<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 createSafe(self , grid):
        n = len(grid)
        m = len(grid[0])

        safe = [[float('inf')]*m for _  in range(n)]

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

        while len(q):
            sfe += 1
            sze = len(q)
            for _ in range(sze):
                ci , cj = q.popleft()

                if sfe >= safe[ci][cj]:
                    continue

                safe[ci][cj] = sfe
                neigh = [(ci+1 , cj) , (ci-1 , cj) , (ci , cj+1) , (ci , cj-1)]    
                
                for ni , nj in neigh:
                    if 0<=ni<len(grid) and 0<=nj<len(grid):
                        q.append((ni , nj))

        return safe

    def predicate(self , safe , mid):
        if safe[0][0] < mid:
            return False

        n = len(safe)
        m = len(safe[0])

        q = deque()
        visited = set()

        q.append((0 , 0))
        visited.add((0 , 0))
        while len(q):
            ci , cj = q.popleft()
            
            if ci == n-1 and cj == m-1:
                return True

            neigh = [(ci+1 , cj) , (ci-1 , cj) , (ci , cj+1) , (ci , cj-1)]    
            for ni , nj in neigh:
                if 0<=ni<n and 0<=nj<m and (ni , nj) not in visited and safe[ni][nj]>=mid:
                    q.append((ni , nj))
                    visited.add((ni , nj))
        
        return False
        
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid)

        safe = self.createSafe(grid)

        low = 0
        high = 1000000000

        while low<high:
            mid = (low+high)//2 + 1

            if self.predicate(safe , mid):
                low = mid
            
            else:
                high = mid-1

        return low

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