<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:
        numRows, numCols = len(grid), len(grid[0])
        
        neighbor_dir = [(0, 1), (0, -1), (-1, 0), (1, 0)]
        
        # find minDist from theives
        multi_queue = deque()
        # go through each element in grid, and put the thieves onto multi queues
        for r in range(numRows):
            for c in range(numCols):
                if grid[r][c] == 1:
                    multi_queue.append((r, c))
                    grid[r][c] = 0
                else:
                    grid[r][c] = -1

        # calculate safeness factors for each cell
        while multi_queue:
            sz = len(multi_queue)
            while sz > 0:
                r, c = multi_queue.popleft()
                for rd, cd in neighbor_dir:
                    nr, nc = r + rd, c + cd
                    if 0 <= nr < numRows and 0 <= nc < numCols and grid[nr][nc] == -1:
                        grid[nr][nc] = grid[r][c] + 1
                        multi_queue.append((nr, nc))
                sz -= 1

        maxHeap = [(-grid[0][0], (0, 0), (0, 0))]
        shortest = {}

        while maxHeap:
            currDist, currCoords, prevCoords = heapq.heappop(maxHeap)
            r, c = currCoords
            pr, pc = prevCoords
            if (r, c) in shortest:
                continue
            shortest[(r,c)] = (-currDist, (pr, pc)) 
            # go to each neighbor
            for rd, cd in neighbor_dir:
                nr, nc = r + rd, c + cd
                if 0 <= nr < numRows and 0 <= nc < numCols and (nr, nc) not in shortest:
                    heapq.heappush(maxHeap, (-grid[nr][nc], (nr, nc), (r, c)))
        
        # for key, val in shortest.items():
        #     print(key, val)
        # backtrace the path
        result = float('inf')
        currCoords = (numRows - 1, numCols - 1)
        while True:
            dist, prevCoords = shortest[(currCoords[0], currCoords[1])]
            result = min(result, dist)
            if result == 0:
                return 0
            if currCoords == (0, 0):
                return result
            currCoords = prevCoords




        
        
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