<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:
        n = len(grid)
        queue = collections.deque()
        new_grid = [[0 for i in range(len(grid[0]))] for j in range(len(grid))]
        visited = set()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    queue.append([i,j,0])
                    visited.add((i,j))
        ans = 0
        # if grid[0][0] == 1 or grid[n-1][n-1] == 1:
        #     return 0
        
        while len(queue)>0:
            i,j,dist = queue.popleft()
            new_grid[i][j] = dist
            
            if i+1 < n and (i+1,j) not in visited and grid[i+1][j] == 0:
                visited.add((i+1,j))
                queue.append([i+1,j,dist+1])
            
            if i-1 >= 0 and (i-1,j) not in visited and grid[i-1][j] == 0:
                visited.add((i-1,j))
                queue.append([i-1,j,dist+1])
            
            if j-1 >= 0 and (i,j-1) not in visited and grid[i][j-1] == 0:
                visited.add((i,j-1))
                queue.append([i,j-1,dist+1])
            
            if j+1 < n and (i,j+1) not in visited and grid[i][j+1] == 0:
                visited.add((i,j+1))
                queue.append([i,j+1,dist+1])
        
        
        def good(val):
            queue = collections.deque()
            queue.append([0,0])
            visited = set()
            visited.add((0,0))
            if new_grid[0][0] < val:
                return False
            while len(queue)>0:
                i,j = queue.popleft()
                if (i,j) == (n-1,n-1):
                    return True
                if i +1 < n and new_grid[i+1][j] >= val and (i+1,j) not in visited:
                    queue.append([i+1,j])
                    visited.add((i+1,j))
                if i -1 >= 0 and new_grid[i-1][j] >= val and (i-1,j) not in visited:
                    queue.append([i-1,j])
                    visited.add((i-1,j))
                
                if j +1 < n and new_grid[i][j+1] >= val and (i,j+1) not in visited:
                    queue.append([i,j+1])
                    visited.add((i,j+1))
                
                if j - 1 >= 0 and new_grid[i][j-1] >= val and (i,j-1) not in visited:
                    queue.append([i,j-1])
                    visited.add((i,j-1))
            
            return False
        
        l,h,ans = 0,10**3,-1
        while l <= h:
            mid = (l+h)//2 
            if good(mid):
                ans = mid
                l = mid + 1
            else:
                h = mid -1
        
        return ans
        
        
            
      
    
    
            
        
        
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