<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 UnionFind:
    """A class that implements the Union-Find (Disjoint Set) data structure."""
    # https://github.com/doocs/leetcode/tree/main/solution/2800-2899/2812.Find%20the%20Safest%20Path%20in%20a%20Grid
  
    def __init__(self, size):
        # Parent pointers initialised to point to themselves and sizes initialised to 1
        self.parent = list(range(size))
        self.set_size = [1] * size

    def find(self, x):
        """Finds the representative (root) of the set containing 'x'. Implements path compression."""
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, a, b):
        """Unites the sets containing 'a' and 'b'. Return False if they are already in the same set, True otherwise."""
        root_a, root_b = self.find(a), self.find(b)
        if root_a == root_b:
            return False
          
        # Union by size, making the smaller root point to the larger one
        if self.set_size[root_a] > self.set_size[root_b]:
            self.parent[root_b] = root_a
            self.set_size[root_a] += self.set_size[root_b]
        else:
            self.parent[root_a] = root_b
            self.set_size[root_b] += self.set_size[root_a]
        return True


class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        """Calculates the maximum safeness factor in a grid avoiding unsafe cells."""
        # Initialize variables
        n = len(grid)
        # If start or end cells are unsafe, return 0
        if grid[0][0] or grid[n - 1][n - 1]:
            return 0
          
        # BFS to find distances from unsafe cells
        queue = deque()
        dist = [[float('inf')] * n for _ in range(n)]
        # Seed initial distances for unsafe cells
        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    queue.append((i, j))
                    dist[i][j] = 0
        # Directions for moving to adjacent cells
        directions = (-1, 0, 1, 0, -1)
        while queue:
            ci, cj = queue.popleft()
            for da, db in pairwise(directions):
                ni, nj = ci + da, cj + db
                if 0 <= ni < n and 0 <= nj < n and dist[ni][nj] == float('inf'):
                    dist[ni][nj] = dist[ci][cj] + 1
                    queue.append((ni, nj))

        # Sort cells based on their distance from unsafe cells in descending order
        candidates = sorted(((dist[i][j], i, j) for i in range(n) for j in range(n)), reverse=True)
        uf = UnionFind(n * n)
        for d, i, j in candidates:
            # Attempt to connect current cell with all its neighbors
            for da, db in pairwise(directions):
                x, y = i + da, j + db
                if 0 <= x < n and 0 <= y < n and dist[x][y] >= d:
                    uf.union(i * n + j, x * n + y)
            # Check if start and end cells are connected
            if uf.find(0) == uf.find(n * n - 1):
                return int(d)
        return 0
3660