<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:
        thieves = set()
        n = len(grid[0])
        for row in range(n):
            for col in range(n):
                if grid[row][col] == 1:
                    thieves.add((row, col))
        
        if (0, 0) in thieves or (n-1, n-1) in thieves:
            return 0

        def manhattan(pos1, pos2):
            return abs(pos1[0] - pos2[0]) + abs(pos1[1] - pos2[1])

        def safe_dist(pos, thieves):
            nonlocal n
            min_dist = n
            for t in thieves:
                min_dist = min(min_dist, manhattan(pos, t))

            return min_dist

        def greedy_bfs(grid, thieves):
            i = 0
            j = 0
            to_visit = []

            min_dist = safe_dist((i, j), thieves)
            heapq.heappush(to_visit, (-min_dist, i, j))
            visited = set()
            visited.add((i, j))

            while len(to_visit) > 0:
                next = heapq.heappop(to_visit)

                dist, i, j = -next[0], -next[1], -next[2]

                # print(dist)

                if i == n-1 and j == n-1:
                    return dist

                if i < n-1 and (i+1, j) not in visited and (i+1, j) not in thieves:
                    heapq.heappush(to_visit, (-min(dist, safe_dist((i+1, j), thieves)), -(i+1), -j))
                    visited.add((i+1, j))
                if j < n - 1 and (i, j+1) not in visited and (i, j+1) not in thieves:
                    heapq.heappush(to_visit, (-min(dist, safe_dist((i, j+1), thieves)), -i, -(j+1)))
                    visited.add((i, j+1))
                if i > 0 and (i-1, j) not in visited and (i-1, j) not in thieves:
                    heapq.heappush(to_visit, (-min(dist, safe_dist((i-1, j), thieves)), -(i-1), -j))
                    visited.add((i-1, j))
                if j > 0 and (i, j-1) not in visited and (i, j-1) not in thieves:
                    heapq.heappush(to_visit, (-min(dist, safe_dist((i, j-1), thieves)), -i, -(j-1)))
                    visited.add((i, j-1))

            return 0
                
        return greedy_bfs(grid, thieves)
32996