<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>

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        grid_size = len(grid)
        distance = [[float('inf')] * grid_size for _ in range(grid_size)]
        pos_queue = collections.deque()
        self._record_thief_pos(grid, pos_queue, distance)
        self._set_thief_distance(pos_queue, distance, directions)
        safeness_max_heap = [(-distance[0][0], 0, 0)]
        max_safeness: List[List[float]]
        max_safeness = [[-1] * grid_size for _ in range(grid_size)]
        max_safeness[0][0] = distance[0][0]

        while safeness_max_heap:
            neg_safe, r, c = heappop(safeness_max_heap)
            safeness = - neg_safe
            if r == grid_size - 1 and c == grid_size - 1:
                return int(safeness)
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                if self._out_of_bound(grid_size, nr, nc):
                    continue
                distance_new = distance[nr][nc]
                new_safe = min(safeness, distance_new)
                if new_safe > max_safeness[nr][nc]:
                    max_safeness[nr][nc] = new_safe
                    heappush(safeness_max_heap, (-new_safe, nr, nc))
        return -1

    def _set_thief_distance(self, pos_queue: collections.deque, distance: List[List[float]],
                            directions: List[tuple[int, int]]) -> None:
        grid_size = len(distance)
        while pos_queue:
            r, c = pos = pos_queue.popleft()
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                if self._out_of_bound(grid_size, nr, nc):
                    continue
                if distance[nr][nc] != float('inf'):
                    continue
                distance[nr][nc] = distance[r][c] + 1
                pos_queue.append((nr, nc))

    @staticmethod
    def _record_thief_pos(grid: List[List[int]], pos_queue: collections.deque, distance: List[List[float]]) -> None:
        THIEF = 1
        grid_size = len(grid)
        # start from thief, visit all position
        for row in range(grid_size):
            for col in range(grid_size):
                # start from thief, just record all thieves
                if grid[row][col] == THIEF:
                    pos_queue.append((row, col))
                    distance[row][col] = 0

    @staticmethod
    def _out_of_bound(grid_size: int, row: int, col: int) -> bool:
        if row < 0 or col < 0:
            return True
        if row >= grid_size or col >= grid_size:
            return True
        return False
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