<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:
        """
        calculate the safeness factor for each cell
        and then perform dijkstra from (0, 0) to (n - 1, n - 1)
        """
        safeness = dict()
        N = len(grid)

        def get_manhattan_distance(src: Tuple[int], dst: Tuple[int]) -> int:
            a, b = src
            x, y = dst
            return abs(a - x) + abs(b - y)
        
        def get_adjacents(pos: Tuple[int]) -> Generator[Tuple[int], None, None]:
            x, y = pos
            if x > 0:
                yield x - 1, y
            if y > 0:
                yield x, y - 1
            if x + 1 < N:
                yield x + 1, y
            if y + 1 < N:
                yield x, y + 1

        def propagate(thief_positions: List[Tuple[int]]):
            queue = deque(thief_positions)
            visited = set()

            while queue:
                pos, dist = queue.popleft()
                visited.add(pos)
                if pos not in safeness or safeness[pos] > dist:
                    safeness[pos] = dist
                    for adj in get_adjacents(pos):
                        if adj not in visited:
                            queue.append((adj, dist + 1))
                            visited.add(adj)

        thieves = []        
        for r in range(N):
            for c in range(N):
                if grid[r][c] == 1:
                    thieves.append(((r, c), 0))
        propagate(thieves)
        
        def can_satisfy(desired: int) -> bool:
            q = deque([(0, 0)])
            visited = set()

            while q:
                pos = q.popleft()
                visited.add(pos)
                if safeness[pos] < desired:
                    continue
                if pos == (N - 1, N - 1):
                    return True
                for adj in get_adjacents(pos):
                    if adj not in visited:
                        q.append(adj)
                        visited.add(adj)
            
            return False
        
        lo, hi = min(safeness.values()), max(safeness.values())

        while lo <= hi:
            mid = (lo + hi) // 2
            if can_satisfy(mid):
                lo = mid + 1
            else:
                hi = mid - 1
        
        return lo - 1
                
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