<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 heapq import heappop, heappush
class Solution:
    def maximumSafenessFactor(self, grid):
        n, cans = len(grid), []
        tg, ctr = n-1, -1
        if grid[0][0] or grid[tg][tg]:return 0
        for y, row in enumerate(grid):
            for x, val in enumerate(row):
                if val:
                    cans.append((x, y))
                    row[x] = 0
                else:row[x] = -800
        while cans:
            _cans = []
            for x, y in cans:
                if x > 0 and grid[y][_x := x-1] < ctr:
                    grid[y][_x] = ctr
                    _cans.append((_x, y))
                if (_x := x+1) < n and grid[y][_x] < ctr:
                    grid[y][_x] = ctr
                    _cans.append((_x, y))
                if y > 0 and grid[_y := y-1][x] < ctr:
                    grid[_y][x] = ctr
                    _cans.append((x, _y))
                if (_y := y+1) < n and grid[_y][x] < ctr:
                    grid[_y][x] = ctr
                    _cans.append((x, _y))
            cans = _cans
            ctr -= 1
        hp = [(sft if (sft := grid[0][0]) > (eft := grid[tg][tg]) else eft, 0, 0)]
        while hp:
            ft, x, y = heappop(hp)
            if ft == 0 or (x == tg and y == tg):return -ft
            if grid[y][x] == 1:continue
            grid[y][x] = 1
            if x > 0 and (nft := grid[y][_x := x-1]) != 1:heappush(hp, (nft if nft > ft else ft, _x, y))
            if (_x := x+1) < n and (nft := grid[y][_x]) != 1:heappush(hp, (nft if nft > ft else ft, _x, y))
            if y > 0 and (nft := grid[_y := y-1][x]) != 1:heappush(hp, (nft if nft > ft else ft, x, _y))
            if (_y := y+1) < n and (nft := grid[_y][x]) != 1:heappush(hp, (nft if nft > ft else ft, x, _y))
22908