<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)
        queue = collections.deque()
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    queue.append((i,j,0))
    
        seen = set()
        max_man_distance = 0

        while queue:
            i,j,level = queue.popleft()
            if (i,j) not in seen:
                seen.add((i,j))
                grid[i][j] = level if grid[i][j] == 0 else float('-inf')
                max_man_distance = max(max_man_distance, level)
                for inc_i, inc_j in {(0,1), (1,0), (0,-1), (-1,0)}:
                    new_i, new_j = i + inc_i, j + inc_j
                    if 0 <= new_i < n and 0 <= new_j < n and grid[new_i][new_j] != 1:
                        queue.append((new_i, new_j, level + 1))
        

        lo, hi = 0, max_man_distance
        res = 0
        while lo <= hi:
            mid = (lo + hi) // 2
            stack = [(0,0)]
            seen = set()
            while stack:
                i,j = stack.pop()
                if (i,j) not in seen and 0 <= i < n and 0 <= j < n and grid[i][j] >= mid:
                    if (i,j) == (n-1, n-1):
                        res = mid
                        lo = mid + 1
                        break
                    seen.add((i,j))
                    for inc_i, inc_j in {(0,1), (1,0), (0,-1), (-1,0)}:
                        new_i, new_j = i + inc_i, j + inc_j
                        stack.append((new_i, new_j))
            if not stack: 
                hi = mid - 1
        return res

# perform BFS from all the thieves to determine minimum manhattan distance from all thieves
# iterate through grid
    # if thief: add to queue with level 0
#while queue:
    # pop from queue
    # if not seen
        # add to seen set
        # change grid value to level (if thief, change it to "thief")
        # update max minimum manhattan distance we've seen
        # iterate through neighbors
            #check if valid and not equal to 1
                # add to queue with level + 1

# lo = 1, hi = max min manhattan distance
#while lo <= hi:
    # mid = (lo + hi) // 2
    #initialize stack, seen set
    # perform DFS, but only ging to elements with values less than or equal to mid
    # while stack:
        # pop from stack
        #if equals bottom right:
            # lo = mid
            # break
        # if not seen
            # add to seen set
            #iterate through neighbors
                # check if valid and less than or equal to mid:
                    # add to stack
    # if not stack: hi = mid - 1


# safeness factor of a path: minimum manhattan distance from any cell in the path to a thief
# return maximum safeness factor of all possible paths

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