<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)
        distance = [[0] * n for _ in range(n)]

        visited = set()
        q = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append([i,j])
                    visited.add((i,j))
        cur = 1
        while q:
            for _ in range(len(q)):
                x,y = q.popleft()
                for dx, dy in [[1,0],[0,1],[-1,0],[0,-1]]:
                    xx,yy = x+dx,y+dy
                    if 0<=xx<=n-1 and 0<=yy<=n-1 and (xx,yy) not in visited:
                        distance[xx][yy] = cur
                        visited.add((xx,yy))
                        q.append([xx,yy])
            cur += 1
        ######### min(distance[0][0], distance[n-1][n-1]) #########

        # factor = -float("inf")
        # q = deque()
        # q.append([0,0, distance[0][0]])
        # while q:
        #     x,y,d = q.popleft()

        #     if (x,y) == (n-1, n-1):
        #         if d == min(distance[0][0], distance[n-1][n-1]):
        #             return min(distance[0][0], distance[n-1][n-1])
        #         factor = max(factor, d)
        #         continue

        #     for dx,dy in [[1,0],[0,1]]:
        #         min_d = d
        #         xx, yy = x+dx, y+dy
        #         if 0<=xx<=n-1 and 0<=yy<=n-1 and (factor == -float("inf") or distance[xx][yy] > factor):
        #             min_d = min(min_d, distance[xx][yy])
        #             q.append([xx,yy,min_d])
        
        # return factor 

        factor = min(distance[0][0], distance[n-1][n-1])
        for d in range(factor, -1, -1):
            q = deque()
            q.append([0,0])

            visited  = set()
            visited.add((0,0))

            while q:
                x,y = q.popleft()
                if (x,y) == (n-1, n-1):
                    return d

                for dx,dy in [[1,0],[0,1],[0,-1],[-1,0]]:
                    xx, yy = x+dx, y+dy
                    if 0<=xx<=n-1 and 0<=yy<=n-1 and distance[xx][yy] >= d and (xx,yy) not in visited:
                        q.append([xx,yy])
                        visited.add((xx,yy))
        




                    



        
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