Category: Kotlin coding test

In modern programming environments, coding tests are becoming an essential element.
Especially, they are establishing themselves as one of the various methods to find outstanding talents in the software development field.
This article will explore the algorithm problem-solving process using the Kotlin language,
along with how to effectively solve problems using debugging.

Algorithm Problem: Two Sum

Given an integer array nums and an integer target,
the problem is to select two numbers from nums such that their sum returns the indices that equal target.
You cannot use the same element twice, and it is assumed that there is exactly one solution.
The problem can be summarized as follows.

Problem Description

• Input: nums = [2, 7, 11, 15], target = 9
• Output: [0, 1] (nums[0] + nums[1] = 2 + 7 = 9)

How to Solve the Problem

There can be several methods to solve this problem, but one of the most efficient ways is to use a hash map.
Using a hash map allows us to find elements with an average time complexity of O(1).

Algorithm Approach

• Initialize a hash map.
• Iterate through the array and calculate the differences for each element.
• Check if this difference already exists in the hash map.
• If it exists, return the corresponding index; otherwise, store the current element and its index in the hash map.

Code Implementation

        
fun twoSum(nums: IntArray, target: Int): IntArray {
    val map = mutableMapOf()

    for ((index, num) in nums.withIndex()) {
        val complement = target - num
        if (map.containsKey(complement)) {
            return intArrayOf(map[complement]!!, index)
        }
        map[num] = index
    }
    throw IllegalArgumentException("No two sum solution")
}
        
    

Using Debugging

The code above implements the basic logic; however, in an actual coding test, various errors that may occur during execution need to be considered.
Here, we will explain how to use debugging tools.

Kotlin Debugging Methods

• Utilizing the IDE’s Debugging Features: In IDEs like IntelliJ IDEA, you can step through the code execution step by step to find out where issues occur.
• Log Output: You can use log outputs, such as println, to print the value of specific variables or track the progress of the code. For example, you might print the contents of the hash map to find overlapping values.
• Exception Handling: In case of exceptions, you can determine where the issue occurred by using appropriate error messages.
  For instance, if the problem’s data differs from what is expected, an IllegalArgumentException-related error message can be printed.

Problem Solving Through Debugging Process

1. First, write the code, set a few input values, and run it in debugging mode.
2. Check the value of variables at each step and compare them against expected results.
3. Inspect the state values or changes in the hash map to ensure that the correct index is being tracked.
4. If it does not function correctly, change the conditions to find the root cause and add logs to test various scenarios.
5. Once the issue is resolved, review the final code and check for any areas that can be optimized.

Conclusion

Solving algorithm problems using Kotlin begins with choosing efficient data structures and algorithms.
Debugging techniques are crucial skills for developers, allowing them to resolve issues through appropriate approaches when problems arise.
Additionally, through practice with various problems, debugging skills can also be developed.

If you are preparing for a coding test, it is important to not only understand the problem but also to gain experience in tracking and solving the root causes of issues using debugging.
I hope you enjoy and efficiently prepare for coding tests by utilizing the various features provided by the Kotlin language!

This article deals with an algorithm problem to find the minimum number of coins required, and it will explain the Kotlin code and process to solve it in detail.

Problem Description

When we have various coin values, we need to find the minimum number of coins required to make a specific amount. The goal is to minimize the number of coins needed to make the given amount.

Problem Input:

• Number of coin types N (1 ≤ N ≤ 100)
• Value of each coin Aᵢ (1 ≤ Aᵢ ≤ 10,000)
• Target amount M (1 ≤ M ≤ 1,000,000)

Problem Output:

Print the minimum number of coins required to make the target amount M. If it is not possible to make the amount, print -1.

Example

Input Example:

        3
        1
        3
        4
        6
        

Output Example:

        2
        

In the above example, there are 3 types of coins with values 1, 3, and 4. To make the target amount 6, we can use 2 coins by either (3 + 3) or (4 + 1 + 1).

Problem Solving Process

To solve this problem, we can use Dynamic Programming. Dynamic Programming is a technique to solve problems by reusing previous results, which can be very effectively applied in this problem.

1. Setting up the DP array

First, we declare a DP array to store the number of coins. DP[i] signifies the minimum number of coins needed to make amount i. During initialization, we set DP[0] to 0 and the rest to infinity (INF).

2. Defining the recurrence relation

We update the DP array by iterating over the values of each coin up to the amount M. Given a coin’s value, we start from that value and update the minimum number of coins for every possible amount.

The recurrence relation is as follows:

        DP[j] = min(DP[j], DP[j - coin] + 1)
        

Here, coin is the value of the currently considered coin.

