Count Unguarded Cells in the Grid

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Arrays

You are given two integers m and n representing a 0-indexed m x n grid. You are also given two 2D integer arrays guards and walls where guards[i] = [row_i, col_i] and walls[j] = [row_j, col_j] represent the positions of the i^th guard and j^th wall respectively.

A guard can see every cell in the four cardinal directions (north, east, south, or west) starting from their position unless obstructed by a wall or another guard. A cell is guarded if there is at least one guard that can see it.

Return the number of unoccupied cells that are not guarded.

Example 1:

Input: m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
Output: 7
Explanation: The guarded and unguarded cells are shown in red and green respectively in the above diagram.
There are a total of 7 unguarded cells, so we return 7.

Example 2:

Input: m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
Output: 4
Explanation: The unguarded cells are shown in green in the above diagram.
There are a total of 4 unguarded cells, so we return 4.

Constraints:

1 <= m, n <= 10⁵
2 <= m * n <= 10⁵
1 <= guards.length, walls.length <= 5 * 10⁴
2 <= guards.length + walls.length <= m * n
guards[i].length == walls[j].length == 2
0 <= row_i, row_j < m
0 <= col_i, col_j < n
All the positions in guards and walls are unique.

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 dimensions of the grid (m, n), and what are the possible ranges for these values?
What values are possible within the 'guards' and 'walls' arrays? Are the coordinates guaranteed to be within the grid boundaries?
If a cell is guarded by multiple guards, should it be counted multiple times or only once?
Is the grid guaranteed to be rectangular, or could it be irregular?
If there are no unguarded cells, what value should I return?

Brute Force Solution

Approach

The problem involves figuring out which spots on a grid are safe from being seen by guards. A brute force method involves checking every single spot on the grid to determine if it's guarded or not.

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

Start with the first empty spot on the grid.
Imagine a beam of light extending from each guard in all four directions: up, down, left, and right.
Check if any of these light beams hit the current empty spot you are considering. Consider that the light beams stop when they hit a wall or another guard.
If any of the beams hit the spot, then the spot is guarded.
If none of the beams hit the spot, then the spot is unguarded.
Repeat this process, checking every single empty spot on the entire grid, one at a time.
After checking all the spots, count how many spots were determined to be unguarded.

Code Implementation

def count_unguarded_cells_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])
    unguarded_count = 0

    for row_index in range(rows):
        for col_index in range(cols):
            if grid[row_index][col_index] == 0:
                is_guarded = False

                # Iterate through all guards
                for guard_row in range(rows):
                    for guard_col in range(cols):
                        if grid[guard_row][guard_col] == 1:

                            # Check upwards direction
                            blocked = False
                            for k in range(guard_row - 1, -1, -1):
                                if grid[k][guard_col] == 1:
                                    blocked = True
                                    break
                                if k == row_index and guard_col == col_index:
                                    is_guarded = True
                                    blocked = True
                                    break
                            if is_guarded:
                                break

                            # Check downwards direction
                            blocked = False
                            for k in range(guard_row + 1, rows):
                                if grid[k][guard_col] == 1:
                                    blocked = True
                                    break
                                if k == row_index and guard_col == col_index:
                                    is_guarded = True
                                    blocked = True
                                    break
                            if is_guarded:
                                break

                            # Check left direction
                            blocked = False
                            for k in range(guard_col - 1, -1, -1):
                                if grid[guard_row][k] == 1:
                                    blocked = True
                                    break
                                if guard_row == row_index and k == col_index:
                                    is_guarded = True
                                    blocked = True
                                    break
                            if is_guarded:
                                break

                            # Check right direction
                            for k in range(guard_col + 1, cols):
                                if grid[guard_row][k] == 1:
                                    break
                                if guard_row == row_index and k == col_index:
                                    is_guarded = True
                                    break
                            if is_guarded:
                                break
                    if is_guarded:
                        break

                if not is_guarded:
                    unguarded_count += 1

    return unguarded_count

Big(O) Analysis

Time Complexity

O(m * n * (m + n)) – The brute force approach iterates through each cell of the grid, which is of size m x n, contributing a factor of m * n. For each cell, we iterate through each guard to simulate the beams. In the worst case, we might need to traverse up to m cells vertically and n cells horizontally from the guard to the wall or another guard. Therefore, the total time complexity is O(m * n * (number of guards * (m + n))). Assuming the maximum number of guards can be m*n, the time complexity is O(m * n * (m * n * (m + n))), if we assume a constant number of guards, the complexity becomes O(m * n * (m + n)). If the number of guards is proportional to m*n, the runtime is O((m*n)^2 * (m+n)), but we simplify it to O(m * n * (m + n)) based on the prompt assuming a smaller number of guards.

