<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

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        
        # Find all thieves
        thieves = [(i, j) for i in range(m) for j in range(n) if grid[i][j] == 1]
        
        # Calculate distance from each cell to nearest thief
        def bfs_distance():
            queue = deque(thieves)
            distance = [[float('inf')] * n for _ in range(m)]
            for i, j in thieves:
                distance[i][j] = 0
            
            while queue:
                x, y = queue.popleft()
                for dx, dy in [(1,0), (-1,0), (0,1), (0,-1)]:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < m and 0 <= ny < n and distance[nx][ny] > distance[x][y] + 1:
                        distance[nx][ny] = distance[x][y] + 1
                        queue.append((nx, ny))
            return distance
        
        distance = bfs_distance()
        
        def validPath(v):
            visited = [[False] * n for _ in range(m)]
            
            def dfs(i, j):
                if i == m - 1 and j == n - 1:
                    return True
                visited[i][j] = True
                for dx, dy in [(0, 1), (1, 0), (-1, 0), (0, -1)]:
                    ni, nj = i + dx, j + dy
                    if 0 <= ni < m and 0 <= nj < n and not visited[ni][nj] and distance[ni][nj] >= v:
                        if dfs(ni, nj):
                            return True
                return False

            return dfs(0, 0) if distance[0][0] >= v else False

        # Binary search
        left, right = 0, max(max(row) for row in distance)
        while left < right:
            mid = (left + right + 1) // 2
            if validPath(mid):
                left = mid
            else:
                right = mid - 1
        
        return left
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