<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:
        def isSafe(i,j):
            if i<0 or j<0 or i>=n or j>=m:
                return False
            return True

        n,m=len(grid),len(grid[0])
        q=deque()
        distance={}
        directions=[[-1,0],[0,-1],[1,0],[0,1]]
        for i in range(n):
            for j in range(m):
                if grid[i][j]==1:
                    q.append((i,j,0))
                    distance[(i,j)]=0
        
        while q:
            # print(q)
            x,y,dist=q.popleft()
            for dx,dy in directions:
                if isSafe(x+dx,y+dy) and (x+dx,y+dy) not in distance:
                    distance[(x+dx,y+dy)]=dist+1
                    q.append((x+dx,y+dy,dist+1))
        # print(distance)
        
        
        visited=set()
        def dijiktra():
            maxHeap=[(-distance[(0,0)],0,0)]
            safeness=defaultdict(int)
            safeness[(0,0)]=distance[(0,0)]
            while maxHeap:
                dist,i,j=heapq.heappop(maxHeap)
                # if i==n-1 and j==m-1:
                #     return -dist
                for di,dj in directions:
                    x,y=i+di,j+dj
                    if isSafe(x,y) and grid[x][y]==0:
                        new_dist=min(safeness[(i,j)],distance[(x,y)])
                        # print(new_dist,x,y)
                        if new_dist>safeness[(x,y)]:
                            safeness[(x,y)]=new_dist
                            heapq.heappush(maxHeap,(-new_dist,x,y))

            # print(safeness)
            # if safeness[(n-1,m-1)]==float('inf'):
            #     return 0
            return safeness[(n-1,m-1)]

        
        return dijiktra()
                

        
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