Problem Description

You are given a problem related to tiling a 2 x n board with two types of shapes: a 2 x 1 domino and an L shaped tromino. The shapes can be rotated as needed to fit into the tiling of the board. The task is to find out how many different ways you can completely cover the 2 x n board using these tiles. It's important to note that two tilings are considered different only if there's at least one pair of adjacent cells in the board that is covered in one tiling and not covered in the other. Since the number of ways can be very large, the answer should be returned modulo 10^9 + 7.

Intuition

To solve this problem, we can use dynamic programming because the problem has an optimal substructure and overlapping subproblems. Each state in our dynamic programming represents the number of ways to tile a board up to positions (i, j), where i is the position on the upper row and j is the position on the lower row.

Starting from a fully untiled board (0, 0), we consider all possible moves we can make with either of our two types of tiles—one that takes us to (i + 1, j + 1) which represents placing a domino tile vertically, to (i + 2, j + 2) representing placing two domino tiles horizontally next to each other, to (i + 1, j + 2) or (i + 2, j + 1) representing placing a tromino tile. Every move is a transition to a new state, and the number of ways to tile the board for that state is the sum of ways of reaching it from all previous states.

The recursion ends when we consider the entire board tiled (i == n and j == n), which is our base case and we return 1 as there is exactly one way to have a whole tiled board from an entirely tiled board—do nothing.

Since there are overlapping subproblems (we could reach a state from different previous moves), memoization via the @cache decorator is used to avoid redundant calculations. Lastly, every addition of ways is done modulo 10^9 + 7 to prevent integer overflow and adhere to the problem constraints of returning the count of ways modulo 10^9 + 7.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution to this problem makes use of a recursive depth-first search (DFS) approach along with memoization to count the number of ways to tile the 2 x n board. The DFS function dfs(i, j) is calculating the number of tilings for the board up to the point (i, j), bearing in mind that i and j represent the ending positions on the upper and lower rows, respectively.

The solution uses the @cache decorator to memorize intermediate results. This is because the DFS function will be called many times with the same arguments, and we want to avoid recalculating those results each time. The memoization optimizes our approach from an exponential time complexity to a polynomial time complexity.

Here is the step-by-step approach:

1. If either i or j exceeds n, we've gone beyond the board, which means this is not a valid tiling, so return 0.
2. If i == n and j == n, we succeeded in tiling the entire board and thus return 1.
3. If i == j, both rows are equally tiled thus far. We can place either two dominoes side by side (horizontally), one domino vertically, or a tromino (which will cover 2 x 1 area and one additional square either above or below). We accordingly call the dfs function for the following possible next steps: dfs(i + 2, j + 2), dfs(i + 1, j + 1), dfs(i + 2, j + 1), and dfs(i + 1, j + 2).
4. If i > j, the upper row is ahead, so we can only place a domino horizontally on the lower row or a tromino that covers two cells on the lower row and one on the upper. Thus, the new states to go to are dfs(i, j + 2) and dfs(i + 1, j + 2).
5. If i < j, the lower row is ahead, so we can place a domino vertically on the upper row to reach dfs(i + 2, j) or place a tromino in a way that covers two cells on the upper row and one on the lower to reach dfs(i + 2, j + 1).

In each of these cases, we add up the counts returned by all possible step(s) taken from the current state, and this sum is then returned modulo 10^9 + 7 as required by the problem statement. This result is then used for further calculations as we backtrack through the recursion.

When the initial call to dfs(0, 0) completes, we have computed the total number of valid tilings for the entire board, and this result is returned as the final answer.

Example Walkthrough

Let's walk through a small example using a 2 x 3 board to illustrate the solution approach with the method dfs(i, j).

1. Start by calling dfs(0, 0) since no tiles have been placed yet.

2. Since i == j, there are three moves we can consider:

   • Place two horizontal dominoes to cover the whole first column, dfs(2, 2).
   • Place one vertical domino to cover the first cells of both rows, dfs(1, 1).
   • Place one tromino (L-shaped), which will leave one cell in the second column on one of the rows uncovered, so we call both dfs(1, 2) and dfs(2, 1).

3. For each recursive call dfs(2, 2), dfs(1, 1), dfs(1, 2), and dfs(2, 1), repeat steps similar to step 2.

   When calling dfs(2, 2) from dfs(0, 0):

   • i == j and they are equal to n, so it's a complete tiling and returns 1.

   When calling dfs(1, 1) from dfs(0, 0):

   • i == j, do the similar three steps:
     • dfs(3, 3): Goes beyond the board, thus returns 0.
     • dfs(2, 2): Complete tiling, returns 1.
     • dfs(2, 3) and dfs(3, 2): Goes beyond the board, thus both return 0.

   When calling dfs(1, 2) from dfs(0, 0):

   • i < j, only two moves are possible:
     • dfs(3, 2): Goes beyond the board, thus returns 0.
     • dfs(3, 3): Goes beyond the board, thus returns 0.

   When calling dfs(2, 1) from dfs(0, 0):

   • i > j, only two moves are possible:
     • dfs(2, 3): Goes beyond the board, thus returns 0.
     • dfs(3, 3): Goes beyond the board, thus returns 0.

4. To get the result for dfs(0, 0) add up all the ways from the recursive calls:

   • From dfs(2, 2) we got 1 way.
   • From dfs(1, 1) we consider the result from dfs(2, 2), which is 1 way.
   • Calls dfs(1, 2) and dfs(2, 1) do not add more ways as they result in 0 valid tilings.

5. So, dfs(0, 0) would return 1 (from dfs(2, 2)) + 1 (from dfs(1, 1), which includes the result from dfs(2, 2)) modulo 10^9 + 7, giving us a total of 2 ways to tile the 2 x 3 board.

The result from dfs(0, 0) is the final answer for the problem, which is 2 different ways to completely cover the 2 x 3 board using the given domino and tromino shapes.

Solution Implementation

Time and Space Complexity

The provided Python function numTilings is intended for finding the number of ways to tile a 2xN board with 2x1 dominos and trominoes (L-shaped tiles). This is a dynamic programming problem often posed on platforms like LeetCode.

Time Complexity

The time complexity of the function is difficult to frame precisely due to the complex recursive structure. On every call to dfs, potentially up to 4 recursive calls can be made, and since this algorithm calls the dfs with arguments up to n (both i and j), the time complexity is loosely bounded by a function of the form O(4^m), where m represents the number of recursive steps needed to reach the base case from the current step.

However, due to memoization (caching of results), repeated states are not recalculated, significantly reducing the number of recursive calls. Memoization effectively ensures that each state of (i, j) where 0 <= i, j <= n is computed only once. Since i and j can take on n+1 possible values each, the actual time complexity can be approximated as O(n^2).

Space Complexity

The space complexity consists of the space used by the memoization cache and the stack space used by the recursion.

The cache will have an entry for each unique (i, j) pair, so it will consume O(n^2) space. The stack space, in the worst case, could be as deep as the maximum depth of recursive calls, which is O(n) because in the worst case we increase one of i or j by 1 until n is reached.

Therefore, the overall space complexity of the numTilings function can be considered as O(n^2) due to the memoization cache, which is the dominating factor here.

So, the final analysis is:

• Time Complexity: O(n^2), considering memoization
• Space Complexity: O(n^2), due to the caching and the call stack

Learn more about how to find time and space complexity quickly using problem constraints.