Stamping The Grid - Leetcode Solution

The problem "Stamping The Grid" on LeetCode asks us to find a way to stamp a given grid of '?' characters with a given stamp that consists entirely of 'X' characters. We are allowed to perform the following operation any number of times:

• Select a non-overlapping subset of the stamp's cells to fill in a contiguous group of same-sized squares within the grid.

The goal of the problem is to find a sequence of operations that will completely fill the grid with 'X' characters and return the sequence of indices of each operation performed.

Approach:

We can solve this problem using a combination of DFS (Depth First Search) and Greedy Algorithm. We start by performing DFS on the grid to check if the given stamp can be used to stamp any region of the grid. We perform the DFS in such a way that we try to 'stamp' the grid using the largest possible stamp that will fit into a given empty cell.

Once we find a suitable region that can be stamped with the given stamp, we fill it with 'X' characters and mark the indices of the cells that have been stamped. We then continue our search for other sub-regions in the grid that can be stamped using the same approach. We then repeat this process until the entire grid is stamped.

To improve our efficiency, we can also use a memoization technique to keep track of previously processed sub-regions of the grid. This way, we do not need to repeat the same search on already processed sub-regions.

Algorithm:

1. Start by performing DFS on the grid to search for a sub-region that can be stamped with the given stamp. We start the search from the bottom-right corner of the grid to maximize the area of the sub-region that can be stamped.

2. After finding a suitable sub-region, replace the '?' characters with 'X' characters and mark the indices of the cells that have been stamped.

3. Repeat step 1 and 2 until the entire grid is stamped.

4. Return the sequence of indices of each operation performed.

We can also use a priority queue to keep track of the sub-regions that can be stamped, prioritizing the regions with larger area.

Time Complexity:

The DFS can take O(N * M) time in the worst case, where N and M are the dimensions of the grid. However, since we use memoization to avoid repeating the same search on previously processed sub-regions, the actual time complexity is much lower.

Since we iterate through the entire grid for every stamping operation, the overall time complexity of the algorithm is O(K * N * M), where K is the maximum number of stamping operations needed to completely fill the grid.

Space Complexity:

The main space usage of the algorithm is for the memoization table that stores the processed sub-regions of the grid. This table can take up to O(N * M) space. Additionally, we also need to store the temporary stamping indices, which can take up to O(N * M) space. Therefore, the overall space complexity of the algorithm is O(N * M).

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