<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

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # If the start (0,0) or end (n-1, n-1) cells are 1, return 0 as it's unsafe to start or end there.
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return 0
        
        n = len(grid)  # Get the size of the grid.
        q = deque()  # Queue for BFS.

        ###### Step 1: Compute the distance to the nearest '1' in the grid for each cell #######
        dis = [[-1] * n for _ in range(n)]  # dis[i][j] will store the distance from (i,j) to the nearest '1'
        
        # Initialize the queue with all the cells that contain a '1'
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append((i, j, 0))  # Append (x, y, distance = 0)
        
        # Multi-source BFS to compute the shortest distance from each cell to the nearest '1'
        while q:
            x, y, t = q.popleft()
            if dis[x][y] != -1:  # If this cell has already been processed, skip it
                continue
            dis[x][y] = t  # Set the distance to the current value 't'
            # Explore the four possible directions (right, down, left, up)
            for nx, ny in (x+1, y), (x, y+1), (x-1, y), (x, y-1):
                if 0 <= nx < n and 0 <= ny < n and dis[nx][ny] == -1:
                    q.append((nx, ny, t + 1))  # Add the neighboring cells to the queue with incremented distance

        ###### Step 2: Use binary search to find the maximum safeness factor #######
        vis = [[-1] * n for _ in range(n)]  # Visited array to track whether a cell has been visited

        # Helper function to check if we can safely travel from (0,0) to (n-1, n-1) with a given safeness factor
        def check(tar):
            q = deque()
            q.append((0, 0))  # Start from the top-left corner
            if dis[0][0] < tar:  # If the starting cell is not safe enough, return False
                return False
            while q:
                x, y = q.popleft()
                if x == n - 1 and y == n - 1:  # If we reached the bottom-right corner, return True
                    return True
                if vis[x][y] == tar:  # If the cell has been visited during this iteration, skip it
                    continue
                vis[x][y] = tar  # Mark the cell as visited for this iteration
                # Explore the four possible directions
                for nx, ny in (x+1, y), (x, y+1), (x-1, y), (x, y-1):
                    if 0 <= nx < n and 0 <= ny < n and vis[nx][ny] != tar and dis[nx][ny] >= tar:
                        q.append((nx, ny))  # Add the neighboring cells if they are safe enough
            return False

        l, r = 0, 2 * n  # Initialize the binary search bounds

        ###### Step 3: Perform binary search to find the largest safeness factor #######
        while l < r:
            mid = (l + r + 1) >> 1  # Calculate the middle value
            if check(mid):  # Check if we can safely traverse the grid with this safeness factor
                l = mid  # If yes, move to the right half (higher safeness factor)
            else:
                r = mid - 1  # If no, move to the left half (lower safeness factor)
        
        return r  # Return the maximum safeness factor found

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