<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 heapq

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

        r = len(grid)
        c = len(grid[0])

        t_m = [[float('inf') for i in range(c)] for j in range(r)]

        mx = [1, 0, -1, 0]
        my = [0, 1, 0, -1]

        t = []
        for i in range(r):
            for j in range(c):
                if grid[i][j] == 1:
                    t.append((i, j))
                    t_m[i][j] = 0 

        def bfs(t_lst):
            visited = [[0 for i in range(c)] for j in range(r)]
            dq = deque(t_lst)
            
            while dq:
                cy, cx = dq.popleft()

                for i in range(4):
                    ny = cy + my[i]
                    nx = cx + mx[i]

                    if 0 <= ny < r and 0 <= nx < c:
                        if t_m[ny][nx] == float('inf'):
                            t_m[ny][nx] = t_m[cy][cx] + 1
                            dq.append((ny, nx))

        bfs(t)

        visited = [[0 for i in range(c)] for j in range(r)]
        visited[0][0] = 1
        max_heap = [(-t_m[0][0], 0, 0)] 

        while max_heap:
            m_scr, cy, cx = heapq.heappop(max_heap)
            m_scr = -m_scr
            
            if cy == r - 1 and cx == c - 1:
                flag = True
                return m_scr

            for i in range(4):
                ny = cy + my[i]
                nx = cx + mx[i]

                if 0 <= ny < r and 0 <= nx < c:
                    if visited[ny][nx] == 0:
                        visited[ny][nx] = 1
                        heapq.heappush(max_heap, (-min(m_scr, t_m[ny][nx]), ny, nx))
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