Category: Kotlin coding test

1. What is the Euler Problem?

Leonardo Euler was a mathematician who made many contributions to various fields of mathematics. The Euler problems, named after him, are primarily problems that require mathematical thinking and algorithmic approaches. These problems often focus on discovering ways to perform various calculations based on simple mathematical concepts.

2. Example of Euler Problem – Problem 1: Find the sum of all natural numbers from 1 to 1000 that are multiples of 3 or 5

To clearly understand the problem, let’s restate it.
We want to find the numbers from 1 to 1000 that are divisible by 3 or 5 and calculate their sum.

3. Problem-solving Approach

This problem can be solved using simple loops and conditional statements.
The following approach can be considered.

1. Iterate through all natural numbers from 1 to 1000.
2. Check if each number is divisible by 3 or 5.
3. Sum the numbers that meet the criteria.
4. Print the final result.

4. Implementing in Kotlin

Now, let’s solve the problem with Kotlin code according to the above approach.

fun main() {
        var sum = 0
        for (i in 1 until 1000) {
            if (i % 3 == 0 || i % 5 == 0) {
                sum += i
            }
        }
        println("The sum of natural numbers from 1 to 1000 that are multiples of 3 or 5 is: $sum")
    }

4.1 Explanation of the Code

The above code is structured as follows:

• var sum = 0: Initializes a variable to store the sum.
• for (i in 1 until 1000): Iterates from 1 to 999. The until keyword is used so that 1000 is not included.
• if (i % 3 == 0 || i % 5 == 0): Checks if the current number is a multiple of 3 or 5.
• sum += i: Adds the current number to the sum if it meets the condition.
• println(...): Prints the final result.

5. Execution Result

When the program is run, you can obtain the following result:

    The sum of natural numbers from 1 to 1000 that are multiples of 3 or 5 is: 233168
    

6. Additional Lessons Learned

Through this problem, we can learn a few important points:

• Step-by-step thought process for problem-solving.
• Effective use of conditional statements and loops.
• How to output results.

7. Extension of the Problem

This problem can be extended in several different ways. For example, you can change 1000 to 10000 or try using different numbers other than 3 and 5. Changing the scope of the problem can also provide more practice.

8. Conclusion

In this article, we explored how to solve the Euler problem using Kotlin. Algorithmic problems are great practice for improving various problem-solving skills. I hope you continue to solve various problems and gain experience.

9. References

• Project Euler
• Kotlin Documentation
• GeeksforGeeks

Problem Description

The problem of finding the sum of consecutive natural numbers is a common topic in coding tests.
Given the input integer N, the problem is to find M such that the sum of consecutive natural numbers from 1 to M equals N. Here, M must be a natural number greater than or equal to 1, and the sum of consecutive natural numbers can be calculated using the following formula:

S = 1 + 2 + 3 + ... + M = M * (M + 1) / 2

Input

– Integer N (1 ≤ N ≤ 1,000,000) – the sum of natural numbers to be found

Output

– Output M when the sum of consecutive natural numbers from 1 to M equals N.
– If M does not exist, output -1.

Approach to Problem

To solve this problem, it is useful to first rewrite the condition S = N into an equation.
When setting S = 1 + 2 + ... + M equal to N, rearranging gives us:

N = M * (M + 1) / 2

The above equation can be transformed to 2N = M * (M + 1).
We can then solve for M using this equation and check if the value is a natural number.

Solution Process

1. To find possible M for N, start with M equal to 1 and increase it while calculating S.
2. If S equals N, print M and terminate.
3. If S exceeds N, there is no need to search further, print -1 and terminate.

Code Implementation

Below is the code that implements the above logic in Kotlin.

fun findContinuousNaturalSum(N: Int): Int {
        var sum = 0
        for (M in 1..N) {
            sum += M
            if (sum == N) {
                return M
            } else if (sum > N) {
                return -1
            }
        }
        return -1
    }

    fun main() {
        val N = 15
        val result = findContinuousNaturalSum(N)
        println("The value of M such that the sum of consecutive natural numbers equals $N is: $result")
    }

Code Explanation

– The findContinuousNaturalSum function finds M such that the sum of consecutive natural numbers equals
N.
– The sum variable stores the sum from 1 to M.
– Using a for loop, M is incremented from 1 to N, while calculating sum.
– If sum equals N, M is returned, and if sum exceeds N, -1 is returned to terminate.

