<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>

from collections import deque
from typing import List

class Solution:
    def getMinDistance(self, grid, d):
        queue = deque()
        m = len(grid)
        n = len(grid[0])

        # Add all thief locations to the queue as starting points
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    queue.append((i, j, 0))  # (x, y, distance)
                    d[i][j] = 0  # Distance to self is 0

        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Down, up, right, left

        while queue:
            x, y, distance = queue.popleft()
            for direction in directions:
                new_x, new_y = x + direction[0], y + direction[1]
                if 0 <= new_x < m and 0 <= new_y < n and grid[new_x][new_y] == 0 and d[new_x][new_y] > distance + 1:
                    queue.append((new_x, new_y, distance + 1))
                    d[new_x][new_y] = distance + 1  # Update distance

    def get_path(self, i, j, grid, d, vis, v):
        m = len(grid)
        n = len(grid[0])

        # If we've reached the bottom-right corner
        if i == m - 1 and j == n - 1:
            return True

        vis[i][j] = 1  # Mark as visited

        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Down, up, right, left

        # Try moving in all four directions
        for direction in directions:
            new_x, new_y = i + direction[0], j + direction[1]
            if 0 <= new_x < m and 0 <= new_y < n and grid[new_x][new_y] == 0 and vis[new_x][new_y] == 0 and d[new_x][new_y] >= v:
                if self.get_path(new_x, new_y, grid, d, vis, v):
                    return True

        return False

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])

        # Initialize the distance array to infinity
        d = [[float("inf") for _ in range(n)] for _ in range(m)]

        # Compute minimum distances using multi-source BFS from all thieves (1s)
        self.getMinDistance(grid, d)

        # Binary search for maximum safeness factor `v`
        left, right = 0, min(d[0][0], d[m-1][n-1])  # Safeness can't be higher than the smallest value between start and end

        while left <= right:
            mid = (left + right) // 2
            vis = [[0 for _ in range(n)] for _ in range(m)]  # Reset visited array for each trial
            if self.get_path(0, 0, grid, d, vis, mid):  # Check if there's a valid path with safeness `mid`
                left = mid + 1  # Try for a higher safeness
            else:
                right = mid - 1  # Try for a lower safeness


        if right == -1:
            right = 0
        return right

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