Solution Explanation

This problem asks to find the number of ways to build a sturdy brick wall with given height, width, and brick sizes. A sturdy wall means that adjacent rows don't join bricks at the same location (except at the ends).

The solution uses a combination of Depth-First Search (DFS) and Dynamic Programming.

1. Generating Possible Row Configurations (DFS):

The dfs function generates all possible ways to arrange bricks in a single row to fill the given width. It does this recursively:

• It explores all possible brick choices from the bricks array.
• For each brick, it adds the brick's width to the current v (current width filled).
• If v exceeds the width, it backtracks.
• If v equals width, a valid row configuration is found, and it's added to the res list.

2. Checking for Sturdy Wall Condition:

The check function determines if two rows (a and b) can be placed adjacently to form a sturdy wall. It simulates building the two rows. If at any point the cumulative width from the left is the same for both rows (other than at the start or end of the row), they are not sturdy and it returns false. Otherwise it returns true.

3. Dynamic Programming for Counting Sturdy Walls:

The core idea here is to use dynamic programming to efficiently count the number of sturdy walls.

• dp[i][j] represents the number of ways to build an i-row-high wall ending with row configuration j.
• The base case dp[0][j] = 1 for all valid row configurations j (because there is one way to build a 0-row-high wall).
• The recurrence relation is: dp[i][j] = Σ dp[i-1][k], where the sum is over all k such that rows j and k can form a sturdy combination (i.e., check(res[j], res[k]) returns true).

This means that the number of ways to build an i-row-high wall ending with configuration j is the sum of the number of ways to build an (i-1)-row-high wall that can be stacked on top of configuration j.

The final answer is the sum of dp[height-1][j] for all j, representing the total number of ways to build a sturdy wall of the desired height.

Time Complexity Analysis:

• DFS to generate row configurations: The time complexity is exponential in the worst case, O(bw), where 'b' is the number of brick types and 'w' is the width. This is because in the worst case each position can have any of the b brick types.

• Dynamic Programming: The time complexity is O(h * n2), where 'h' is the height and 'n' is the number of valid row configurations generated by the DFS. The nested loops iterate through each height and each pair of row configurations.

• Therefore, the overall time complexity is dominated by the DFS part in the worst case, making it exponential, O(bw) if n is proportional to b<sup>w</sup>

Space Complexity Analysis:

• res: Stores all possible row configurations. In the worst case, this can be O(bw).
• dp: The dynamic programming table has dimensions h x n, so its space complexity is O(h * n), where n is in O(bw) in the worst case.
• Overall space complexity is also dominated by the row configurations, making it O(bw) in the worst case.

The provided code implements this solution in Python, Java, C++, and Go. Note that the exponential time complexity of the DFS can make this solution impractical for large values of width and a large number of brick types. For larger inputs, more advanced optimization techniques might be necessary.