Block Placement Queries

Hard

Block Placement Queries

Hard

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:

Can you please describe the specific dimensions and constraints for the blocks and the placement area? What are the minimum and maximum values for these dimensions?
Are the block dimensions always integers, or can they be floating-point numbers?
If a block cannot be placed according to the query, what should the function return? Should I return a specific error code or a boolean value indicating failure?
Are there any restrictions on the orientation of the blocks? Can they be rotated, or must they be placed in their original orientation?
Is it possible for blocks to overlap when placed, or are they required to be placed without any overlap?

Brute Force Solution

Approach

The brute force method for block placement involves trying every single possible arrangement of blocks. We check if each arrangement meets the requirements. If it does, we count it; otherwise, we discard it.

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

Consider each possible starting position for placing a block.
For each starting position, check if placing the block there is valid, meaning does it fit within the given space and does it overlap with any existing blocks.
If it's a valid placement, mark the block as placed in that position.
If it's not valid, move on to the next possible starting position.
Repeat this process for every single block you need to place.
Count the number of valid arrangements you were able to create.
Return the total number of valid arrangements you found.

Code Implementation

def count_valid_arrangements_brute_force(grid_size, block_sizes):
    total_valid_arrangements = 0

    def is_valid_placement(current_grid, block_size, start_position):
        row_start, col_start = start_position

        # Check if the block fits within the grid
        if row_start + block_size > grid_size or col_start + block_size > grid_size:
            return False

        # Check for overlaps with existing blocks
        for row_index in range(row_start, row_start + block_size):
            for col_index in range(col_start, col_start + block_size):
                if current_grid[row_index][col_index] == 1:
                    return False

        return True

    def place_block(current_grid, block_size, start_position):
        row_start, col_start = start_position
        new_grid = [row[:] for row in current_grid]

        for row_index in range(row_start, row_start + block_size):
            for col_index in range(col_start, col_start + block_size):
                new_grid[row_index][col_index] = 1

        return new_grid

    def backtrack(current_grid, remaining_blocks):
        nonlocal total_valid_arrangements

        if not remaining_blocks:
            # All blocks have been placed, so it's a valid arrangement
            total_valid_arrangements += 1
            return

        block_size = remaining_blocks[0]

        # Iterate through all possible starting positions
        for row_start in range(grid_size):
            for col_start in range(grid_size):
                start_position = (row_start, col_start)

                # Check if placing the block at this position is valid
                if is_valid_placement(current_grid, block_size, start_position):
                    # Place the block and recursively call backtrack
                    new_grid = place_block(current_grid, block_size, start_position)
                    backtrack(new_grid, remaining_blocks[1:])

    initial_grid = [[0] * grid_size for _ in range(grid_size)]

    # Initiate the backtracking process with an empty grid
    backtrack(initial_grid, block_sizes)

    return total_valid_arrangements

Big(O) Analysis

Time Complexity

O(n^k) – The brute force algorithm explores all possible block placements. Let n be the size of the available space and k be the number of blocks. For each of the k blocks, we iterate through all n possible starting positions. In the worst case, we need to consider all possible combinations of these placements. This process leads to n choices for the first block, n choices for the second block, and so on, resulting in approximately n * n * ... * n (k times) which equals n^k. Therefore, the time complexity is O(n^k).

Space Complexity

O(N) – The brute force approach, as described, essentially tests every possible arrangement of blocks. To track which positions are occupied by already placed blocks, a data structure representing the space, such as a 2D array or a list of booleans, is needed. The size of this data structure directly depends on the size of the available space, which we can denote as N. Therefore, the auxiliary space required grows linearly with N, to keep track of block placements during the arrangement checks.

Optimal Solution

Approach

The problem asks us to efficiently find the first spot in a grid where we can place a block of a specific size. Instead of checking every single spot, we will keep track of already occupied areas to skip over them quickly. This allows us to find the valid spot without unnecessary checks.

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

Imagine the grid as a series of rows.
For each row, keep track of the rightmost point that is already occupied by a block.
When asked to place a new block, start checking from the left of the row, skipping over any part of the row that is already occupied.
If the block fits entirely within the empty space in the row, place the block there and update the rightmost occupied point for that row.
If the block doesn't fit, move to the next row and repeat the process.
If we can't find a spot for the block in any of the rows, it means the block cannot be placed.

Code Implementation

def solve():
    grid_rows = 5
    grid_columns = 10
    rightmost_occupied = [0] * grid_rows

    def find_block_placement(block_width):
        for row_index in range(grid_rows):
            # Start checking from the leftmost available spot in the row.
            start_position = rightmost_occupied[row_index]
            if start_position + block_width <= grid_columns:

                #Block fits, so place it and update the occupied tracker.
                rightmost_occupied[row_index] = start_position + block_width

                return (row_index, start_position)

        return None

    #Example Usage
    block_width_1 = 3
    placement_1 = find_block_placement(block_width_1)
    print(f"Placement for block of width {block_width_1}: {placement_1}")

    block_width_2 = 7
    placement_2 = find_block_placement(block_width_2)
    print(f"Placement for block of width {block_width_2}: {placement_2}")

    #Attempt to place a block that is too large, tests the none case.
    block_width_3 = 15
    placement_3 = find_block_placement(block_width_3)
    print(f"Placement for block of width {block_width_3}: {placement_3}")

solve()

Big(O) Analysis

Time Complexity

O(m*w) – Let m be the number of rows in the grid, and w be the maximum width occupied in any row (which can be up to the number of columns). The algorithm iterates through each row (up to m rows) until it finds a suitable placement. For each row, in the worst-case scenario, it may need to check positions up to the rightmost occupied point w to find a free space to place the block. Therefore, in the worst case, the search within each row takes O(w) time. Thus, in the worst case, we might need to check all rows, leading to a total time complexity of O(m*w).

Space Complexity

O(R) – The solution maintains a data structure to keep track of the rightmost occupied point for each row. This requires an array (or similar data structure) of size R, where R is the number of rows in the grid. The space used by this array is auxiliary space, as it's used in addition to the input grid itself. Therefore, the auxiliary space complexity is directly proportional to the number of rows. This means the space complexity is O(R).

Edge Cases

Case	How to Handle
Null or empty input array of blocks	Return an empty list indicating no placements are possible, or throw an IllegalArgumentException based on requirements.
Empty requirements array	Return an array of 0s, indicating all blocks are needed, or a different default based on requirements.
Requirements that cannot be satisfied (e.g., block with required feature doesn't exist)	Return -1, null, or throw an exception to indicate that placement is impossible, based on requirements.
Blocks array with only one block	Check if the single block satisfies all requirements; if so, return 0; otherwise, return -1 (or alternative error indication).
Very large blocks array (potential memory or time complexity issues)	Ensure the algorithm uses space and time efficiently, considering potential optimizations like early termination or using appropriate data structures.
Requirements array with duplicate features	Handle duplicate requirements appropriately, possibly by using a Set to represent requirements or by only considering the first occurrence.
Blocks with no features at all	Treat these blocks as not satisfying any requirements; they will not contribute to fulfilling the requirements.
Integer overflow potential when calculating distances or indices	Use appropriate data types (e.g., long) to store indices and distances, and handle potential overflow cases.