<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 find_distance(self, grid: List[List[int]]) -> List[List[int]]:
        # Initialization
        n = len(grid)
        dist_grid = [[-1 for _ in range(n)] for _ in range(n)]
        queue = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    queue.append((i, j))
                    dist_grid[i][j] = 0
 
        # BFS
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        while queue:
            r, c = queue.popleft()
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                if 0 <= nr < n and 0 <= nc < n and dist_grid[nr][nc] == -1:
                    dist_grid[nr][nc] = dist_grid[r][c] + 1
                    queue.append((nr, nc))
        return dist_grid

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # Compute Manhattan distance of all cells
        dist_grid = self.find_distance(grid)
        
        # Now do Dijkstra's algorithm to find shortest path
        n = len(grid)
        dp = [[float('inf') for _ in range(n)] for _ in range(n)]
        dp[0][0] = dist_grid[0][0]
        directions = ((1, 0), (0, 1), (-1, 0), (0, -1))
        queue = [(-dp[0][0], (0, 0))]
        heapq.heapify(queue)
        while queue:
            # Get next node and check if already been processed
            distance, (i, j) = heapq.heappop(queue)
            if distance > dp[i][j]:
                continue

            # Check every neighbor and add them to queue if necessary
            for dr, dc in directions:
                # Get new distance to thieve
                new_i, new_j = i + dr, j + dc
                if 0 <= new_i < n and 0 <= new_j < n:
                    neighbor_distance = max(distance, -dist_grid[new_i][new_j])
                    if neighbor_distance < dp[new_i][new_j]:
                        dp[new_i][new_j] = neighbor_distance
                        heapq.heappush(queue, (neighbor_distance, (new_i, new_j)))

        return -dp[n-1][n-1]
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