Category: Kotlin coding test

Problem Description

Given an integer n and a sequence of n integers, your task is to implement a method to sort the given sequence
in ascending order using a stack. You need to process the sequence taking advantage of the stack’s LIFO (Last In, First Out) property,
and the maximum number of elements that can enter the stack is n.

Input Format

• The first line contains an integer n (1 ≤ n ≤ 100,000).
• The second line contains n integers separated by spaces (1 ≤ each number ≤ 1,000,000), and the sequence may contain duplicates.

Output Format

Output the series of stack operations (push, pop) used to sort the sequence in ascending order.
If it is not possible to sort the sequence using the stack, you should output “NO”.

Example Input

    5
    4 3 6 8 5
    

Example Output

    PUSH 4
    PUSH 3
    POP 3
    PUSH 6
    POP 6
    PUSH 5
    POP 5
    POP 4
    

Problem Solution

This problem requires using a stack to sort the input sequence in ascending order.
Let’s approach the problem with the following steps.

Algorithm Design

1. **Input Handling**: Read the integer n and the sequence.
2. **Initialize Stack and Output List**: Initialize the list to store the sequence and the stack.
3. **Initialize Target Value**: Set the next target value that should be in order as 1.
4. **Traverse the Sequence**: Loop through the input sequence,
– If the current number is equal to the target value, perform the pop operation.
– If it is different from the target value, push it onto the stack.
5. **Empty the Stack**: If there are values left in the stack, pop and output them.
6. **Output**: Finally, display the sorted sequence or output “NO”.

Coding Implementation

If the above algorithm is implemented in Kotlin, it looks like this:

fun main() {
    val n = readLine()!!.toInt()
    val sequence = readLine()!!.split(" ").map { it.toInt() }
    val stack = mutableListOf()
    val output = mutableListOf()
    var current = 1
    var index = 0

    while (current <= n) {
        while (index < n && (stack.isEmpty() || stack.last() != current)) {
            stack.add(sequence[index++])
            output.add("PUSH ${stack.last()}")
        }
        if (stack.isNotEmpty() && stack.last() == current) {
            stack.removeAt(stack.size - 1)
            output.add("POP $current")
        } else {
            println("NO")
            return
        }
        current++
    }

    output.forEach { println(it) }
}

Key Considerations

There are several important considerations in solving this problem. First, you need to understand the concept of the stack
and implement it in code, utilizing the stack effectively in certain situations. Additionally,
since the input numbers must be sorted in ascending order, a logic that guarantees this is needed.
The maximum size of the stack is n, allowing for an O(n) space complexity.

Conclusion

In this lecture, we addressed the problem of creating an ascending sequence using a stack. We could confirm
our understanding of basic stack operations and their utilization through actual code implementation.
The stack data structure is covered in many algorithm problems, so I hope you learn how to utilize stacks through this problem.

Coding tests play an important role in modern hiring processes. In the preparation process, solving algorithm problems is a very significant part. In this course, I will detail the process of solving an algorithm problem that calculates the sum of given numbers using Kotlin.

Problem Description

There are N given integers. Write a program that takes the N integers as input and prints their sum.

Input Format:
The first line contains an integer N (1 ≤ N ≤ 1000).
The second line contains N integers separated by spaces. Each integer is between -1000 and 1000.

Output Format:
Print the sum of the N integers provided as input.

Example Problem

Input:
5
1 2 3 4 5
Output:
15

Approach to the Problem

This problem is a very basic one, where we store the given integers in an array and then calculate their sum. Since Kotlin allows us to easily handle lists and various collections, we can use it to solve the problem.

The general approach is as follows:

1. Read the two lines of data provided as input.
2. Read the integer N from the first line.
3. Read N integers from the second line and store them in a list.
4. Sum all values in the list to find the total.
5. Print the sum.

Kotlin Implementation

Now, based on the method above, let’s implement it in Kotlin. Below is the implemented code:

        
        import java.util.Scanner

        fun main() {
            val scanner = Scanner(System.`in`)
            
            // Read N from the first line
            val N = scanner.nextInt()
            var sum = 0
            
            // Read N numbers from the second line
            for (i in 0 until N) {
                sum += scanner.nextInt()
            }
            
            // Print the sum
            println(sum)
        }
        
    

Code Explanation

The above code functions in the following order:

• Uses Scanner to read input: Scanner helps to receive input.
• Reads N: Reads the integer N from the first line.
• Uses a for loop to read and add numbers: Through N iterations, it collects each integer and adds it to sum.
• Outputs the result: Prints the final sum.

Code Testing

You can try various inputs to test the written code. For example:

        
        Input:
        3
        10 -5 2
        
    

When given the above input, the program will display the following output:

        
        Output:
        7
        
    

Conclusion

In this course, we learned how to solve a simple problem of calculating the sum of numbers using Kotlin. By effectively utilizing basic algorithms and Kotlin’s syntax, problems can be easily solved.

