<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:
        #cur stores (x, y) coordinates
        def bfs(distances, stack):
            visited = set()
            curDist = 0
            neighbors = [(1, 0), (0, 1), (-1, 0), (0, -1)]
            while stack:
                numNodes = len(stack)
                for i in range(numNodes):
                    cur = stack.popleft()
                    x, y = cur[0], cur[1]

                    if (x, y) in visited: continue
                    distances[x][y] = curDist
                    visited.add(cur)

                    for dx, dy in neighbors:
                        if x+dx in range(0, len(grid)) and y+dy in range(0, len(grid)) and (x+dx, y+dy) not in visited:
                            stack.append((x+dx, y+dy))
                curDist += 1

        distances = [[0 for i in range(len(grid))] for j in range(len(grid))]
        thiefs = deque()
        for i in range(len(grid)):
            for j in range(len(grid)):
                if grid[i][j] == 1:
                    thiefs.append((i, j))
        print(thiefs)
        bfs(distances, thiefs)
        print(distances, thiefs)

        #Perform dijkstras but take min(node value, next node value) to update nodes
        pq = [(-distances[0][0], 0, 0)]
        visited = set()
        neighbors = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        while pq:
            cur = heapq.heappop(pq)
            dist = cur[0]
            x, y = cur[1], cur[2]
            if (x, y) in visited: continue
            visited.add((x, y))
            if x == len(grid)-1 and y == len(grid)-1:
                return -dist
            for dx, dy in neighbors:
                if x+dx in range(len(grid)) and y+dy in range(len(grid)) and (x+dx, y+dy) not in visited:
                    heapq.heappush(pq, (max(dist, -distances[x+dx][y+dy]), x+dx, y+dy))

7187