Stamping the Grid

Hard

Rubrik

1 view

Topics:

ArraysSliding Windows

You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied).

You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements:

Cover all the empty cells.
Do not cover any of the occupied cells.
We can put as many stamps as we want.
Stamps can overlap with each other.
Stamps are not allowed to be rotated.
Stamps must stay completely inside the grid.

Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false.

Example 1:

Input: grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3
Output: true
Explanation: We have two overlapping stamps (labeled 1 and 2 in the image) that are able to cover all the empty cells.

Example 2:

Input: grid = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]], stampHeight = 2, stampWidth = 2 
Output: false 
Explanation: There is no way to fit the stamps onto all the empty cells without the stamps going outside the grid.

Constraints:

m == grid.length
n == grid[r].length
1 <= m, n <= 10⁵
1 <= m * n <= 2 * 10⁵
grid[r][c] is either 0 or 1.
1 <= stampHeight, stampWidth <= 10⁵

Try coding it in LeetCode

Stamping the Grid

Hard

Rubrik

1 view

Topics:

ArraysSliding Windows

You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied).

You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements:

Cover all the empty cells.
Do not cover any of the occupied cells.
We can put as many stamps as we want.
Stamps can overlap with each other.
Stamps are not allowed to be rotated.
Stamps must stay completely inside the grid.

Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false.

Example 1:

Input: grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3
Output: true
Explanation: We have two overlapping stamps (labeled 1 and 2 in the image) that are able to cover all the empty cells.

Example 2:

Input: grid = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]], stampHeight = 2, stampWidth = 2 
Output: false 
Explanation: There is no way to fit the stamps onto all the empty cells without the stamps going outside the grid.

Constraints:

m == grid.length
n == grid[r].length
1 <= m, n <= 10⁵
1 <= m * n <= 2 * 10⁵
grid[r][c] is either 0 or 1.
1 <= stampHeight, stampWidth <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on m, n, stampHeight, and stampWidth? Specifically, what are the maximum and minimum values?
If it's impossible to cover all 0s with the stamp, should I return `false`, or is there a different value I should return in that case?
Can the `grid` matrix be empty (m=0 or n=0), and what should I return in that scenario?
Is `stampHeight` or `stampWidth` ever 0? If so, what is the expected behavior?
If there are multiple valid ways to place the stamps, am I required to find a specific arrangement, or can I return `true` as soon as I find *any* valid arrangement?

Brute Force Solution

Approach

The brute force approach to solving this stamping problem involves trying all possible placements of the stamp on the grid. We systematically check every potential location and orientation to see if the stamping constraints are satisfied. If it works, it's a valid stamping, and we can record it.

Here's how the algorithm would work step-by-step:

Consider the stamp. Imagine picking it up and placing it down on the grid.
Start at the very top-left corner and try stamping it there. Check if it covers any forbidden areas or goes outside the grid.
If the stamp placement is valid, remember that location. If not, forget about it.
Move the stamp slightly to the right and repeat the checking process.
Continue moving the stamp across the row, checking for validity at each spot.
Once you reach the end of the row, move the stamp down one position and start again from the left.
Keep doing this until you have tried stamping the grid at every possible position.
After trying all placements, analyze the valid stamp locations to see if all the necessary areas are covered.

Code Implementation

def stamp_the_grid_brute_force(grid, stamp_height, stamp_width):
    grid_height = len(grid)
    grid_width = len(grid[0])
    valid_stamps = []

    for row_start in range(grid_height):
        for col_start in range(grid_width):
            # Check if the stamp placement is valid given the grid size
            if row_start + stamp_height <= grid_height and col_start + stamp_width <= grid_width:
                is_valid = True

                for row in range(row_start, row_start + stamp_height):
                    for col in range(col_start, col_start + stamp_width):
                        # Check if stamping here would cover a forbidden area
                        if grid[row][col] == 1:
                            is_valid = False
                            break
                    if not is_valid:
                        break

                if is_valid:
                    valid_stamps.append((row_start, col_start))

    # After finding all valid stamp placements, check if entire grid is covered
    if not valid_stamps:
        return False
    
    covered_grid = [[0] * grid_width for _ in range(grid_height)]
    for row_start, col_start in valid_stamps:
        for row in range(row_start, row_start + stamp_height):
            for col in range(col_start, col_start + stamp_width):
                covered_grid[row][col] = 1

    for row in range(grid_height):
        for col in range(grid_width):
            # Verify all required areas are covered
            if grid[row][col] == 0 and covered_grid[row][col] == 0:
                return False

    return True

Big(O) Analysis

Time Complexity

O(m*n*sr*sc) – Let m be the number of rows in the grid, n be the number of columns in the grid, sr be the number of rows in the stamp, and sc be the number of columns in the stamp. The brute force approach iterates through all possible placements of the stamp on the grid. There are (m - sr + 1) possible row positions and (n - sc + 1) possible column positions for the top-left corner of the stamp. For each of these placements, the algorithm checks if the stamp is valid, which requires examining all cells covered by the stamp, taking O(sr * sc) time. Therefore, the overall time complexity is O((m - sr + 1) * (n - sc + 1) * sr * sc), which simplifies to O(m*n*sr*sc) when sr and sc are relatively small compared to m and n.

