<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 collections import deque
class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        def compute_distance_to_thieves():
            # use bfs to find the distance between thieves to any cells
            q = deque()
            dist = [[float("inf")] * cols for _ in range(rows)]
            
            # Add all thief positions to the queue with initial distance 0
            for r in range(rows):
                for c in range(cols):
                    if grid[r][c] == 1:
                        q.append((r, c, 0))
                        dist[r][c] = 0  # Thief cells have a distance of 0
            
            # BFS to find the shortest distance from any thief
            while q:
                curr_r, curr_c, cost = q.popleft()
                
                # Explore adjacent cells
                for dy, dx in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                    nr, nc = curr_r + dy, curr_c + dx
                    n_cost = cost + 1
                    
                    # Check boundaries and if the cell hasn't been visited yet
                    if 0 <= nr < rows and 0 <= nc < cols and dist[nr][nc] == float("inf"):
                        dist[nr][nc] = n_cost  # Set the distance to this cell
                        q.append((nr, nc, n_cost))  # Add to the queue with updated cost
            
            return dist

          
            # Perform BFS to compute the distance to the closest thief for all cells
            while q:
                curr_r, curr_c, cost = q.popleft()
                for dy, dx in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                    nr, nc = curr_r + dy, curr_c + dx
                    n_cost = cost + 1
                    if 0 <= nr < rows and 0 <= nc < cols and dist[nr][nc] == float("inf"):
                        dist[nr][nc] = n_cost
                        q.append((nr, nc, n_cost))
            return dist
        def can_reach_safely(k, rows, cols, dist):
            # returns true if we can reach every where by maximum k 
            visited = set([(0, 0)])
            q = deque([(0, 0)])
            if dist[0][0] < k:
                return False
            
            while q:
                curr = q.popleft()
                if curr == (rows - 1, cols - 1):
                    return True
                for dy, dx in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                    ny, nx = curr[0] + dy, curr[1] + dx
                    if (ny, nx) not in visited and 0 <= ny < rows and 0 <= nx < cols and dist[ny][nx] >= k:
                        visited.add((ny, nx))
                        q.append((ny, nx))
            return False
        
        dist = compute_distance_to_thieves()
        l, r = 0, rows + cols 
        while l <= r:
            m = l + (r - l) // 2
            if can_reach_safely(m, rows, cols, dist):
                l = m + 1 
            else:
                r = m - 1 
        return r
                        
                
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