Space Complexity

O(1) – The given brute force solution iterates through each empty cell in the grid and simulates beams of light from each guard. It doesn't create any auxiliary data structures that scale with the input grid size to store intermediate results or visited cells. Only a few variables are used to keep track of the current cell being checked and to determine if the cell is guarded, taking constant extra space irrespective of the grid dimensions. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The most efficient strategy involves simulating the movement of guards from each guard location and marking all the cells they cover. Cells that are guarded are marked, and the unguarded cells can then be counted easily. This approach avoids redundant checks and directly determines which cells are protected.

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

First, imagine the grid and the initial positions of the walls and guards.
Then, simulate the 'vision' of each guard moving in all four directions (up, down, left, right).
As a guard moves in a direction, mark every cell it can 'see' as 'guarded' until it hits a wall or another guard.
Repeat this process for every guard in the grid.
Once all guards have extended their vision and marked the grid, count the number of cells that are NOT marked as 'guarded'. These are the unguarded cells.

Code Implementation

def count_unguarded(rows, cols, guards, walls):
    grid = [['unguarded' for _ in range(cols)] for _ in range(rows)]

    for row_index, col_index in walls:
        grid[row_index][col_index] = 'wall'

    for row_index, col_index in guards:
        grid[row_index][col_index] = 'guard'

    for start_row, start_col in guards:
        # Simulate guard vision in each direction.
        for direction_row, direction_col in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
            current_row = start_row
            current_col = start_col

            while True:
                current_row += direction_row
                current_col += direction_col

                if 0 <= current_row < rows and 0 <= current_col < cols:
                    cell_type = grid[current_row][current_col]
                    
                    if cell_type == 'wall' or cell_type == 'guard':
                        break

                    # Mark the cell as guarded.
                    grid[current_row][current_col] = 'guarded'

                else:
                    break

    # Count unguarded cells after marking all guarded cells.
    unguarded_count = 0
    for row in grid:
        for cell in row:
            if cell == 'unguarded':
                unguarded_count += 1

    return unguarded_count

Big(O) Analysis

Time Complexity

O(m * n * (g + w)) – Let m be the number of rows and n be the number of columns in the grid. The algorithm iterates through each guard (g) location. From each guard location, it simulates vision in four directions (up, down, left, right). In the worst case, each guard can traverse the entire row or column until it hits a wall (w) in each direction. Therefore, the time complexity is proportional to the product of the number of rows, the number of columns, the number of guards, and the number of walls, so the time complexity becomes O(m * n * (g + w)).

Space Complexity

O(M*N) – The algorithm simulates the guard's vision on the grid, requiring us to potentially mark each cell as guarded or not. This marking is typically done using an auxiliary data structure (like a boolean matrix) of the same size as the input grid, which is M rows by N columns where M is the number of rows, and N is the number of columns in the grid. Therefore, the space required for this auxiliary grid scales linearly with the grid's dimensions. The auxiliary space used is proportional to M*N, thus the space complexity is O(M*N).

Edge Cases

Case	How to Handle
Empty grid (m=0 or n=0)	Return 0 because there are no cells at all.
Grid with only guards or only walls	The algorithm should still correctly identify unguarded cells when all cells are initially marked as guarded or blocked.
Guard and wall on same cell	Prioritize the guard on that cell, meaning it becomes a guarding point, not a blockage.
Large grid dimensions (m and n are large)	Ensure the time complexity is optimal, possibly avoiding brute force iteration of every cell for each guard.
Guards clustered together	Multiple guards guarding the same cell should not cause incorrect counting.
Walls completely block guards from some sections of the grid	The algorithm should correctly identify unguarded cells in sections isolated by walls.
Grid is a single row or single column	Ensure correct handling of edge cases when the grid's dimensions are severely skewed.
Integer overflow when calculating grid size or cell counts	Use appropriate data types (e.g., long) to prevent potential overflow errors.

Empty grid (m=0 or n=0)

How to Handle:

Return 0 because there are no cells at all.

Grid with only guards or only walls

How to Handle:

The algorithm should still correctly identify unguarded cells when all cells are initially marked as guarded or blocked.

Guard and wall on same cell

How to Handle:

Prioritize the guard on that cell, meaning it becomes a guarding point, not a blockage.

Large grid dimensions (m and n are large)

How to Handle:

Ensure the time complexity is optimal, possibly avoiding brute force iteration of every cell for each guard.

Guards clustered together

How to Handle:

Multiple guards guarding the same cell should not cause incorrect counting.

Walls completely block guards from some sections of the grid

How to Handle:

The algorithm should correctly identify unguarded cells in sections isolated by walls.

Grid is a single row or single column

How to Handle:

Ensure correct handling of edge cases when the grid's dimensions are severely skewed.

Integer overflow when calculating grid size or cell counts

How to Handle:

Use appropriate data types (e.g., long) to prevent potential overflow errors.