<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] or grid[n-1][n-1]:
            return 0

        INF = 10**6
        mD = [[INF] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                mD[i][j] = min(
                    mD[i-1][j] + 1 if i > 0 else INF,
                    mD[i][j-1] + 1 if j > 0 else INF,
                    0 if grid[i][j] == 1 else INF
                )
        for i in range(n-1, -1, -1):
            for j in range(n-1, -1, -1):
                mD[i][j] = min(
                    mD[i+1][j] + 1 if i < n - 1 else INF,
                    mD[i][j+1] + 1 if j < n - 1 else INF,
                    mD[i][j]
                )
        
        # q = [(-mD[0][0], (0, 0))]
        # print(mD)
        # mSF = [[-1] * n for _ in range(n)]
        # mSF[0][0] = mD[0][0]
        # while q:
        #     sf, p = heappop(q)
        #     sf = -sf
        #     print(sf, p)
        #     currD = min(sf, mD[p[0]][p[1]])
        #     if p == (n-1, n-1):
        #         return currD
        #     for nb in [(p[0]+1, p[1]), (p[0], p[1]+1), (p[0]-1, p[1]), (p[0], p[1]-1)]:
        #         if nb[0] < 0 or nb[0] >= n or nb[1] < 0 or nb[1] >= n or (mSF[nb[0]][nb[1]]!= -1 and mSF[nb[0]][nb[1]] >= currD):
        #             continue
        #         heappush(q, (-currD, nb))
        #         mSF[nb[0]][nb[1]] = currD

        q = deque([(mD[0][0], (0, 0))])
        mSF = INF
        vis = set()
        vis.add((0, 0))
        while q:
            sf, (i, j) = q.popleft()  # get the cell with maximum safeness distance
            mSF = min(sf, mSF)
            if i == n - 1 and j == n - 1:
                return mSF
            for (y, x) in [(i+1, j), (i, j+1), (i-1, j), (i, j-1)]:
                if x < 0 or x >= n or y < 0 or y >= n or (y, x) in vis:
                    continue
                vis.add((y, x))
                if mD[y][x] >= mSF:
                    q.appendleft((mSF, (y, x)))
                else:
                    q.append((mD[y][x], (y, x)))

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