<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
        N = len(grid)
        thief_locs = deque()

        for r in range(N):
            for c in range(N):
                if grid[r][c] == 1:
                    grid[r][c] = 0
                    thief_locs.append((r,c))
                else:
                    grid[r][c] = float("inf") 
        
        def pre_compute_safety_score():
            #max_safety_score = 0
            visited = set()

            while thief_locs:
                r,c = thief_locs.popleft()

                if (r,c) in visited: continue
                visited.add((r,c))

                curr_safety_score = grid[r][c]
                for i, j in ((r-1,c), (r+1,c), (r,c-1), (r,c+1)):
                    if not (0<= i < N and 0 <= j < N) or (i,j) in visited: continue
                    grid[i][j] = min(curr_safety_score+1, grid[i][j])
                    #max_safety_score = max(max_safety_score, grid[i][j])

                    thief_locs.append((i,j))
            #return max_safety_score
        

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

            while q:
                r,c = q.popleft()
                if (r,c) == (N-1, N-1):
                    return True

                if (r,c) in visited: continue
                visited.add((r,c))

                for i,j in ((r-1,c), (r+1,c), (r,c-1), (r,c+1)):
                    if not (0 <= i < N and 0 <= j < N) or (i,j) in visited or grid[i][j] < min_safety_score: continue
                    q.append((i,j))
            
            return False

        # start, end = 0, pre_compute_safety_score()
        pre_compute_safety_score()
        start, end = 0, min(grid[0][0], grid[-1][-1])
        max_safety = 0

        while start <= end:
            mid = (start+end)//2
            if path_is_within_safety_score(mid):
                max_safety = max(max_safety, mid)
                start = mid + 1
            else:
                end = mid - 1

        return max_safety





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