Category: Kotlin coding test

This course delves deeply into the 2*N tile filling problem and details the algorithms and Kotlin implementation methods to solve it.

Problem Description

This is a problem of calculating the number of ways to fill a rectangle of size 2*N with tiles of size 1*2. Tiles can be placed either horizontally or vertically. The goal is to find the number of possible tile placements for a given N value.

Examples:
– N=1: 1 way (1 tile placed vertically)
– N=2: 2 ways (1 tile placed horizontally/vertically)
– N=3: 3 ways
– N=4: 5 ways
– N=5: 8 ways

As the value of N increases, the number of possible arrangements also increases. This problem is closely related to the Fibonacci sequence.

Approach

To solve this problem, the initial approach is as follows:

1. Recursive Approach: This involves exploring the ways to place the tiles to find possible combinations. Starting from 1 way when N is 0 and 1 way when N is 1, we explore all cases.
2. Dynamic Programming: This method uses the Fibonacci sequence to store previous results and solve larger problems based on those results.

Solution Using Dynamic Programming

The most efficient way is to use Dynamic Programming. This method avoids redundant calculations and solves the problem with a time complexity of O(N).

Algorithm Explanation

We define a DP array for the calculations:

        dp[i] = number of ways to fill a rectangle of size i
        - dp[0] = 1 (filling nothing)
        - dp[1] = 1 (1*2 tile placed vertically)
        - dp[2] = 2 (two 1*2 tiles placed horizontally or vertically)
        - dp[n] = dp[n-1] + dp[n-2]
        

Now we fill the dp array sequentially according to the range of N. dp[n-1] accounts for the case where the last tile is placed vertically, while dp[n-2] accounts for the case where the last two tiles are placed horizontally.

Kotlin Implementation

Code Example

        fun tileCount(n: Int): Int {
            if (n == 0) return 1
            if (n == 1) return 1

            val dp = IntArray(n + 1)
            dp[0] = 1
            dp[1] = 1

            for (i in 2..n) {
                dp[i] = dp[i - 1] + dp[i - 2]
            }

            return dp[n]
        }

        fun main() {
            val n = 5 // Example with N=5
            println("Number of ways to fill a 2*$n tile: ${tileCount(n)}")
        }
        

When you run the above code, it outputs the number of ways to fill a rectangle of size 2*5.

Time Complexity Analysis

The time complexity of this algorithm is O(N), and the space complexity is also O(N). As N increases, the computation time and memory usage increase linearly. This level of complexity is efficient for most real-world problems.

Conclusion

The 2*N tile filling problem can be easily and efficiently solved through recursive approaches and dynamic programming. I hope that understanding the problem structure through Kotlin solutions provides a good experience that can be applied in real situations.

This concludes the problem-solving process for the 2*N tile filling problem. I encourage you to practice various algorithms by working on different problems!

Problem Description

This problem is an algorithmic problem of sorting a given sequence. Specifically,
you need to write a program that receives N integers as input, sorts
them in ascending order, and then prints each integer on a separate line.

Input Format

The first line contains an integer N (1 ≤ N ≤ 1,000,000).
The second line contains N integers, each of which is
between -1,000,000,000 and
1,000,000,000.

Output Format

Sort the input integers in ascending order and print each integer on
a separate line.

Approach to the Problem

To solve this problem, you can use an algorithm to sort the array.
Kotlin provides a powerful library function for array sorting by default.
However, due to the large size of numbers, attention must be paid to time complexity.
A special sorting algorithm should be applied for optimal performance.

Algorithm Selection

Given that the input size of the problem can go up to 1,000,000,
it is advisable to use efficient sorting algorithms like quicksort or
mergesort, which have a time complexity of O(N log N).
Additionally, in Kotlin, you can easily sort by utilizing the
sort() method from the standard library.

Code Implementation

Below is the solution code using Kotlin:

fun main() {
    val n = readLine()!!.toInt()  // Input N from the first line
    val numbers = IntArray(n)      // Create an integer array

    for (i in 0 until n) {
        numbers[i] = readLine()!!.toInt()  // Input the next N integers
    }

    numbers.sort()  // Perform sorting

    numbers.forEach { println(it) }  // Output the sorted numbers
}
    

Code Explanation

1. Use readLine()!!.toInt() to read integers from standard input.
Read the value of N from the first line and store all subsequent N integers in the
numbers array.

2. Use the numbers.sort() method to sort
the array in ascending order. This method uses
an efficient sorting algorithm internally.

3. Finally, print the sorted numbers line by line using
forEach.

Example of Results

Below is an example input and output after running this program:

Input:

5
5
2
3
1
4
    

Output:

1
2
3
4
5
    

Memory and Time Complexity

This algorithm has a time complexity of O(N log N). This is because
it sorts by dividing the input array size logarithmically.
Memory usage requires O(N) to store the array.

Summary of Problem-Solving Process

• Store all numbers given as input in an array.
• Sort the array in ascending order.
• Output the sorted result.

Conclusion

In this tutorial, we learned how to sort numbers using Kotlin.
You should practice solving algorithmic problems systematically to
derive correct results even with various inputs and conditions.

Keep enhancing your skills through different algorithmic problems!