<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 UF:
    def __init__(self, n):
        self.parents = {i: i for i in range(n)}
        self.size = {i: 1 for i in range(n)}
    
    def find(self, x):
        if self.parents[x] != x:
            self.parents[x] = self.find(self.parents[x])
        return self.parents[x]
    
    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        
        if rootX != rootY:
            if self.size[rootX] < self.size[rootY]:
                self.parents[rootX] = rootY
                self.size[rootY] += self.size[rootX]
            else:
                self.parents[rootY] = rootX
                self.size[rootX] += self.size[rootY]

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        total_cells = rows * cols
        uf = UF(total_cells)
        
        def to_index(x, y):
            return x * cols + y
        
        mp = [[]]
        for x in range(rows):
            for y in range(cols):
                if grid[x][y]:
                    mp[0].append([x, y])
                    grid[x][y] = None

        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        while grid[0][0] is not None and grid[-1][-1] is not None:
            que = mp[-1]
            que2 = []
            mp.append(que2)

            for x, y in que:
                for dx, dy in directions:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] is not None:
                        que2.append([nx, ny])
                        grid[nx][ny] = None

        que = mp.pop()
        for x, y in que:
            grid[x][y] = 0

        for x in range(rows):
            for y in range(cols):
                if grid[x][y] is not None:
                    idx = to_index(x, y)
                    if x + 1 < rows and grid[x + 1][y] is not None:
                        uf.union(idx, to_index(x + 1, y))
                    if y + 1 < cols and grid[x][y + 1] is not None:
                        uf.union(idx, to_index(x, y + 1))

        if uf.find(0) == uf.find(total_cells - 1):
            return len(mp)

        while mp:
            que = mp.pop()
            for x, y in que:
                grid[x][y] = 0
                idx = to_index(x, y)
                for dx, dy in directions:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] is not None:
                        uf.union(idx, to_index(nx, ny))
            if uf.find(0) == uf.find(total_cells - 1):
                break

        return len(mp)
60568