<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:
        manhattan_dist = [[sys.maxsize]*len(grid[0]) for _ in range(len(grid))]
        r,c = len(grid), len(grid[0])
        q = []
        if grid[0][0] or grid[r-1][c-1]:
            return 0

        # find thiefs, for bfs to build manhattan dist
        for i in range(r):
            for j in range(c):
                if grid[i][j] == 1:
                    manhattan_dist[i][j] = 0
                    q.append((i,j))
        # build manhattan distance grid
        while q:
            (i,j) = q[0]
            q = q[1:]

            for I,J in [(i+1, j), (i-1, j), (i,j+1), (i, j-1)]:
                if 0 <= I < r and 0 <= J < c and manhattan_dist[I][J] > manhattan_dist[i][j] + 1:
                    q.append((I,J))
                    manhattan_dist[I][J] = manhattan_dist[i][j] + 1
       
        # checks for a following factor v there exists a path in grid from 0,0 to r-1,c-1
        def isSafe(v):
            if manhattan_dist[0][0] < v: return False
            path_q = [(0,0)]
            visited_path = set()
            while path_q:
                i,j = path_q[0]
                # reached the end of grid, path exists!
                if i == r - 1 and j == c - 1:
                    return True
                path_q = path_q[1:]
                for I,J in [(i+1, j), (i-1, j), (i,j+1), (i, j-1)]:
                    if (I,J) in visited_path or I >= r or I < 0 or J >= c or J < 0:
                        continue
                    # path doesnt exist for this node
                    if manhattan_dist[I][J] < v:
                        continue
                    path_q.append((I,J))
                    visited_path.add((I,J))


        
        # binary search to find minimum safe distance
        low, high = 0, 400 * 400
        ans = 0
        while low <= high:
            mid =  low + (high - low) // 2
            if isSafe(mid):
                ans = mid
                low = mid + 1
                
            else:
                high = mid - 1
            
        return ans
        
        
        
39721