Examples

Input: 15
Output: 5
Explanation: 1+2+3+4+5 = 15, so M is 5.

Input: 14
Output: -1
Explanation: There is no case where the sum from 1 to M equals 14.

Conclusion

The problem of finding the sum of consecutive natural numbers is an important one that effectively utilizes basic loops and conditional statements in programming.
Through this problem, you can enhance your understanding of basic syntax and logic construction in Kotlin, which is beneficial for learning approaches to algorithmic problem-solving.
It is recommended to practice these fundamental problems sufficiently to solve a variety of future problems.

Problem Description

When planning a trip, we need to consider the time and cost required to travel between several cities. Let’s solve the problem of creating an optimal travel plan based on the information provided below.

Problem: Travel Route Optimization

You are given a graph representing N cities and the travel time/cost between each pair of cities. Write a program to calculate the optimal route that starts from the initial city, visits all cities in minimum time or cost, and returns to the starting city.

Input

• The first line contains the number of cities N (1 ≤ N ≤ 12).
• Next, an N x N matrix is provided where each element represents the travel time (or cost) between two cities. If travel is not possible, it is represented by -1.

Output

Print the minimum travel time (or cost).

Example

Problem Solving Process

1. Understanding the Problem

This problem involves finding a path that minimizes the travel time between the given cities. This is commonly known as the Traveling Salesman Problem (TSP), which is an NP-hard problem. The input graph is represented as an adjacency matrix, and we need to explore the optimal route based on this matrix.

2. Approach

To solve this problem, we will combine Depth-First Search (DFS) with a Dynamic Programming (DP) technique using bit masking. Bit masking allows us to efficiently represent the state of visited cities and helps avoid redundant calculations.

3. Kotlin Implementation

        
        fun findMinCost(graph: Array, visited: Int, pos: Int, n: Int, memo: Array): Int {
            // If all cities have been visited
            if (visited == (1 shl n) - 1) {
                return graph[pos][0] // Return to the starting city
            }

            // Memoization check
            if (memo[visited][pos] != -1) {
                return memo[visited][pos]
            }

            var ans = Int.MAX_VALUE

            // Explore all cities
            for (city in 0 until n) {
                // If not visited and travel is possible
                if ((visited and (1 shl city)) == 0 && graph[pos][city] != -1) {
                    val newCost = graph[pos][city] + findMinCost(graph, visited or (1 shl city), city, n, memo)
                    ans = Math.min(ans, newCost)
                }
            }

            // Save in memoization
            memo[visited][pos] = ans
            return ans
        }

        fun tsp(graph: Array, n: Int): Int {
            // Initialization
            val memo = Array(1 shl n) { IntArray(n) { -1 } }
            return findMinCost(graph, 1, 0, n, memo)
        }

        fun main() {
            val graph = arrayOf(
                intArrayOf(0, 10, 15, 20),
                intArrayOf(10, 0, -1, 25),
                intArrayOf(15, -1, 0, 30),
                intArrayOf(20, 25, 30, 0)
            )
            val n = graph.size
            val result = tsp(graph, n)
            println("Minimum travel cost: $result")
        }
        
    

4. Code Explanation

The above code shows the overall flow of the algorithm for optimizing the travel route. The main parts are as follows:

• findMinCost: Takes the current city and the state of visited cities as parameters and recursively calculates the minimum cost.
• tsp: Calculates the overall minimum cost using bit masking and memoization.

5. Performance Analysis

This code has a time complexity of O(N^2 * 2^N), where N is the number of cities. This is due to the nature of searching all paths with DFS and the feature of storing visited states using bit masking. It is practically executable for inputs with up to 12 cities, making it efficient for reasonably sized inputs.

Conclusion

Through the travel planning algorithm problem, we have learned how to utilize Dynamic Programming techniques with DFS and bit masking. This is a valuable skill for coding tests and algorithm challenges, so be sure to master it. Keep improving your algorithm skills through various problems!