3. Deriving the final result

After calculating the DP array for all coins and amounts, we check DP[M] to determine if that value is infinity (INF) or not. If it is infinity, we cannot make amount M, so we print -1; otherwise, we print the value of DP[M].

Code Implementation

fun minCoins(coins: IntArray, target: Int): Int {
    // Initialize the DP array to infinity
    val dp = IntArray(target + 1) { Int.MAX_VALUE }
    dp[0] = 0 // 0 coins are needed to make 0

    // Update the DP array for each coin value
    for (coin in coins) {
        for (j in coin..target) {
            if (dp[j - coin] != Int.MAX_VALUE) {
                dp[j] = Math.min(dp[j], dp[j - coin] + 1)
            }
        }
    }

    // Check the result
    return if (dp[target] == Int.MAX_VALUE) -1 else dp[target]
}

fun main() {
    // Input values
    val coins = intArrayOf(1, 3, 4)
    val target = 6
    val result = minCoins(coins, target)
    println(result) // Output: 2
}
        

Comparison and Optimization

The time complexity of this algorithm is O(N * M) and the space complexity is O(M). This algorithm is efficient in terms of time, especially when there are many coin types or the target amount is very large. However, there are ways to optimize the code further. For example, sorting the coin values in descending order and processing larger coins first can lead to faster results.

Various Cases

Here, we validate the performance of the algorithm through various input values. Testing different combinations of coin types and amounts to derive results is also an important step.

Case 1:

Input: 4, 2, 5, 7, target amount = 14

Output: 2 (using 2 coins of 7)

Case 2:

Input: 1, 3, 4, target amount = 11

Output: 3 (using 2 coins of 4 and 1 coin of 3)

1. Overview of Dynamic Programming

Dynamic programming is a methodology for solving complex problems by breaking them down into simpler subproblems.
This technique is especially useful for solving optimization problems and addresses them repeatedly using well-defined rules.
Generally, dynamic programming must meet two conditions:

• Optimal Substructure: An optimal solution to the problem must be composed of optimal solutions to its subproblems.
• Overlapping Subproblems: There must be cases where subproblems are solved multiple times.

2. When to Use Dynamic Programming

Dynamic programming is primarily used in the following cases:

• Calculating Fibonacci sequences
• Longest Common Subsequence
• Minimum Edit Distance
• Knapsack Problem

3. Algorithm Problem: Knapsack Problem

Now, let’s take a deeper look at the knapsack problem using dynamic programming. This problem involves selecting items to achieve the maximum value within a given weight limit.

Problem Description

Given a weight limit for the bag, the problem is to find the maximum value achievable with a combination of items when their weights and values are known.
The input is in the following format:

• n: Number of items
• W: Maximum weight of the bag
• weight[]: Array of weights of each item
• value[]: Array of values of each item

Example Input

    n = 4
    W = 7
    weight[] = {1, 2, 3, 2}
    value[] = {20, 5, 10, 40}
    

Example Output

    Maximum Value: 60
    

4. Solving Process Using Dynamic Programming

To solve this problem, we will apply the basic idea of dynamic programming.
By selecting items one by one, we will calculate the maximum value considering the current weight that can be carried in the bag.

4.1 Initializing the DP Table

The DP table will be initialized as a two-dimensional array. Here, dp[i][j] stores the maximum value for a maximum weight j considering the first i items.
The initial value will be set to 0.

4.2 State Transition

We will define state transitions based on whether to include an item or not.
For each item i, the maximum value will be calculated under the following two conditions:

• When item i is not included: dp[i][j] = dp[i-1][j]
• When item i is included: dp[i][j] = dp[i-1][j-weight[i-1]] + value[i-1] (if j >= weight[i-1])

4.3 Final Implementation

We will now implement the above logic using Kotlin.

    fun knapsack(n: Int, W: Int, weight: IntArray, value: IntArray): Int {
        val dp = Array(n + 1) { IntArray(W + 1) }

        for (i in 1..n) {
            for (w in 0..W) {
                if (weight[i - 1] <= w) {
                    dp[i][w] = maxOf(dp[i - 1][w],
                                    dp[i - 1][w - weight[i - 1]] + value[i - 1])
                } else {
                    dp[i][w] = dp[i - 1][w]
                }
            }
        }
        return dp[n][W]
    }
    

5. Complexity Analysis

The time complexity of this algorithm is O(nW), and the space complexity is also O(nW).
Here, n is the number of items, and W is the maximum weight of the bag.
However, there is a way to optimize space complexity.

