matrices, it is important to suitably select the ‘order of matrix multiplication’ to enhance operational efficiency. For a given list of matrices, we need to compute the number of operations for all possible multiplication orders and find the minimum.

Input Format

The first line contains a natural number N (1 ≤ N ≤ 30). The next N lines provide two integers R and C representing the size of each matrix. Here, R refers to the number of rows and C refers to the number of columns. It is assumed that the multiplication of each matrix is given in the form R×C.

Output Format

Print the minimum number of matrix multiplication operations on one line.

Example Input

    3
    10 20
    20 30
    30 40
    

Example Output

    3000
    

Solution Process

The algorithm used to solve this problem is the ‘Dynamic Programming Method for Deciding the Order of Matrix Multiplication’. By using this method, we can efficiently compute the multiplication operation cost for each matrix combination.

Dynamic Programming Approach

To determine the order of matrix multiplication for the given problem, we proceed with the following steps:

  1. Create a DP table based on the entered matrix sizes.
  2. Recursively calculate the multiplication ratio and update to the minimum value.
  3. Finally, print the final value from the DP table.

DP Array and Initialization

First, we declare a two-dimensional array dp and initialize all values to 0. This array will store the minimum multiplication cost from matrix i to j in dp[i][j].

Operation Count Calculation Logic

To calculate the number of multiplication operations for the matrix, we proceed with the DP as follows:

  1. Iterate through all possible intervals from i to j.
  2. Select a midpoint k for the current interval (i ≤ k < j).
  3. Calculate the costs of the two parts based on the current DP value and the midpoint, and sum them up.
  4. Update the minimum value.

Code Implementation

Below is the code implemented in Kotlin for the above process:

    fun matrixChainOrder(dims: IntArray): Int {
        val n = dims.size - 1
        val dp = Array(n) { IntArray(n) }

        for (chainLength in 2..n) {
            for (i in 0..n - chainLength) {
                val j = i + chainLength - 1
                dp[i][j] = Int.MAX_VALUE
                for (k in i until j) {
                    val cost = dp[i][k] + dp[k + 1][j] + dims[i] * dims[k + 1] * dims[j + 1]
                    dp[i][j] = minOf(dp[i][j], cost)
                }
            }
        }
        return dp[0][n - 1]
    }

    fun main() {
        val n = readLine()!!.toInt()
        val dims = IntArray(n + 1)
        for (i in 0 until n) {
            val (r, c) = readLine()!!.split(" ").map { it.toInt() }
            dims[i] = r
            if (i == n - 1) {
                dims[i + 1] = c
            } else {
                dims[i + 1] = c
            }
        }
        println(matrixChainOrder(dims))
    }
    

Conclusion

Through this problem, we learned the optimization process of matrix multiplication, namely how to determine the optimal multiplication order. It is important to calculat the trivial parts step by step using dynamic programming to reduce the overall number of operations. Prepare adequately for these types of problems in coding tests, and practice repeatedly to gain an advantage. Next time, we will tackle even more in-depth problems!

kotlin coding test course, Floyd-Warshall

Algorithm Problem Solving and Theory Explanation

Introduction to the Floyd-Warshall Algorithm

The Floyd-Warshall Algorithm is an algorithm used to find the shortest paths between all pairs of vertices in a graph.
This algorithm is suitable for calculating the shortest distance for all pairs of a given graph,
and it works even in the presence of negative weights.
Its complexity is O(V^3), where V represents the number of vertices.
The Floyd-Warshall Algorithm utilizes dynamic programming techniques to compute the shortest paths,
incrementally building up the optimal solution.

Problem Description

Given the paths and distances between certain cities,
we will solve the problem of finding the shortest distance between all cities.
This can be very useful for finding optimal routes in heavily trafficked cities.

Problem

There are N cities, and given the distances between the cities,
write a program to output the shortest distance between all cities.
An example input is as follows:

3
0 4 2
float(inf) 0 5
float(inf) float(inf) 0
        

An example output is as follows:

0 4 2
float(inf) 0 5
float(inf) float(inf) 0
        

Solution Process

1. Input Processing

We process the input to solve the problem.
The first line gives the number of cities N, and the following N lines provide the distance information between each city.
‘float(inf)’ indicates there is no path between the two cities.

2. Initial Matrix Creation

Based on the given distance information, we create a 2D array to form the initial matrix.
This matrix contains distance information from city A to city B,
with initial values set to the given distances, and ‘float(inf)’ for cases where there is no path.

3. Applying the Floyd-Warshall Algorithm

The next step is to apply the Floyd-Warshall Algorithm.
The algorithm considers each city K as an intermediate vertex,
determining whether the path from city I to J is shorter by going through city K (i.e.,
if distance[I][J] > distance[I][K] + distance[K][J]),
and updates to the shorter path accordingly.

4. Outputting the Result

We output the shortest distance matrix for all cities.
The result is formatted to distinguish between each city’s shortest distances and ‘float(inf)’.

5. Kotlin Implementation

fun floydWarshall(graph: Array) {
    val n = graph.size
    for (k in 0 until n) {
        for (i in 0 until n) {
            for (j in 0 until n) {
                if (graph[i][j] > graph[i][k] + graph[k][j]) {
                    graph[i][j] = graph[i][k] + graph[k][j]
                }
            }
        }
    }
}

// Main function to execute the program
fun main() {
    val n = readLine()!!.toInt()
    val graph = Array(n) { IntArray(n) { Int.MAX_VALUE } }
    
    // Initialize the graph with input
    for (i in 0 until n) {
        val distances = readLine()!!.split(" ").map { it.toInt() }
        for (j in 0 until n) {
            graph[i][j] = distances[j]
        }
        graph[i][i] = 0 // Distance from a city to itself is always 0
    }
    
    floydWarshall(graph)
    
    // Output the final distance matrix
    for (i in 0 until n) {
        for (j in 0 until n) {
            if (graph[i][j] == Int.MAX_VALUE) {
                print("float(inf) ")
            } else {
                print("${graph[i][j]} ")
            }
        }
        println()
    }
}
        

Conclusion

The Floyd-Warshall Algorithm is an extremely useful tool for finding the shortest paths between all pairs.
By using this algorithm, optimal routes between cities can be found efficiently,
and it can be applied in various fields.
This lecture aims to enhance understanding of algorithm implementation using Kotlin.