<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 collections
import heapq

class Solution:
    def getNeighbors(self, row, col):
        for drow, dcol in [(-1,0), (0,-1), (0,1), (1,0)]:
            nrow = row + drow
            ncol = col + dcol
            if 0 <= nrow < self.ROWS and 0 <= ncol < self.COLS:
                yield nrow, ncol

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        self.ROWS, self.COLS = len(grid), len(grid[0])
        distances = [[float('inf')] * self.COLS for _ in range(self.ROWS)]
        queue = collections.deque()
        seen = set()
        
        # First BFS to compute distances from each cell to the nearest thief
        for row in range(self.ROWS):
            for col in range(self.COLS):
                if grid[row][col] == 1:
                    distances[row][col] = 0
                    queue.append((row, col, 0))
                    seen.add((row, col))
        
        while queue:
            row, col, distance = queue.popleft()
            for nrow, ncol in self.getNeighbors(row, col):
                if (nrow, ncol) not in seen:
                    seen.add((nrow, ncol))
                    queue.append((nrow, ncol, distance + 1))
                    distances[nrow][ncol] = distance + 1

        # Second BFS using max-heap to find maximum safeness factor
        visited = set()
        minDistanceSoFar = {}
        minDistanceSoFar[(0,0)] = distances[0][0]
        maxheap = []
        heapq.heappush(maxheap, (-distances[0][0], 0 ,0))
        while maxheap:
            minDistance, row, col = heapq.heappop(maxheap)
            # mark as visited
            visited.add((row, col))
            minDistance = -minDistance

            if row == self.ROWS - 1 and col == self.COLS - 1:
                return minDistance
            
            for nrow, ncol in self.getNeighbors(row, col):
                if not (nrow, ncol) in visited:
                    if (not (nrow, ncol) in minDistanceSoFar) or min(distances[nrow][ncol], minDistance) > minDistanceSoFar[(nrow, ncol)]:
                        minDistanceSoFar[(nrow,ncol)] = min(distances[nrow][ncol], minDistance)
                        heapq.heappush(maxheap, (-min(distances[nrow][ncol], minDistance), nrow, ncol))
            

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