<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:

    dir = [(0,1), (1,0), (0,-1), (-1,0)]

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

        q = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append((i, j))
                    grid[i][j] = 0
                else:
                    grid[i][j] = -1

        while q:
            l = len(q)
            while l > 0:
                curr = q.popleft()
                for d in self.dir:
                    x, y = curr[0] + d[0], curr[1] + d[1]
                    val = grid[curr[0]][curr[1]]
                    if 0 <= x < n and 0 <= y < n and grid[x][y] == -1:
                        grid[x][y] = val + 1
                        q.append((x, y))
                l -= 1

        start, end, res = 0, 0, -1
        for i in range(n):
            for j in range(n):
                end = max(end, grid[i][j])
        
        while start <= end:
            mid = (end + start) // 2
            if self.isValidSafeness(grid, mid):
                res = mid
                start = mid + 1
            else:
                end = mid - 1
        
        return res


    def isValidSafeness(self, grid, min_safeness):
        n = len(grid)

        if grid[0][0] < min_safeness or grid[n-1][n-1] < min_safeness:
            return False
        
        q = deque([(0,0)])
        visited = set([(0,0)])

        while q:
            curr = q.popleft()
            if curr[0] == n - 1 and curr[1] == n - 1:
                return True
            
            for d in self.dir:
                x, y = curr[0] + d[0], curr[1] + d[1]
                if (x,y) not in visited and 0 <= x < n and 0 <= y < n and grid[x][y] >= min_safeness:
                    visited.add((x, y))
                    q.append((x,y))
        
        return False



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