<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:
    dirs = [(0,1), (1,0), (0,-1), (-1, 0)]

    def is_valid(self, grid, x, y):
        m, n = len(grid), len(grid[0])
        return 0 <= x < m and 0 <= y < n

    def calculateSafeness(self, grid):
        multi_soure_deque = deque()
        m, n = len(grid), len(grid[0])
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    multi_soure_deque.append((i,j))
                else:
                    grid[i][j] = -1
        
        while multi_soure_deque:
            i, j = multi_soure_deque.popleft()

            for d in self.dirs:
                x, y = i+d[0], j+d[1]

                if not self.is_valid(grid, x, y) or grid[x][y] != -1:
                    continue
                
                grid[x][y] = grid[i][j] + 1
                multi_soure_deque.append((x, y))

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # Note: we are doing dijsitra but we do not need the dist array here
        # as the dist array contains the min val to reach that node
        # which keeps on changing as new paths reaching that node are traversed
        # CHECK: minimim effort leetcode problem -- there the dist array
        # represents the max abs height diff of a path and if we get a path
        # passing through (i,j) which has less max abs height diff, we update the
        # dist(i,j) for the same
        # here, we do not need to maintain any such thing as the values are fixed
        # these values denote the distance to the nearest theif which is not changing

        m, n = len(grid), len(grid[0])

        self.calculateSafeness(grid)

        # max heap -- need max distance from theif everytime
        heap = [(-grid[0][0], 0, 0)]
        visited = set()
        visited.add((0, 0))

        while heap:
            dist, i, j = heapq.heappop(heap)

            if i==m-1 and j==n-1:
                return -dist

            for d in self.dirs:
                x, y = i+d[0], j+d[1]

                if not self.is_valid(grid, x, y) or (x, y) in visited:
                    continue
                
                visited.add((x, y))
                next_dist = min(-dist, grid[x][y])
                heapq.heappush(heap, (-next_dist, x, y))
        
        return -1

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