Such types of problems are often presented in coding tests. Practice processing various data to improve your coding skills.

In the next session, we will address more complex problems, so stay tuned!

1. Introduction

Coding tests are one of the important processes for finding modern programming jobs. As technology advances and the importance of data structures and algorithms comes to the fore, these skills have become a major criteria for evaluating a programmer’s capabilities. In this course, we will understand the concepts of Stack and Queue using Kotlin and solve algorithmic problems based on these concepts.

2. Basic Concepts of Stack and Queue

2.1. Stack

A stack is a data structure that stores data and operates on a Last In First Out (LIFO) basis. That is, the most recently added data is the first to be removed. Stacks are commonly used in function call stacks, browser back buttons, and string parentheses validation algorithms.

2.2. Queue

A queue is another data structure that stores data and operates on a First In First Out (FIFO) basis. That is, the data that enters first is the data that exits first. Queues are primarily used in printer job management, process scheduling, and breadth-first search (BFS) algorithms.

3. Problem Description

Now let’s solve a problem using stacks and queues. Let’s take a look at the following problem:

Problem: Parentheses Validity Check

Write a function to determine whether a given string consists of valid parentheses. Valid parentheses mean that every open parenthesis must be closed, and every closed parenthesis must form a pair with its corresponding open parenthesis.

For example, the string “{[()]}” is valid, but “{[(])}” is not valid.

4. Problem Analysis

To solve this problem, it is advisable to follow these steps:

1. Traverse the string and store open parentheses in a stack.
2. When encountering a closed parenthesis, check if it pairs with the most recently stored open parenthesis from the stack.
3. If the stack is not empty after processing all characters, the string is invalid.

5. Algorithm Design

Therefore, we can design an algorithm as follows:

1. Initialize the stack.
2. Iterate through each character of the string.
3. If the character is an open parenthesis, add it to the stack.
4. If it is a closed parenthesis, pop from the stack and check if it matches the corresponding open parenthesis.
5. If the stack is not empty, the string is invalid, and return false.
6. After processing all characters, return true if the stack is empty.

6. Kotlin Implementation

Now, let’s implement the algorithm described above in Kotlin.

fun isValidParentheses(s: String): Boolean {
    val stack = mutableListOf()
    val mapping = mapOf('(' to ')', '{' to '}', '[' to ']')

    for (char in s) {
        if (mapping.containsKey(char)) {
            stack.add(char)
        } else if (stack.isNotEmpty() && mapping[stack.last()] == char) {
            stack.removeAt(stack.size - 1)
        } else {
            return false
        }
    }
    return stack.isEmpty()
}

7. Code Explanation

Each part of the code has the following roles:

• val stack = mutableListOf(): Initializes the stack.
• val mapping = mapOf('(' to ')', '{' to '}', '[' to ']'): Stores the mapping of open and closed parentheses.
• for (char in s): Iterates through each character of the string.
• if (mapping.containsKey(char)): Adds the character to the stack if it is an open parenthesis.
• else if (stack.isNotEmpty() && mapping[stack.last()] == char): If it is a closed parenthesis, check if the matching open parenthesis is on top of the stack, and remove it if it matches.
• After processing the string, return true if the stack is empty, otherwise return false.

8. Complexity Analysis

The time complexity of this algorithm is O(n), where n is the length of the string. This is because each character is processed only once. The space complexity is also O(n) since, in the worst case, all open parentheses may be stored in the stack.

9. Test Cases

You can use the following test cases to verify that the implementation is correct.

fun main() {
    println(isValidParentheses("{[()]}")) // true
    println(isValidParentheses("{[(])}")) // false
    println(isValidParentheses("{{[[(())]]}}")) // true
    println(isValidParentheses("{")) // false
    println(isValidParentheses("}")) // false
}

10. Conclusion

In this course, we explored the basic concepts of stacks and queues, along with solving the problem of validating parentheses using a stack. This approach is a common type of question in coding tests, so it is important to practice sufficiently to become familiar with the use of stacks and queues.

11. References

• LeetCode: Valid Parentheses
• GeeksforGeeks: Stack Data Structure
• Wikipedia: Stack (abstract data type)

Now, readers, you can understand the concepts of stack and queue using Kotlin and try solving various problems.

Hello! Today we will learn how to determine the order of permutations using Kotlin. Solving algorithm problems is an important part of coding tests, and understanding permutations and combinations will help solve various problems.

Problem Description

This problem is to find the order of a specific permutation from the second index to the first index of a given sequence. The elements of the sequence consist of integers from 1 to N, and there are no duplicates. For example, when the sequence is [1, 2, 3], all permutations are as follows:

• 1, 2, 3
• 1, 3, 2
• 2, 1, 3
• 2, 3, 1
• 3, 1, 2
• 3, 2, 1

In the above example, [2, 1, 3] is the 3rd permutation. The goal is to output the order number in this manner.

