<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>

# 0-1 BFS
# https://leetcode.com/problems/find-the-safest-path-in-a-grid/solutions/5158996/c-rust-o-n-2-no-dijkstra-pure-bfs-explained-beats-97-44/

# we still use negative distance to be consistent with other versions of solutions in python althougth we won't use heapq in this solution.

import collections
import heapq

# distance is negative because heap in python is min-heap
class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        distance = collections.deque()

        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    distance.append( (i,j) )
                else:
                    grid[i][j] = 1  # mark as undefined
        distance.append( (-1,-1) ) 

        count = -1
        while distance:
            x,y = distance.popleft()

            if x == -1:
                if distance:
                    distance.append( (-1,-1) )
                    count -= 1
                else:
                    break
            else:
                if x+1 < rows and grid[x+1][y] == 1:
                    grid[x+1][y] = count
                    distance.append( (x+1, y) )
                if y+1 < cols and grid[x][y+1] == 1:
                    grid[x][y+1] = count
                    distance.append( (x,y+1) )
                if 0 < x and grid[x-1][y] == 1:
                    grid[x-1][y] = count
                    distance.append( (x-1,y) )
                if 0 < y and grid[x][y-1] == 1:
                    grid[x][y-1] = count
                    distance.append( (x,y-1) )
        
        # use 0-1 BFS to search the minimun of safeness
        directions = [ (1,0), (0,1), (-1,0), (0,-1) ]
        explore = collections.deque()
        min_safeness = grid[0][0]
        explore.append( (min_safeness, 0,0) ) # (safeness, x, y)
        grid[0][0] = 1
        while explore:
            safeness, x, y = explore.popleft()

            min_safeness = max(safeness, min_safeness)

            if x == rows-1 and y == cols-1:
                break

            for dx,dy in directions:
                new_x = x + dx
                new_y = y + dy
                if 0 <= new_x < rows and 0 <= new_y < cols and grid[new_x][new_y] != 1:
                    if grid[new_x][new_y] > min_safeness:
                        explore.append( (grid[new_x][new_y], new_x, new_y) )
                    else:
                        explore.appendleft( (grid[new_x][new_y], new_x, new_y) )
                    grid[new_x][new_y] = 1
        
        return -min_safeness
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