<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>

"""
input: matrix n x n; 1 contains a thief
output: maximum safeness factor from all paths

start at (0,0) -> (n-1, n-1)
can only move to adjacent cells: top, bottom, left, right
safeness factor of a path is minimum manhattan distance from a cell to any thief in the grid
manhattan distance between (a,b) and (x,y) is |a - x| + |b - y|

[1, 0, 0],
[0, 0, 0],
[0, 0, 1]

process manhattan distances from cells with thieves
keep auxillary 2D matrix of same size

[ 0, 1, 2]
[ 1, 2, 3]
[ 2, 3, 0]

[0, 0, 0, 1],
[0, 0, 0, 0],
[0, 0, 0, 0],
[1, 0, 0, 0]

[ 3, 2, 1, 0]
[ 2, 3, 2, 1]
[ 1, 2, 3, 2]
[ 0, 1, 2, 3]

now that we have cells with min safeness factor at each cell, 
there exists a path with safeness factor v if all cells have manhattan distance >= v
test for each safeness factor (0,0) using binary search

[ 0, 1, 2, 3 ]
     l
     r
     m
"""
from collections import deque
import heapq

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        n = len(grid)
        # keep auxillary 2D matrix of same size
        safety = [ [-1]*n for _ in range(n) ]
        adjacent = [(1,0), (0,1), (-1,0), (0,-1)]

        def manhattan_dist(node1, node2):
            return abs(node1[0] - node2[0]) + abs(node1[1] + node2[1])

        def bfs(thieves):
            queue = deque(thieves)

            while queue:
                for _ in range(len(queue)):
                    i, j = queue.popleft()
                    safe_factor = safety[i][j]

                    for move in adjacent:
                        next_i, next_j = i + move[0], j + move[1]

                        if (0 <= next_i < n) and (0 <= next_j < n) and safety[next_i][next_j] == -1:
                            safety[next_i][next_j] = safe_factor + 1
                            queue.append((next_i, next_j))


        # process manhattan distances from cells with thieves
        thieves = []
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    thieves.append((i,j))
                    safety[i][j] = 0

        bfs(thieves)

        # now that we have cells with min safeness factor at each cell, 
        # use Dijkstra to greedily find path with largest safeness factor
        priority_queue = [(-safety[0][0], 0, 0)]
        safety[0][0] = -1

        while priority_queue:
            safety_factor, i, j = heapq.heappop(priority_queue)

            if i == n -1 and j == n - 1:
                return -safety_factor

            for move in adjacent:
                next_i, next_j = i + move[0], j + move[1]
                if (0 <= next_i < n) and (0 <= next_j < n) and safety[next_i][next_j] != -1:
                    heapq.heappush(priority_queue, (max(safety_factor, -safety[next_i][next_j]), next_i, next_j))
                    safety[next_i][next_j] = -1

        return 0
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