<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>n x n</code>, where <code>(r, c)</code> represents:</p>

<ul>
	<li>A cell containing a thief if <code>grid[r][c] = 1</code></li>
	<li>An empty cell if <code>grid[r][c] = 0</code></li>
</ul>

<p>You are initially positioned at cell <code>(0, 0)</code>. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.</p>

<p>The <strong>safeness factor</strong> of a path on the grid is defined as the <strong>minimum</strong> manhattan distance from any cell in the path to any thief in the grid.</p>

<p>Return <em>the <strong>maximum safeness factor</strong> of all paths leading to cell </em><code>(n - 1, n - 1)</code><em>.</em></p>

<p>An <strong>adjacent</strong> cell of cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> and <code>(r - 1, c)</code> if it exists.</p>

<p>The <strong>Manhattan distance</strong> between two cells <code>(a, b)</code> and <code>(x, y)</code> is equal to <code>|a - x| + |b - y|</code>, where <code>|val|</code> denotes the absolute value of val.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example1.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example2.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[0,0,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 2) is cell (0, 0). The distance between them is | 0 - 0 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example3.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 3) is cell (1, 2). The distance between them is | 0 - 1 | + | 3 - 2 | = 2.
- The closest cell of the path to the thief at cell (3, 0) is cell (3, 2). The distance between them is | 3 - 3 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length == n &lt;= 400</code></li>
	<li><code>grid[i].length == n</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one thief in the <code>grid</code>.</li>
</ul>

from collections import deque
import math

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        N = len(grid)
        if N == 1:
            return 1 - grid[0][0]
        maximum_safeness = [[N * 2] * N for _ in range(N)]
        for row_index in range(N - 1, -1, -1):
            for column_index in range(N - 1, -1, -1):
                if grid[row_index][column_index] == 1:
                    maximum_safeness[row_index][column_index] = 0
                else:
                    if row_index + 1 < N:
                        maximum_safeness[row_index][column_index] = min(maximum_safeness[row_index][column_index], 1 + maximum_safeness[row_index + 1][column_index])
                    if column_index + 1 < N:
                        maximum_safeness[row_index][column_index] = min(maximum_safeness[row_index][column_index], 1 + maximum_safeness[row_index][column_index + 1])
        for row_index in range(N):
            for column_index in range(N):
                if grid[row_index][column_index] != 1:
                    if row_index - 1 >= 0:
                        maximum_safeness[row_index][column_index] = min(maximum_safeness[row_index][column_index], 1 + maximum_safeness[row_index - 1][column_index])
                    if column_index - 1 >= 0:
                        maximum_safeness[row_index][column_index] = min(maximum_safeness[row_index][column_index], 1 + maximum_safeness[row_index][column_index - 1])
        queue = deque([(0, 0)])
        current_max = [[-1] * N for _ in range(N)]
        current_max[0][0] = maximum_safeness[0][0]
        while queue:
            current_row, current_column = queue.popleft()
            for row_delta, column_delta in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                new_row = current_row + row_delta
                new_column = current_column + column_delta
                if new_row < 0 or new_column < 0 or new_row >= N or new_column >= N:
                    continue
                new_path_min = min(current_max[current_row][current_column], maximum_safeness[new_row][new_column])
                if new_row == N - 1 and new_column == new_row:
                    return new_path_min
                if new_path_min > current_max[new_row][new_column]:
                    current_max[new_row][new_column] = new_path_min
                    if queue and current_max[new_row][new_column] >= current_max[queue[0][0]][queue[0][1]]:
                        queue.appendleft((new_row, new_column))
                    else:
                        queue.append((new_row, new_column))
        return -1
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