Space Complexity

O(1) – The described brute force approach iterates through possible stamp placements and remembers only if a location is valid. It doesn't explicitly create large data structures to store intermediate results or visited locations. The algorithm primarily uses a few variables to track the current stamp position and potentially a boolean to indicate validity which take constant space. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The trick is to avoid redundant checks by working backward and strategically filling in the grid. We start by figuring out where stamps *must* be placed and then checking if this arrangement satisfies the initial constraints. If it does, we're golden!

Here's how the algorithm would work step-by-step:

First, let's identify all the cells that absolutely need to be covered by a stamp. These are the marked cells in our grid.
Now, pretend we have placed stamps to cover all those necessary cells. Specifically, place them so their top-left corner touches each necessary cell.
Next, imagine the resulting grid after all these stamps have been applied. Check if there are now any marked cells left uncovered. If there are, it's impossible, and we're done.
After that, we need to check for the opposite problem: that we didn't accidentally stamp any empty cells. If we did, it's also impossible.
Finally, if we made it this far, it means we successfully covered all the marked cells without stamping any empty ones, and we've found a solution.

Code Implementation

def stamping_the_grid(grid, stamp_height, stamp_width):
    grid_height = len(grid)
    grid_width = len(grid[0])
    stamped_grid = [[0] * grid_width for _ in range(grid_height)]

    # Mark cells that must be covered.
    necessary_cells = []
    for row in range(grid_height):
        for col in range(grid_width):
            if grid[row][col] == 1:
                necessary_cells.append((row, col))

    # Place stamps at each necessary cell.
    for row, col in necessary_cells:
        for stamp_row in range(stamp_height):
            for stamp_col in range(stamp_width):
                current_row = row + stamp_row
                current_col = col + stamp_col
                if 0 <= current_row < grid_height and 0 <= current_col < grid_width:
                    stamped_grid[current_row][current_col] = 1

    # Check if any marked cells are now uncovered.
    for row in range(grid_height):
        for col in range(grid_width):
            if grid[row][col] == 1 and stamped_grid[row][col] == 0:
                return False

    # Avoid stamping empty cells
    for row in range(grid_height):
        for col in range(grid_width):
            if grid[row][col] == 0 and stamped_grid[row][col] == 1:
                return False

    return True

Big(O) Analysis

Time Complexity

O(m * n) – Identifying the necessary cells in the grid takes O(m * n) time, where m and n are the dimensions of the grid. Placing stamps based on these necessary cells also requires iterating through the necessary cells, with each placement taking constant time since the stamp size is fixed, therefore the placement portion is bounded by the number of necessary cells, and this is no more than O(m*n). Checking if all marked cells are covered involves iterating through the entire grid once, taking O(m * n) time. Similarly, checking if any empty cells are stamped requires another iteration through the entire grid, which is also O(m * n). Thus, the dominant operation is the grid traversals, each occurring within O(m*n) time, which results in an overall time complexity of O(m*n).

Space Complexity

O(N*M) – The algorithm implicitly creates an auxiliary grid of size N*M to represent the stamped grid, where N is the number of rows and M is the number of columns, as it simulates placing stamps and checking for uncovered/overstamped cells. This stamped grid stores intermediate results of stamp placements. The operations of filling in the stamps contributes to this N*M space, as we must modify each element of the N*M grid when needed. Therefore, the auxiliary space complexity is O(N*M).

Edge Cases

Case	How to Handle
Null or empty grid	Return true if grid is null or empty as there are no 0s to cover.
Stamp dimensions larger than grid dimensions	If stamp dimensions exceed grid, valid placement isn't possible unless the entire grid is 0, so check if all grid elements are 0.
Stamp dimensions are 1x1	Check if all 0s in the grid can be covered with 1x1 stamps which is always possible if grid[i][j]==0.
Grid filled entirely with 1s	Return true, as there are no 0s to cover.
Grid filled entirely with 0s and stamp covers the entire grid	Return true if the stamp covers the entire grid, implying only one stamp placement is needed and valid.
Grid with single 0 surrounded by 1s, stamp larger than 1x1	If the stamp is bigger than 1x1, and a single 0 is surrounded by 1's, it cannot be covered, return false.
Large grid dimensions (m, n are large) leading to potential memory issues	Optimize space complexity by using difference arrays to track overlapping stamp placements efficiently.
Integer overflow when calculating stamp positions or coverage area	Ensure that intermediate calculations like grid indices or stamp counts are within integer bounds to avoid unexpected results.