<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>

import heapq
class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])

        visited = [[-1 for j in range(m)] for i in range(n)]
        queue = collections.deque()

        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    queue.append((i, j))

        k = [(-1, 0), (0, -1), (1, 0), (0, 1)]

        level = 0
        while queue:
            l = len(queue)
            for i in range(l):
                popped_value = queue.popleft()
                x = popped_value[0]
                y = popped_value[1]

                if visited[x][y] == -1:
                    visited[x][y] = level

                    for l in k:
                        a = x + l[0]
                        b = y + l[1]
                        if self.isValid(a, b, n, m) and grid[a][b] == 0:
                            queue.append((a, b))
            level += 1

        # print(visited)


        visited_heap = [[False for j in range(m)] for i in range(n)]
        k = [(-1, 0), (0, 1), (1, 0), (0, -1)]
        i = 0
        j = 0
        heap = []
        ans = inf
        heapq.heappush(heap, (-1 * visited[i][i], i, j))
        while heap:
            # print(heap, "simli")
            popped_value = heapq.heappop(heap)
            value = popped_value[0] * -1
            i = popped_value[1]
            j = popped_value[2]
            ans = min(ans, value)
            if i == n-1 and j == m-1:
                return ans
            if visited_heap[i][j] == False:
                visited_heap[i][j] = True
                for l in k:
                    a = i + l[0]
                    b = j + l[1]
                    if self.isValid(a, b, n, m):
                        heapq.heappush(heap, (-1 * visited[a][b], a, b))






    def isValid(self, a, b, n, m):
        if a >= 0 and a < n and b >= 0 and b < m:
            return True
        return False






        
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