<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:
        self.n = len(grid)

        thief_cells = []
        for row in range(self.n):
            for col in range(self.n):
                if grid[row][col] == 1:
                    thief_cells.append([row,col])

        self.thief_grid = [[math.inf] * self.n for _ in range(self.n)]
        queue = thief_cells.copy()
        dist = 0
        while queue:
            new_queue = []
            for row, col in queue:
                result = self.traverse(row, col, dist)
                new_queue += result
            queue = [cell for cell in new_queue if cell]
            dist += 1

        low = 0
        high = min(self.thief_grid[0][0], self.thief_grid[-1][-1])
        while low < high - 1:
            mid = int((low + high) // 2)
            if self.path(mid, set()):
                low = mid
            else:
                high = mid
        if self.path(high, set()):
            return high
        else:
            return low

    def path(self, dist: int, seen: set):
        queue = [[0,0]]
        while queue:
            new_queue = []
            for row, col in queue:
                if row >= 0 and row < self.n and col >= 0 and col < self.n and f'{row}_{col}' not in seen and self.thief_grid[row][col] >= dist:
                    seen.add(f'{row}_{col}')
                    if row == self.n-1 and col == self.n-1:
                        return True
                    new_queue += [[row,col-1], [row-1,col], [row,col+1], [row+1,col]]
            queue = new_queue
        return False

    def traverse(self, row: int, col: int, dist: int):
        if row >= 0 and row < self.n and col >= 0 and col < self.n and self.thief_grid[row][col] > dist:
            self.thief_grid[row][col] = dist
            return [[row,col-1], [row-1,col], [row,col+1], [row+1,col]]
        else:
            return [None]

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