<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 get_neighbors(self, x, y):
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
    
        # Generate not visited orthogonal neighbors
        neighbors = [
            (x + dx, y + dy) 
            for dx, dy in directions
            if 0 <= x+dx < self.n and 0 <= y+dy < self.n and not self.visited[x+dx][y+dy]
        ]
        
        return neighbors


    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # maybe first collect all theives
        # n x n grid
        n = len(grid)
        self.n = n
        
        # if thief is at start/dest, then we're screwed
        if grid[0][0] or grid[-1][-1]:
            return 0        

        # keep track of the safeness of each square in the grid
        # in an n x n grid, the maximal safeness is n-1 + n-1 (i.e thief is in the opposite corner)
        safeness = [[(n-1)*2 for _ in range(n)] for _ in range(n)]
        q = deque()
        for i in range(n):
            for j in range(n):
                # safety of a thief square is 0
                if grid[i][j]:
                    safeness[i][j] = 0                    
                    q.append((i,j))
        
        # now that we've the positions of the thieves (q), we know their neighbors are safety = 1...
        # and neighbors' neighbors are safety + 2, and so on...
        # keep track of visited nodes while we bfs
        self.visited = [[False] * n for _ in range(n)]
        while q:
            x, y = q.popleft()
            # get_neighbors returns unvisited neighbors            
            for i, j in self.get_neighbors(x, y):
                # overwrite safeness if we have a neighbor whose + 1 is less than our current safeness
                if safeness[i][j] > safeness[x][y] + 1:
                    safeness[i][j] = safeness[x][y] + 1
                    q.append((i,j))
        
        # now starting from 0,0, we make our way to 1,1 by always choosing safer square to travel to
        best = safeness[0][0]
        best_pointer = best
        # k,v = int, {} = candidates for next step in path w/ safety 'k'
        self.candidates = defaultdict(set)
        self.candidates[best].add((0,0))
        while best_pointer:
            while self.candidates[best_pointer]:
                i, j = self.candidates[best_pointer].pop()
                self.visited[i][j] = True
                best = min(best, safeness[i][j])
                
                # if we've reached the end
                if (i, j) == (n-1, n-1):
                    return best
                
                # get unvisited neighbors and add them to the priority queue
                for x, y in self.get_neighbors(i,j):
                    self.candidates[safeness[x][y]].add((x,y))
                    # increase our "priority" pointer if we find something "safer" than what we're currently on
                    if safeness[x][y] > best_pointer:
                        best_pointer = safeness[x][y]
                    self.visited[x][y] = True
            best_pointer -= 1

        return 0
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