<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:
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return 0
        distance = [ [math.inf] * len(grid[0]) for _ in range(len(grid))]
        dirs = [[1, 0], [-1, 0], [0, -1], [0, 1]]
        queue = deque()
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    distance[i][j] = 0
                    queue.append((i, j))
        while queue:
            i, j = queue.popleft()
            for dx, dy in dirs:
                x = i + dx
                y = j + dy
                if x >= 0 and x < len(grid) and y >= 0 and y < len(grid[0]) and distance[x][y] == math.inf:
                    distance[x][y] = distance[i][j] + 1
                    queue.append((x, y))
        heap = [[-distance[0][0], 0, 0]] #dis, i, j
        seen = set()
        seen.add((0, 0))
        while heap:
            dis, i, j = heapq.heappop(heap)
            if i == len(grid) - 1 and j == len(grid[0]) - 1:
                return -dis
            for dx, dy in dirs:
                x = i + dx
                y = j + dy
                if x >= 0 and x < len(grid) and y >= 0 and y < len(grid[0]) and (x, y) not in seen:
                    heapq.heappush(heap, [-min(-dis, distance[x][y]), x, y])
                    seen.add((x, y))
        return -1

            

'''
[[0,0,0,1],
 [0,0,0,0],
 [0,0,0,0],
 [1,0,0,0]]

#distance
[[3,2,1,0],
 [2,3,2,1],
 [1,2,3,2],
 [0,1,2,3]]
heap = [ [-2, 0, 1] , [-2, 1, 1], [-1, 2, 0]]
 i: 1, j: 0
 2

 [[0,1,1],
  [0,1,1], 
  [1,1,0]]

  [[1, 0, 1], 
   [2, 1, 1], 
   [0, 1, 2]]
'''
38376