<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:
        # theif if 1
        # empty if 0
        # initially at (0,0)
        # safeness - for any cell to any cell with theif, min(manhattan distance)
        # - the closer, the smaller the number

        # return the max safeness factor of all paths from 0, 0 to corner
        # path where min distance to theif is maximized - return the safeness
        
        # cell closest to the theif dictates

        # from all theif, calculate the manhattan distance using bfs
        # - if find smaller number then replace

        # find path where the min is maximized
        # - if find greater, then keep min <- prefer this path since max of min
        # - if find smaller, then make smaller

        # using bfs to calculate all distance
        # cell - distance
        distanceMap = dict()
        queue = deque()
        ROW = len(grid)
        COL = len(grid[0])
        for r in range(ROW):
            for c in range(COL):
                if grid[r][c] == 1:
                    queue.append((r, c))
                    distanceMap[(r, c)] = 0
        
        DIRS = [[0, 1], [1, 0], [-1, 0], [0, -1]]
        while queue:
            row, col = queue.popleft() # neighbor, theif loc
            cur_dist = distanceMap[(row, col)]
            for rowOff, colOff in DIRS:
                newRow = rowOff + row
                newCol = colOff + col
                if newRow < 0 or newRow >= ROW or newCol < 0 or newCol >= COL:
                    continue
                if (newRow, newCol) not in distanceMap:
                    distanceMap[(newRow, newCol)] = float('inf')
                if distanceMap[(newRow, newCol)] > cur_dist + 1:
                    distanceMap[(newRow, newCol)] = cur_dist + 1
                    queue.append((newRow, newCol))

        # then use dijkstra to find path
        heap = [] # pop out the max # maxheap # min distance, row, col
        visited = set()
        heapq.heappush(heap, ((-distanceMap[(0, 0)], 0, 0)))

        while heap:
            cur_dist, row, col = heapq.heappop(heap)
            cur_dist = -cur_dist
            if (row, col) in visited:
                continue
            visited.add((row, col))

            if row == ROW - 1 and col == COL -1:
                return cur_dist

            # put all the neighbors in the path - update 
            for rowOff, colOff in DIRS:
                newRow = rowOff + row
                newCol = colOff + col
                if newRow < 0 or newRow >= ROW or newCol < 0 or newCol >= COL:
                    continue
       
                new_safeness = min(cur_dist, distanceMap[(newRow, newCol)])
                heapq.heappush(heap, (-new_safeness, newRow, newCol))
        

        return 0

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