Category: Kotlin coding test

In this article, we will address ‘card game’ related problems through a Kotlin-based algorithm problem-solving course.
We will explain in detail the theories, code examples, and various approaches necessary to solve algorithm problems.

Problem Description

A card game is a game that starts with two players each holding 5 cards.
Each player calculates their score based on the numbers on the cards.
We will describe a game where the scores of the two players are compared to determine the winner.

Problem Definition

There are two players, A and B. Each player A and B has 5 cards.
The score of the cards is based on their numbers. You need to write a program to determine which player is the winner by comparing the total scores of the two players.

Input

• The first line contains 5 cards of player A separated by spaces.
• The second line contains 5 cards of player B separated by spaces.

Output

• Print the name of the player with the higher total score.
• If the scores are the same, print “Draw”.

Example

    Input:
    3 5 2 1 4
    6 5 4 3 2

    Output:
    B
    

Approach to the Problem

To solve the problem, we go through the following steps:

1. Input the cards of players A and B as lists.
2. Calculate the scores for each player.
3. Compare the scores to determine the winner.

Code Implementation

Below is the code written in Kotlin.

fun main() {
        // Read input.
        val aCards = readLine()!!.split(" ").map { it.toInt() }
        val bCards = readLine()!!.split(" ").map { it.toInt() }

        // Calculate scores.
        val aScore = aCards.sum()
        val bScore = bCards.sum()

        // Determine the winner.
        when {
            aScore > bScore -> println("A")
            aScore < bScore -> println("B")
            else -> println("Draw")
        }
    }

Code Analysis

The above code performs the following functions:

• readLine() reads the two lines of input from the user and splits them into a list of integers based on spaces.
• Each player’s scores are calculated using the sum() method.
• The scores are compared using the when statement, and the name of the player with the higher score is printed.

Testing and Validation

It is essential to test the code in various situations to validate that the correct output is produced based on the input.
You should consider various test cases, including edge cases such as when all cards of player A are zero.

Conclusion

In this lecture, we explored the process of solving a simple card game problem using Kotlin.
Understanding the problem, designing the algorithm, and implementing it in code are vital steps in solving algorithm problems,
which can be improved through practice with various approaches.
Keep practicing consistently and solving various problems to enhance your coding skills.

Additional Practice Problems

Try solving the following additional problems:

• Write a program to calculate how the scoring system changes if each player holds 7 cards.
• Implement a program to add simple rules affecting the final score through card betting for each player.

Consistent practice is essential. Best wishes for successful coding test preparation!

1. Introduction

The key to algorithm problem solving is to develop the ability to understand and solve various problems. Among them, the Longest Common Subsequence (LCS) problem is a fundamental problem that frequently appears in several algorithm tests. In this article, we will take a closer look at the process of solving the LCS problem using Kotlin.

2. Problem Description

Given two strings, the problem is to find the length of the longest subsequence that appears in both strings without deleting any characters. In other words, it is to find the longest subsequence that can be formed while maintaining the order of characters in both strings.

Example

• String 1: “ABCBDAB”
• String 2: “BDCAB”
• Longest Common Subsequence: “BCAB” (Length: 4)

3. Approach to Problem Solving

A common approach to solve the LCS problem is Dynamic Programming. Dynamic Programming is a methodology for solving complex problems by breaking them down into simpler subproblems. Here, we will compare each character of the two strings and calculate the maximum length.

3.1. Defining the Dynamic Programming Table

First, let the lengths of the two strings be m and n, and define dp[i][j] as the length of the LCS of the first i characters of string 1 and the first j characters of string 2. The size of this table is (m+1) x (n+1). The first row and column of the table are initialized to 0.

3.2. Setting the Recurrence Relation

The table will be filled according to the following rules, comparing characters up to the last character of both strings:

• When characters are the same: dp[i][j] = dp[i-1][j-1] + 1
• When characters are different: dp[i][j] = max(dp[i-1][j], dp[i][j-1])

3.3. Extracting the Result

The last cell dp[m][n] of the completed dp table contains the length of the LCS for the two strings.

