<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 heappush, heappop
from collections import deque

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        def safeness(i, j):
            return distances[i][j]
            # return min(abs(i-k) + abs(j-l) for k, l in theifs)
        
        def bfs():
            q = deque([(0, i, j) for i, j in theifs])
            while q:
                dist, i, j = q.pop()
                # print(f"{dist=} {i=} {j=}")
                if distances[i][j] != -1: continue
                distances[i][j] = dist
                for I, J in [(i-1, j), (i, j+1), (i+1, j), (i, j-1)]:
                    if I not in range(h) or J not in range(w) or distances[I][J] != -1:
                        continue
                    q.appendleft((dist+1, I, J))
                    
        
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return 0
        
        w = len(grid[0])
        h = len(grid)
        distances = [[-1]*w for _ in range(h)]
        theifs = [(i, j) for i in range(w) for j in range(h) if grid[i][j] == 1]
        bfs()
        # self.print_grid(grid)
        # print('--------')
        # self.print_grid(distances)

        q = [(-safeness(0, 0), 0, 0)]
        path_safeness = safeness(0, 0)
        visited = set((0, 0))
        while q:
            cur_neg_safeness, r, c = heappop(q)
            path_safeness = min(path_safeness, -cur_neg_safeness)
            if r == h-1 and c == w-1:
                return path_safeness
            neighbors = [(i, j) for i, j in [[r-1, c], [r, c+1], [r+1, c], [r, c-1]] if i in range(h) and j in range(w) and (i, j) not in visited]
            for i, j in neighbors:
                heappush(q, (-safeness(i, j), i, j))
                visited.add((i, j))
    @staticmethod
    def print_grid(g):
        for row in g:
            print(row)
4192