<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:
        n = len(grid)
        q = []
        theifs = deque()
        directions = [(1,0),(0,1),(-1,0),(0,-1)]
        originalTheifSet = theifs.copy()
        visited = set()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    theifs.append([i,j, 0])
                    grid[i][j] = 0
                    visited.add((i,j))
        while theifs:
            curr = theifs.popleft()
            currX = curr[0]
            currY = curr[1]
            currSF = curr[2]
            for dx, dy in directions:
                tempX = currX + dx
                tempY = currY + dy
                tempSF = currSF+1
                if tempX in range(n) and tempY in range(n):
                    if (tempX,tempY) not in visited:
                        visited.add((tempX,tempY))
                        theifs.append([tempX,tempY,tempSF])
                        grid[tempX][tempY] = tempSF
        safest = list()
        minSafeFactor = grid[0][0]
        safest.append([-1* minSafeFactor,0,0])
        heapq.heapify(safest)
        visited = set()
        visited.add((0,0))
        while safest:
            curr = heapq.heappop(safest)
            minSafeFactor = min( -1* curr[0], minSafeFactor)
            currX = curr[1]
            currY = curr[2]
            for dx, dy in directions:
                tempX = currX + dx
                tempY = currY + dy
                if tempX in range(n) and tempY in range(n):
                    if (tempX,tempY) not in visited:
                        tempSF = grid[tempX][tempY]
                        heapq.heappush(safest, [-1* tempSF,tempX,tempY])
                        visited.add((tempX,tempY))
                        if tempX == n-1 and tempY == n-1:
                            return min(minSafeFactor, tempSF)
        return minSafeFactor
                        

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