<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>

from typing import List
from collections import deque
import heapq

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # BFS to find distance of all cells from all thieves

        queue = deque([])
        nrow, ncol = len(grid), len(grid[0])
        distance_grid = [[float("inf")]*len(grid[0]) for _ in range(len(grid))] # minimum manhattan distance from any thief for a cell
        directions = [(-1,0), (1,0) , (0,-1), (0,1)]
        for row in range(len(grid)):
            for col in range(len(grid[0])):
                if grid[row][col] == 1:
                    queue.append((row,col))
                    distance_grid[row][col] = 0 

        while queue: 
            r,c = queue.popleft()
            for di in directions:
                nr,nc = r + di[0], c + di[1]
                if 0 <= nr< nrow and 0 <= nc < ncol and distance_grid[nr][nc] > distance_grid[r][c] + 1:
                    distance_grid[nr][nc]= min(distance_grid[nr][nc], distance_grid[r][c]+1)
                    queue.append((nr,nc))

        # now ewe need todo modified dijkstra 
        #distance_grid represents max saefty for each cell

        pq = [ (-distance_grid[0][0], 0 ,0)]
        heapq.heapify(pq)
        
        safety_grid = [[float("-inf")]* ncol for _ in range(nrow)]  #maximum safety to reach this cell in this array
        safety_grid[0][0] = distance_grid[0][0]

        while pq:
            curr_distance, row,col = heapq.heappop(pq)
            # curr_distace is safety (min manhatan dis) for this celll
            curr_distance *= -1


            # we could encounter the target anytime
            if row == nrow-1 and col == ncol-1:
                return curr_distance

            for di in directions:
                nr,nc = row + di[0], col + di[1]
                if 0 <= nr< nrow and 0 <= nc < ncol:
                    dis = min(curr_distance ,distance_grid[nr][nc])
                    if dis > safety_grid[nr][nc]:
                        safety_grid[nr][nc] = dis
                        heapq.heappush(pq, (-dis, nr, nc))

        return 0
                    

3593