<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:
        matrix_point = self.mark_point(grid)
        l, r = float("inf"), float("-inf")
        for row in matrix_point:
            for v in row:
                l = min(l, v)
                r = max(r, v)
        r += 1
        while l + 1 < r:
            mid = (l + r) >> 1
            if self.is_valid_mid(matrix_point, mid):
                l = mid
            else:
                r = mid
        return l
    
    def is_valid_mid(self, grid, mid) -> bool:
        if grid[0][0] < mid:
            return False
        n = len(grid)
        m = len(grid)
        pool = [(0, 0)]
        visited_set = set(pool)
        while pool:
            tmp = []
            for x, y in pool:
                if (x, y) == (n - 1, m - 1):
                    return True
                for nex, ney in [(x + 1, y), (x, y + 1), (x - 1, y), (x, y - 1)]:
                    if not self.is_valid(grid, nex, ney):
                        continue
                    if (nex, ney) in visited_set:
                        continue
                    if grid[nex][ney] < mid:
                        continue
                    tmp.append((nex, ney))
                    visited_set.add((nex, ney))
            pool = tmp
        return False

    
    def mark_point(self, grid: List[List[int]]) -> List[List[int]]:
        n = len(grid)
        m = len(grid)
        point_matrix = [[0 for _ in range(m)] for _ in range(n)]
        pool = [(i, j) for i, row in enumerate(grid) for j, val in enumerate(row) if val == 1]
        visited_set = set(pool)
        cur_dis = 0
        while pool:
            tmp = []
            for x, y in pool:
                point_matrix[x][y] = cur_dis
                for nex, ney in [(x + 1, y), (x, y + 1), (x - 1, y), (x, y - 1)]:
                    if not self.is_valid(grid, nex, ney):
                        continue
                    if (nex, ney) in visited_set:
                        continue
                    tmp.append((nex, ney))
                    visited_set.add((nex, ney))
            pool = tmp
            cur_dis += 1
        return point_matrix
    
    def is_valid(self, grid, x, y) -> bool:
        return 0 <= x < len(grid) and 0 <= y < len(grid[0])
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