<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:
        cell_to_safeness = {}

        # step 1, modified bfs, similar to walls and gates to calculate
        # the minimum distance each cell has from a thief
        queue = collections.deque()
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    queue.append([i, j, 0])
                    cell_to_safeness[(i, j)] = 0
        
        directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]
        while queue:
            row, col, score = queue.popleft()
            for diff_row, diff_col in directions:
                new_row = row + diff_row
                new_col = col + diff_col

                if (
                    0 <= new_row < len(grid) and
                    0 <= new_col < len(grid[0]) and
                    (new_row, new_col) not in cell_to_safeness
                ):
                    cell_to_safeness[(new_row, new_col)] = score + 1
                    queue.append([new_row, new_col, score + 1])
        
        # initial state, best path score at starting cell 0,0 is
        # the distance that cell 0,0 is from a thief
        max_heap = [[
            -cell_to_safeness[(0, 0)],
            0,
            0,
        ]]

        visited = set([(0,0)])

        while max_heap:
            best_path_score, row, col = heapq.heappop(max_heap)
            # correct the sign
            best_path_score = -best_path_score

            if (row, col) == (len(grid) - 1, len(grid[0]) - 1):
                return best_path_score

            for diff_row, diff_col in directions:
                new_row = row + diff_row
                new_col = col + diff_col

                if (
                    0 <= new_row < len(grid) and
                    0 <= new_col < len(grid[0]) and
                    (new_row, new_col) not in visited
                ):
                    visited.add((new_row, new_col))
                    heapq.heappush(max_heap, [
                        -min(best_path_score, cell_to_safeness[(new_row,new_col)]),
                        new_row,
                        new_col,
                    ])
                    
        
        return -1

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