5.1 Space Complexity Optimization Method

Since dp[i][j] only needs information from the previous row, it can be optimized using a one-dimensional array.

    fun knapsackOptimized(n: Int, W: Int, weight: IntArray, value: IntArray): Int {
        val dp = IntArray(W + 1)

        for (i in 0 until n) {
            for (w in W downTo weight[i]) {
                dp[w] = maxOf(dp[w], dp[w - weight[i]] + value[i])
            }
        }
        return dp[W]
    }
    

6. Conclusion

Dynamic programming is a powerful tool for efficiently solving various problems.
We learned the concepts and implementation methods of dynamic programming through the knapsack problem.
I hope you continue to practice and learn through different problems in the future.

I hope this article helps you prepare for coding tests using Kotlin. Keep learning and practicing to find the optimal solution!

Problem Description

There are N islands given, and plans are in place to build bridges between these islands to connect them. However, bridges can only be built once between each pair of islands, and the length of the bridge is calculated based on the shortest distance between the two islands.
Additionally, in order to build a bridge, the height of each island must be considered according to their terrain.
It is necessary to consider how the height difference between the two islands affects the bridge length.

The problem is to calculate the minimum bridge length required to connect all the islands based on the given height information of the islands.
The length of the bridge is proportional to the height difference between the two islands.
Write a program that finds the minimum bridge length required to connect all islands by building bridges for all possible pairs based on the given heights of the islands.

Input Format

The first line contains the number of islands N (2 ≤ N ≤ 100).
The second line provides N integers separated by spaces, representing the height H (0 ≤ H ≤ 100) of each island.

Output Format

Output the minimum bridge length required to connect all the islands.

Problem Solving Process

1. Understanding the Problem and Designing an Approach

To solve the bridge construction problem, we will approach it through the Minimum Spanning Tree (MST) of a graph. This problem requires calculating the bridge lengths between each pair of islands to connect all islands and minimizing this total,
hence we can use either Kruskal’s algorithm or Prim’s algorithm.
As negative weights are not allowed, when building a bridge between A and B, we define the bridge length as the height difference between A and B, resulting in separate edges in the graph based on the height of the islands.

2. Graph Creation

The edges between each pair of islands correspond to the height difference of the two islands, and the length of the bridge is defined as |H[i] – H[j]|.
We can now calculate the bridge lengths between all pairs of islands to construct the graph.
To achieve this, we will use a nested loop to calculate the bridge lengths for all pairs of islands.

3. Applying the MST Algorithm

Now we will use Kruskal’s algorithm to calculate the minimum bridge length required to connect all islands.
The Kruskal algorithm constructs the MST by first sorting the edges by weight and then adding the edges one by one to avoid forming cycles.

4. Implementation and Validation

Below is the code implemented in Kotlin as described above.
This code takes input to calculate bridge lengths and ultimately outputs the minimum bridge length required to connect all the islands.

        
        import java.util.*

        fun main() {
            val scanner = Scanner(System.`in`)
            val n = scanner.nextInt()
            val heights = IntArray(n)
            for (i in 0 until n) {
                heights[i] = scanner.nextInt()
            }

            val edges = mutableListOf()
            for (i in 0 until n) {
                for (j in i + 1 until n) {
                    val weight = Math.abs(heights[i] - heights[j])
                    edges.add(Edge(i, j, weight))
                }
            }

            edges.sortBy { it.weight }

            val parent = IntArray(n) { it }

            fun find(x: Int): Int {
                if (parent[x] != x) {
                    parent[x] = find(parent[x])
                }
                return parent[x]
            }

            fun union(x: Int, y: Int) {
                val rootX = find(x)
                val rootY = find(y)
                if (rootX != rootY) {
                    parent[rootY] = rootX
                }
            }

            var totalLength = 0
            for (edge in edges) {
                if (find(edge.start) != find(edge.end)) {
                    union(edge.start, edge.end)
                    totalLength += edge.weight
                }
            }

            println("The minimum bridge length required to connect all the islands is: $totalLength")
        }

        data class Edge(val start: Int, val end: Int, val weight: Int)
        
    

Optimization and Considerations

The bridge construction problem can be solved using simple graph and MST algorithms.
However, this algorithm may require relatively high memory usage because it stores all edges.
Therefore, if the height information of each island and the connection information are stored in a more optimized data structure,
better performance can be achieved.

Moreover, due to the nature of the graph, this problem can also work well with more islands or more complex shapes.
In actual coding tests, having a fundamental understanding and grasping the algorithms needed to solve such problems is crucial.
It is also beneficial to practice and familiarize oneself with the patterns of similar problems.

Conclusion

Through this problem, I was able to learn about the use of data structures and algorithms in Kotlin.
The Minimum Spanning Tree problem is one of the topics in coding tests, and it’s important to repeatedly study various modified problems.
A deep understanding of graph theory will be a great help in coding tests.