<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:
        # calculate the distance from 1"s and choose a disatnace with maximum distnace
        # i feel i need to store the mi _distance from 0 to closest 1 using bfs
        paths=deque()
        # i will use a hashmap to store the min_distance from a node to nearesr one
        dic={}
        visited=set()
        max_row=len(grid)
        max_col=len(grid[0])
        for row in range(max_row):
            for col in range(max_col):
                if grid[row][col]==1:
                    dic[(row,col)]=0
                    paths.append([0,row,col])
        neib=[[0,1],[0,-1],[1,0],[-1,0]]
        while paths:
            distance,row,col=paths.popleft()
            if (row,col) in visited:
                continue
            dic[(row,col)]=-1*distance
            visited.add((row,col))
            for nr,nc in neib:
                if min(nr+row,nc+col)<0 or nr+row>=max_row or nc+col>=max_col or (nr+row,nc+col) in visited:
                    continue
                paths.append([distance+1,nr+row,nc+col])
        # i will use a dikstra algortim from 0,0 to end,end
        max_heap=[[dic[0,0],0,0]]
        ans=float("-inf")
        visited=set()
        while max_heap:  #v^2
            far_dis,row,col=heapq.heappop(max_heap)
            ans=max(ans,far_dis)
            if row==max_row-1 and col==max_col-1:
                return -1*ans
            if (row,col) in visited:
                continue
            visited.add((row,col))
            for nr,nc in neib:
                if min(nr+row,nc+col)<0 or nr+row>=max_row or nc+col>=max_col or (nr+row,nc+col) in visited:
                    continue
                heapq.heappush(max_heap,[dic[(nr+row,nc+col)],nr+row,nc+col])







        
        
                
            




        

        
                

        
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