<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:
        # multi-source BFS + binary search / dijkstra(weighted edgs)
        dirs = [[-1, 0], [0, 1], [1, 0], [0, -1]]

        n, m = len(grid), len(grid[0])
        q = collections.deque()

        dists = [[math.inf] * m for _ in range(n)]
        max_dist = 0

        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    q.append((i, j))
                    dists[i][j] = 0

        step = 0
        while q:
            size = len(q)

            step += 1
            for _ in range(size):
                x, y = q.popleft()

                for dx, dy in dirs:
                    nx, ny = x + dx, y + dy

                    if nx >= 0 and nx < n and ny >= 0 and ny < m and dists[nx][ny] > step:
                        dists[nx][ny] = step
                        max_dist = max(max_dist, dists[nx][ny])
                        q.append((nx, ny))

        # Binary search

        def check_path_existence(threshold):
            # dfs
            visited = [[False] * m for _ in range(n)]

            def dfs(x, y):
                visited[x][y] = True
                # edge cases, if dists[0][0] > threshold
                if dists[x][y] < threshold:
                    return False

                if x == n - 1 and y == m - 1:
                    return True

                for dx, dy in dirs:
                    nx, ny = x + dx, y + dy
                    if (
                        nx >= 0
                        and nx < n
                        and ny >= 0
                        and ny < m
                        and not visited[nx][ny]
                        and dists[nx][ny] >= threshold
                    ):
                        if dfs(nx, ny):
                            return True

                return False

            return dfs(0, 0)
        

        left, right = 0, max_dist
        while left <= right:
            mid = (left + right) >> 1
            if check_path_existence(mid):
                left = mid + 1
            else:
                right = mid - 1

        return left - 1

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