4. Kotlin Implementation

Now, based on the theoretical explanation, let’s calculate the LCS using Kotlin.

fun longestCommonSubsequence(s1: String, s2: String): Int {
    val m = s1.length
    val n = s2.length
    val dp = Array(m + 1) { IntArray(n + 1) }

    for (i in 1..m) {
        for (j in 1..n) {
            if (s1[i - 1] == s2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1
            } else {
                dp[i][j] = maxOf(dp[i - 1][j], dp[i][j - 1])
            }
        }
    }

    return dp[m][n]
}

fun main() {
    val s1 = "ABCBDAB"
    val s2 = "BDCAB"
    println("Length of Longest Common Subsequence: ${longestCommonSubsequence(s1, s2)}")
}
            

5. Analysis

The above code has a time complexity of O(m * n) and a space complexity of O(m * n). Therefore, as the length of the strings increases, performance may decrease. However, the LCS problem can be improved through various optimization techniques.

5.1. Improving Space Efficiency

Currently, a 2D array is used, but since only the previous row’s values are needed, it can also be solved with a 1D array. This can reduce the space complexity to O(n).

fun longestCommonSubsequenceOptimized(s1: String, s2: String): Int {
    val m = s1.length
    val n = s2.length
    val dp = IntArray(n + 1)

    for (i in 1..m) {
        val prev = IntArray(n + 1)
        for (j in 1..n) {
            if (s1[i - 1] == s2[j - 1]) {
                dp[j] = prev[j - 1] + 1
            } else {
                dp[j] = maxOf(dp[j - 1], prev[j])
            }
            prev[j] = dp[j]
        }
    }
    return dp[n]
}
            

6. Conclusion

The Longest Common Subsequence problem is a very important problem in algorithm study and code testing. It can be efficiently solved using dynamic programming, and we have learned that there are various optimization techniques. I hope this problem provides an opportunity to develop algorithmic thinking and practice Kotlin programming.

Coding tests and algorithm problems come in various types and difficulties. In this course, we will explore how to arrange parentheses to minimize an expression. This problem asks how efficiently we can calculate an expression through the placement of parentheses.

Problem Description

This is a problem of finding an arrangement of parentheses that minimizes an expression made up of given numbers and operators.
For example, given the expression “2 – 3 + 5”, we need to appropriately place parentheses to compute the minimum value.
Parentheses can be freely placed around each operator, and the goal is to minimize the resulting value.

Input

The expression is given as a string and consists of only the operators + and -. The expression does not contain spaces.
The length of the expression is between 1 and 50 characters.

Output

The output is the minimum value calculated by appropriately placing parentheses.

Problem Solving Process

To solve the problem, we can use either Dynamic Programming or Divide and Conquer techniques.
Here, we will use the Divide and Conquer approach to solve the problem.

Step 1: Parsing the Input

To solve the problem, we first need to separate the input expression into numbers and operators.
The numbers in the expression will be integers, and the operators will be either ‘+’ or ‘-‘.
Thus, by iterating through the string, we will add numbers to a number array and operators to an operator array when encountered.

Step 2: Considering All Possible Cases

Next, we must consider all possible arrangements of parentheses and calculate the minimum value of the expression.
To do this, we will split the expression into left and right parts for each operator and find their respective minimum values.
By applying the operator to the minimum values from both sides, we can derive a new value and find the minimum among them.

Step 3: Recursive Approach

By implementing the above process recursively, we can break down the problem into subproblems and solve them.
At each step, the expression is split based on the operator, and the minimum values for the left and right parts are calculated.

Example Code

        
            fun findMinimumValue(expression: String): Int {
                val numbers = mutableListOf()
                val operators = mutableListOf()
                var num = StringBuilder()

                // Parsing the expression
                for (char in expression) {
                    when {
                        char.isDigit() -> num.append(char)
                        char == '+' || char == '-' -> {
                            numbers.add(num.toString().toInt())
                            operators.add(char)
                            num = StringBuilder()
                        }
                    }
                }
                // Adding the last number
                if (num.isNotEmpty()) {
                    numbers.add(num.toString().toInt())
                }

                return calculateMinimum(numbers, operators, 0, numbers.size - 1)
            }

            fun calculateMinimum(numbers: List, operators: List, start: Int, end: Int): Int {
                if (start == end) return numbers[start]

                var minValue = Int.MAX_VALUE

                for (i in start until end) {
                    val leftMin = calculateMinimum(numbers, operators, start, i)
                    val rightMin = calculateMinimum(numbers, operators, i + 1, end)

                    val result = when (operators[i]) {
                        '+' -> leftMin + rightMin
                        '-' -> leftMin - rightMin
                        else -> throw IllegalArgumentException("Unsupported operator: ${operators[i]}")
                    }

                    minValue = minOf(minValue, result)
                }

                return minValue
            }

            fun main() {
                val expression = "2-3+5"
                val result = findMinimumValue(expression)
                println("Minimum value: $result")
            }
        
    

Conclusion

In this course, we tackled the problem of arranging parentheses to minimize an expression.
Although there are various approaches to solving such problems, we used recursive and divide and conquer methods here.
If you encounter such problems in actual coding tests, it is crucial to thoroughly analyze the problem and approach it with an efficient and clear algorithm.
We encourage you to continue seeking solutions to various problems in the future.

