<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 grid[0][0] == 1 or grid[-1][-1] == 1:
            return 0

        def is_valid(ni,nj):
            if ni < 0 or ni>= len(grid):
                return False
            if nj < 0 or nj >= len(grid[0]):
                return False
            return True

        def path_exists(factor, distance_grid):
            if distance_grid[0][0] < factor:
                return False
            queue = deque()
            queue.append((0,0))
            visited = set()
            while queue:
                i,j = queue.popleft()
                visited.add((i,j))
                if i == len(grid) - 1 and j == len(grid[0]) - 1 and distance_grid[i][j] >= factor:
                    return True
                
                
                
                for ni, nj in [(i+1,j), (i-1,j), (i,j+1), (i,j-1)]:
                    if is_valid(ni,nj) and distance_grid[ni][nj] >= factor and (ni,nj) not in visited:
                        queue.append((ni,nj))
                        visited.add((ni,nj))
            return False


        queue = deque()
        thief_distance_grid = [[float('inf') for _ in range(len(grid[0]))] for _ in range(len(grid))]
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    queue.append((i,j,0))
                    thief_distance_grid[i][j] = 0
        # perform bfs

        
        while queue:
            i,j,distance = queue.popleft()

            for ni,nj in [(i+1,j), (i-1,j), (i,j+1), (i,j-1)]:
                if is_valid(ni,nj) and thief_distance_grid[ni][nj] == float('inf'):
                    thief_distance_grid[ni][nj] = distance+1
                    queue.append((ni,nj,distance+1))
        
        #print(thief_distance_grid)
        low = 0
        high = len(grid) + len(grid[0])
        res = -1
        while low <= high:
            mid = (low+high)//2
            if path_exists(mid, thief_distance_grid):
                res = mid
                low = mid+1
            else:
                high = mid - 1
        
        return res
        
            
                    

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