Input Format

N (1 ≤ N ≤ 20)
P (N integers, unique numbers from 1 to N)

Output Format

Print the order of the given permutation P.

Problem Solving

To solve this problem, we can proceed with the following steps:

1. Generate all possible permutations.
2. Find the given permutation in the generated list.
3. Print the index of that permutation (starting from 1).

Kotlin Implementation

Now, let’s write the code. First, we will use Kotlin’s permutations function to generate all permutations and then find the index of the input permutation.

import kotlin.math.*

fun  List.permutations(): List> {
    if (isEmpty()) return listOf(emptyList())
    val result = mutableListOf>()
    for (i in indices) {
        val item = this[i]
        val remaining = this.filterIndexed { index, _ -> index != i }
        for (permutation in remaining.permutations()) {
            result.add(listOf(item) + permutation)
        }
    }
    return result
}

fun main() {
    val n = 3
    val sequence = listOf(2, 1, 3)

    val totalNumbers = (1..n).toList()
    val allPermutations = totalNumbers.permutations()

    val index = allPermutations.indexOf(sequence) + 1 // Add 1 since the index starts at 0
    println("The permutation $sequence is the ${index}th order.")
}

Code Explanation

The main parts of the code above are as follows:

• The permutations function recursively generates and returns all permutations as a list.
• The main function defines N and the given permutation, and calls permutations to generate all possible permutations.
• Finally, it uses indexOf to find and print the index of the given permutation.

Performance Considerations

When N is 20, the number of possible permutations corresponds to 20!. This is about 240 million cases. Therefore, when N reaches 20, there can be issues with memory and execution speed, so for large inputs, it is necessary to consider other approaches (e.g., the mathematical properties of permutations). Nonetheless, this method works well for smaller numbers.

Conclusion

In this tutorial, we learned how to determine the order of permutations. We showed the process of solving the problem using Kotlin’s features and explained various considerations to aid understanding of algorithm problems. Let’s continue to tackle more algorithm problems and learn various approaches!

Problem Description

Input Format

First line: integer n (size of the array)
Second line: n integers (elements of the array)

Output Format

Output the maximum value.

Solution Process

This problem is an algorithmic challenge that applies a methodology for grouping numbers in order to maximize the product of two numbers. To solve this issue, we need to follow several steps.

1. Data Analysis

First, let’s analyze the properties of the numbers we have. In the case of positive numbers, multiplying the two largest numbers is advantageous for obtaining the maximum product. Conversely, for negative numbers, multiplying two negatives will yield a positive number, so utilizing the two smallest negative numbers is beneficial for finding the maximum product. Therefore, we need to consider the following two cases:

• The product of the two largest positive numbers
• The product of the two smallest negative numbers

2. Algorithm Design

Based on the above cases, we can design an algorithm. First, we will explore all elements of the array to separate the positive and negative numbers, and then find the maximum and minimum values for each.

1. Find the two largest numbers among the positives.
2. Find the two smallest numbers among the negatives.
3. Compare their products to determine the maximum value.

3. Kotlin Implementation

Now, let’s implement the algorithm in Kotlin.

fun maxProduct(numbers: IntArray): Long {
    val positives = mutableListOf()
    val negatives = mutableListOf()

    for (num in numbers) {
        if (num > 0) {
            positives.add(num)
        } else if (num < 0) {
            negatives.add(num)
        }
    }

    positives.sortDescending()
    negatives.sort()

    var maxProduct = Long.MIN_VALUE

    if (positives.size >= 2) {
        maxProduct = maxOf(maxProduct, positives[0].toLong() * positives[1])
    }

    if (negatives.size >= 2) {
        maxProduct = maxOf(maxProduct, negatives[0].toLong() * negatives[1])
    }

    return maxProduct
}

4. Test Cases

Now let’s create a few test cases to validate the function we wrote.

fun main() {
    // Test case 1: All numbers are positive
    println(maxProduct(intArrayOf(1, 2, 3, 4))) // Output: 12

    // Test case 2: All numbers are negative
    println(maxProduct(intArrayOf(-1, -2, -3, -4))) // Output: 6

    // Test case 3: Mixed numbers
    println(maxProduct(intArrayOf(-1, 2, -3, 4))) // Output: 6

    // Test case 4: Combined numbers
    println(maxProduct(intArrayOf(-10, -20, 5, 3, -2))) // Output: 200

    // Test case 5: Including 0
    println(maxProduct(intArrayOf(0, -5, -1, 2))) // Output: 5
}

Conclusion

In this session, we learned about the ‘Maximize Product by Grouping Numbers’ problem, the algorithm design process, and Kotlin programming techniques. This problem provides an opportunity to practice basic data processing methods by analyzing and separating the elements in an array. By implementing this in Kotlin, we hope to gain a better understanding of how theory applies in practice. We encourage you to continue improving your algorithm skills through various problems.