<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>

import queue
class Solution:        
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        safenesss_matrix = [[float("inf")for _ in range(len(grid))] for _ in range(len(grid))]
        
        q = deque()
        for row_idx,row in enumerate(grid):
            for col_idx, col in enumerate(row):
                if col == 1:
                    q.append((0,(row_idx,col_idx)))
        while q:
            dist, point = q.popleft()
            x,y = point
            if min(x,y) < 0 or max(x,y) >= len(grid) or dist >= safenesss_matrix[x][y]:
                continue
            safenesss_matrix[x][y] = dist
            q.append((dist+1,(x+1,y)))
            q.append((dist+1,(x-1,y)))
            q.append((dist+1,(x,y+1)))
            q.append((dist+1,(x,y-1)))
        
        priority_bfs = queue.PriorityQueue()
        priority_bfs.put((-safenesss_matrix[0][0],0,0))
        directions = [[1,0],[-1,0],[0,1],[0,-1]]
        safenesss_matrix[0][0] = -1
        while priority_bfs:
            safeness, x,y = priority_bfs.get()

            if x ==y == len(safenesss_matrix)-1:
                return -safeness
            for x_dir,y_dir in directions:
                new_x,new_y = x+x_dir,y+y_dir
                if min(new_x,new_y) < 0 or max(new_x,new_y) >= len(safenesss_matrix):
                    continue
                priority_bfs.put((-min(-safeness,safenesss_matrix[new_x][new_y]),new_x,new_y))
                safenesss_matrix[new_x][new_y] = -1
            
        return  -1
        
72673