<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 find_distance_matrix(self, grid):
        m, n = len(grid), len(grid[0])
        distances = [
            [float('inf') for _ in range(n)] for _ in range(m)
        ]
        heap = []
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    distances[i][j] = 0
                    heapq.heappush(heap, (0, i, j))
                    
        
        dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        while heap:
            dist, i, j = heapq.heappop(heap)
            if dist > distances[i][j]:
                continue
            for di, dj in dirs:
                if (
                    0 <= i + di < m and
                    0 <= j + dj < n and
                    distances[i + di][j + dj] >  dist + 1
                ):
                    distances[i + di][j + dj] = dist + 1
                    heapq.heappush(heap, (dist + 1, i + di, j + dj))
        
        return distances

    def is_possible(self, distances, threshold):
        m, n = len(distances), len(distances[0])
        target = (m - 1, n - 1)
        visited = [
            [False for _ in range(n)] for _ in range(m)
        ]
        visited[0][0] = True
        considering = [(0, 0)]
        dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        while considering:
            i, j = considering.pop()
            if (i, j) == target:
                return True
            for di, dj in dirs:
                if (
                    0 <= i + di < m and
                    0 <= j + dj < n and
                    not visited[i + di][j + dj] and
                    distances[i + di][j + dj] >= threshold
                ):
                    visited[i + di][j + dj] = True
                    considering.append((i + di, j + dj))
        return False

    def find_maximum_safeness_factor(self, distances):
        low = 0
        high = min(distances[0][-0], distances[-1][-1])
        ans = None
        while low <= high:
            mid = low + ((high - low) >> 1)
            if self.is_possible(distances, mid):
                ans = mid
                low = mid + 1
            else:
                high = mid - 1
        return ans

    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        distances = self.find_distance_matrix(grid)
        maximum_safeness_factor = self.find_maximum_safeness_factor(distances)
        return maximum_safeness_factor if maximum_safeness_factor is not None else -1
        
        




        
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