<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, m = len(grid), len(grid[0])
        nm = n*m
        f = list(range(nm))
        
        def find(x):
            if f[x] != x:
                f[x] = find(f[x])

            return f[x]

        def merge(u, v):
            fu, fv = find(u), find(v)
            if fu < fv:
                f[fu] = fv
            elif fv < fu:
                f[fv] = fu

        # mp is an array, mp[i] stores the [y, x] which has the distance i to the nearest theif
        mp = [[]]
        for y in range(n):
            for x in range(m):
                if grid[y][x]:
                    mp[0].append([y, x])
                    grid[y][x] = None

        # calculate the mp by using BFS
        while grid[0][0] is not None and grid[-1][-1] is not None:
            que = mp[-1]
            que2 = []
            mp.append(que2)

            for y, x in que:
                for y1, x1 in [y+1, x], [y-1, x], [y, x+1], [y, x-1]:
                    if 0 <= y1 < n and 0 <= x1 < m and grid[y1][x1] is not None:
                        que2.append([y1, x1])
                        grid[y1][x1] = None

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

        # Use Union-find to see if there's a router from start to end
        for y in range(n):
            for x in range(m):
                if grid[y][x] is not None:
                    idx = y*m+x
                    if y+1 < n and grid[y+1][x] is not None:
                        merge(idx, idx+m)
                    if x+1 < m and grid[y][x+1] is not None:
                        merge(idx, idx+1)

        if find(0) == nm-1:
            return len(mp)

        # In each step, recovered the [y, x] in mp[-1], which is the nodes with the longest distance to the theif.
        # If there's a valid path, break. Then the answer is this len(mp)
        while mp:
            que = mp.pop()
            for y, x in que:
                grid[y][x] = 0
                idx = y*m+x
                for y1, x1 in [y+1, x], [y-1, x], [y, x+1], [y, x-1]:
                    if 0 <= y1 < n and 0 <= x1 < m and grid[y1][x1] is not None:
                        merge(idx, y1*m+x1)
            if find(0) == nm-1:
                break

        return len(mp)

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