<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>


class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)
        if grid[0][0] == 1 or grid[n - 1][n - 1] == 1: return 0
        q = collections.deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    q.append((i, j))
        dist=[[0]*n for _ in range(n)]
        while q:
            l = len(q)
            for _ in range(l):
                i, j = q.popleft()
                for ni, nj in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]:
                    if 0 <= ni < n and 0 <= nj < n and grid[ni][nj]==0 and dist[ni][nj]==0:
                        dist[ni][nj] = dist[i][j]+1
                        q.append((ni, nj))
        
        heap=[(-dist[0][0], 0, 0)]
        visited=set()
        while heap:
            cost, i,j = heapq.heappop(heap)
            if (i, j) in visited:
                continue
            visited.add((i, j))
            if i == n-1 and j == n-1:
                return -cost
            for ni, nj in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]:
                if 0 <= ni < n and 0 <= nj < n:
                    heapq.heappush(heap, (max(cost, -dist[ni][nj]), ni, nj))
        return -1

        
        


        # n = len(grid)
        # distance = [[0 for _ in range(n)] for _ in range(n)]
        # q = collections.deque()
        # for i in range(n):
        #     for j in range(n):
        #         if grid[i][j]: q.append((i, j))
        
        # dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        # while q:
        #     l = len(q)
        #     for _ in range(l):
        #         i, j = q.popleft()
        #         for di, dj in dirs:
        #             ni, nj = i + di, j + dj
        #             if 0 <= ni < n and 0 <= nj < n and not distance[ni][nj] and not grid[ni][nj]:
        #                 q.append((ni, nj))
        #                 distance[ni][nj] = distance[i][j]+1
        
        # if grid[0][0] == 1 or grid[n - 1][n - 1] == 1: return 0

        # pq = [(-distance[0][0], 0, 0)]

        # while pq:
        #     d, i, j = heapq.heappop(pq)
        #     if grid[i][j] == -1: continue
        #     grid[i][j] = -1
        #     if i == n - 1 and j == n - 1: return -d
        #     for di, dj in dirs:
        #         ni, nj = i + di, j + dj
        #         if 0 <= ni < n and 0 <= nj < n: heapq.heappush(pq, (max(d, -distance[ni][nj]), ni, nj))
        # return -1


        
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