Stamping the Grid - Solution & Explanation

The goal is to cover all empty (0) cells with stamps without covering any occupied (1) cells and without stamps going out of the grid bounds. Using a greedy approach, we try to place the stamps efficiently. We start by marking all valid positions where a stamp can be fully placed, then check for each empty cell if it can be stamped by overlapping valid positions.

This Python solution uses prefix sum arrays to check if stamps can be placed without covering any occupied cells. It iterates over possible top-left positions of the stamp and ensures it covers only empty cells when placed, modifying the grid to simulate stamping.

This approach attempts to place stamps starting from the top left corner, using a recursive backtracking method. If a position can have a stamp placed while satisfying all conditions (not covering any '1's, staying within bounds, etc.), it proceeds recursively to try further placements. If it hits a dead end, it backtracks to try alternative positions.

This JavaScript solution uses backtracking to attempt placing stamps recursively. If an empty cell is encountered, it checks if a stamp can fit without extending out of boundary or covering any occupied (1) cells. Backtracking allows exploration of different paths to try and cover all empty regions.

According to the problem description, every empty cell must be covered by a stamp, and no occupied cell can be covered. Therefore, we can traverse the two-dimensional matrix, and for each cell, if all cells in the area of stampHeight times stampWidth with this cell as the upper left corner are empty (i.e., not occupied), then we can place a stamp at this cell.

To quickly determine whether all cells in an area are empty, we can use a two-dimensional prefix sum. We use s_{i,j} to represent the number of occupied cells in the sub-matrix from (1,1) to (i,j) in the two-dimensional matrix. That is, s_{i, j} = s_{i - 1, j} + s_{i, j - 1} - s_{i - 1, j - 1} + grid_{i-1, j-1}.

Then, with (i, j) as the upper left corner, and the height and width are stampHeight and stampWidth respectively, the lower right coordinate of the sub-matrix is (x, y) = (i + stampHeight - 1, j + stampWidth - 1). We can calculate the number of occupied cells in this sub-matrix through s_{x, y} - s_{x, j - 1} - s_{i - 1, y} + s_{i - 1, j - 1}. If the number of occupied cells in this sub-matrix is 0, then we can place a stamp at (i, j). After placing the stamp, all cells in this stampHeight times stampWidth area will become occupied cells. We can use a two-dimensional difference array d to record this change. That is:

$ \begin{aligned} d_{i, j} &\leftarrow d_{i, j} + 1 \ d_{i, y + 1} &\leftarrow d_{i, y + 1} - 1 \ d_{x + 1, j} &\leftarrow d_{x + 1, j} - 1 \ d_{x + 1, y + 1} &\leftarrow d_{x + 1, y + 1} + 1 \end{aligned}

Finally, we perform a prefix sum operation on the two-dimensional difference array d to find out the number of times each cell is covered by a stamp. If a cell is not occupied and the number of times it is covered by a stamp is 0, then we cannot place a stamp at this cell, so we need to return \texttt{false}. If all "unoccupied cells" are successfully covered by stamps, return \texttt{true}.

The time complexity is O(m times n), and the space complexity is O(m times n). Here, m and n$ are the height and width of the two-dimensional matrix, respectively.