Hello! In this post, we will discuss the topic “Finding Minimum Value 2” as part of the Kotlin coding test course. Among various algorithm problems, finding the minimum value is a common problem with various variations. In this article, we will define the problem and explain the Kotlin algorithm to solve it step by step.

Problem Definition

Let’s solve the following problem:

Problem: Given an integer array nums, write a program that finds the smallest element in the array. The size of the array is assumed to be at least 1, and the integer values are assumed to range from -10⁹ to 10⁹. If the array is empty or contains values that differ from the specification, it should throw an IllegalArgumentException.

Problem Analysis

To solve the problem, we need to consider an algorithm that finds the smallest number in the array. The basic idea to find the minimum is to traverse the array once and update the minimum value by comparing each element. The method of traversing the array once has a time complexity of O(n), which is very efficient. There are various algorithms besides this method, but here we will use the most basic approach.

Algorithm Design

The algorithm consists of the following steps:

1. Exception Handling: Throw an IllegalArgumentException if the array is empty.
2. Initialize Minimum Value: Initialize the minimum value with the first element of the array.
3. Traverse Array: Check each element of the array and update the minimum value if a value less than the current minimum is found.
4. Return Minimum Value: Finally, return the found minimum value.

Kotlin Code Implementation

Now let’s look at the code that implements the above algorithm:

fun findMinimum(nums: IntArray): Int {
    // 1. Check if the array is empty
    if (nums.isEmpty()) {
        throw IllegalArgumentException("The array is empty.")
    }

    // 2. Initialize the minimum value
    var minValue = nums[0]

    // 3. Traverse the array
    for (num in nums) {
        if (num < minValue) {
            minValue = num
        }
    }

    // 4. Return the minimum value
    return minValue
}

Code Explanation

Looking at the above code:

• fun findMinimum(nums: IntArray): Int: This function takes an integer array as a parameter and returns the minimum value.
• if (nums.isEmpty()): If the array is empty, it throws an exception.
• var minValue = nums[0]: Initializes the minimum value with the first element of the array.
• for (num in nums): Traverses all elements of the array
• if (num < minValue): If the current element is less than the minimum value, it updates the minimum value.
• Finally, it returns the minimum value.

Test Cases

Now let’s look at some test cases to test the implemented function:

fun main() {
    val testCases = arrayOf(
        intArrayOf(3, 5, 1, 8, 2),
        intArrayOf(-1, -3, -5, -2),
        intArrayOf(10),
        intArrayOf(0, 1, 0, -1, -2, 2, 1),
        intArrayOf()
    )

    for (testCase in testCases) {
        try {
            println("Test Case: ${testCase.joinToString(", ")} -> Minimum Value: ${findMinimum(testCase)}")
        } catch (e: IllegalArgumentException) {
            println("Test Case: ${testCase.joinToString(", ")} -> Exception Occurred: ${e.message}")
        }
    }
}

Test Results

The results of running the above test cases are as follows:

• Test Case: 3, 5, 1, 8, 2 -> Minimum Value: 1
• Test Case: -1, -3, -5, -2 -> Minimum Value: -5
• Test Case: 10 -> Minimum Value: 10
• Test Case: 0, 1, 0, -1, -2, 2, 1 -> Minimum Value: -2
• Test Case: (Empty Array) -> Exception Occurred: The array is empty.

Conclusion

In this lecture, we reviewed the process of solving the minimum value finding problem using Kotlin. Although it is a basic algorithm to find the smallest value in an array, it allows us to solve the problem efficiently and simply by traversing the array and checking conditions. Such problems are frequently encountered in coding tests, making them very useful for practicing basic algorithms.

I hope this lecture has been helpful, and that it aids you in enhancing your algorithm skills! If you have any additional questions or feedback, feel free to leave a comment.

References

• Kotlin Official Documentation
• LeetCode – Algorithm Problem Practice Platform
• GeeksforGeeks – Programming Problems and Solutions

I hope you continue to learn various problems and algorithms. Thank you!