kotlin coding test course, 2 N tile filling

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!