Python Coding Test Course, 2 N Tile Filling

In today’s world, where programming and algorithms are becoming increasingly important, many companies require algorithm and coding tests as a prerequisite for employment. One of the popular problems is 2*N Tile Filling. In this article, I will analyze the problem and explain the process of solving it in detail. Through this problem, we will also learn the concept of dynamic programming (DP) and look at a Python code implementation using it.

Problem Description

The problem is as follows:

Find the number of ways to fill a 2 x n rectangular space with tiles of size 1 x 2 or 2 x 1.

For example, when n=3, the tiles can be filled in the following ways:

  • 3 vertical 1 x 2 tiles
  • 1 horizontal 1 x 2 tile and 2 vertical tiles
  • 2 horizontal 1 x 2 tiles and 1 vertical tile
  • 1 2 x 1 tile and 2 1 x 2 tiles
  • 3 vertical 2 x 1 tiles

The solution to this problem can be approached through dynamic programming. The key is to recursively divide the problem based on how each tile is placed and avoid duplicate calculations by storing results using the memoization technique.

Approach to Problem Solving

This problem can be solved based on the following rules:

  • Base case: For n=1, only the method of placing a 1 x 2 tile vertically is possible. Therefore, the number of cases is 1.
  • More general case: For n=2, there are two ways: either to place 2 horizontal 1 x 2 tiles or to place 1 vertical 2 x 1 tile. Thus, the number of cases is 2.
  • Recursive relationship: For n > 2, it can be divided in two ways:
    1. Place a 1 x 2 tile and fill the remaining 2 x (n-1) space
    2. Place a 2 x 1 tile and fill the remaining 2 x (n-2) space

    Therefore, the recursive relation can be expressed as:
    f(n) = f(n-1) + f(n-2)

Dynamic Programming Implementation

Now, let’s write a Python code to solve the problem using dynamic programming (DP) based on the recursive relationship above.


def tile_fill(n):
    # Initialize the DP array.
    dp = [0] * (n + 1)
    
    # Set the base cases.
    dp[0] = 1  # The case of filling nothing
    if n >= 1:
        dp[1] = 1  # The case of placing a 1 x 2 tile vertically
    
    # Fill the DP array.
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    
    return dp[n]

Code Explanation

In the above code, we first initialize the DP array from 0 to n. After setting the values for f(0) and f(1) as base cases, we proceed to fill the values from f(2) to f(n) using a loop. Ultimately, the value of dp[n] represents the number of ways to fill the 2 x n rectangle.

Time Complexity Analysis

The time complexity of this algorithm is O(n). This is because in each iteration, the DP array only repeats as many times as the number of elements in the n structure. The space complexity is also O(n).

Summary

In this article, we learned how to utilize dynamic programming techniques through the 2*N Tile Filling Problem. By defining the problem, deriving solutions through repetitive formulas, and implementing this in code, we were able to develop algorithmic thinking. Similar problems may appear in actual coding tests, so it is advisable to practice consistently.

Problem Variations and Applications

If the given rectangular size were to change from 2 x N to another form, for example, to 3 x N, think about how you would approach it. Practicing variations of problems will significantly help in deepening algorithmic thinking.

In conclusion, when solving algorithm problems, it is important to decompose the problem and explore possible solutions. The 2*N tile filling problem is a good example of how to embody that kind of thinking.