Count Unguarded Cells in the Grid

Tested Python solution for LeetCode 2257 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 2257, Medium. Topics: Array, Matrix, Simulation. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 2257   # by problem number
lcpy gen -s count_unguarded_cells_in_the_grid   # by problem name

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Problem

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] = [rowi, coli] and walls[j] = [rowj, colj] represent the positions of the ith guard and jth 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.

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Examples

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.

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.

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Constraints

• 1 <= m, n <= 10^5
• 2 <= m * n <= 10^5
• 1 <= guards.length, walls.length <= 5 * 10^4
• 2 <= guards.length + walls.length <= m * n
• guards[i].length == walls[j].length == 2
• 0 <= rowi, rowj < m
• 0 <= coli, colj < n
• All the positions in guards and walls are unique.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n); every cell is scanned a constant number of times.
    # Space: O(m * n) for the grid.
    def count_unguarded(
        self, m: int, n: int, guards: list[list[int]], walls: list[list[int]]
    ) -> int:
        grid = [[0] * n for _ in range(m)]
        for row, col in guards:
            grid[row][col] = 1
        for row, col in walls:
            grid[row][col] = 2

        for row, col in guards:
            for d_row, d_col in ((-1, 0), (1, 0), (0, -1), (0, 1)):
                next_row, next_col = row + d_row, col + d_col
                while (
                    0 <= next_row < m and 0 <= next_col < n and grid[next_row][next_col] in (0, 3)
                ):
                    grid[next_row][next_col] = 3
                    next_row += d_row
                    next_col += d_col

        return sum(row.count(0) for row in grid)

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Complexity

Time	Space
O(m * n); every cell is scanned a constant number of times.	O(m * n) for the grid.

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Tags

NeetCode All.