<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:
        ### Naive
        # try all paths, for each cell in path: calculate safeness factor (min distance to any thief) -> min safeness factor of all cells
        # find max safeness factor of paths
        # dfs
        # backtrack * n * n
        # repeat: safeness factor of a cell: from all thieves, broaden the radius
        # sf of each cell -> find the path with highest sf cells ? dijkstra

        n = len(grid)

        MAX = n * 2
        q = []
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append((i, j))

        def isValid(r, c):
            return (0 <= r < n and 0 <= c < n)

        sf = [[MAX] * n for _ in range(n)]
        DIRECTIONS = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        point = 0
        while q:
            new_q = []
            for i, j in q:
                if sf[i][j] > point:
                    sf[i][j] = point
                    new_q.extend([(i + d[0], j + d[1]) for d in DIRECTIONS if isValid(i + d[0], j + d[1])])
            q = new_q
            point += 1
        
        def hasValidPath(target):
            # bfs
            if sf[0][0] < target:
                return False
            q = deque([(0, 0)])
            seen = { (0, 0) }
            while q:
                r, c = q.popleft()
                if r == c and r == n - 1:
                    return True

                for d in DIRECTIONS:
                    nr, nc = r + d[0], c + d[1]
                    if isValid(nr, nc) and (nr, nc) not in seen and sf[nr][nc] >= target:
                        q.append((nr, nc))
                        seen.add((nr, nc))

            return False

        # binary search [1 -> point), return max valid, bisect_left
        lo, hi = 1, min(sf[0][0], sf[n - 1][n - 1]) + 1
        
        while lo < hi:
            mid = (lo + hi) // 2
            if hasValidPath(mid): # search on right part
                lo = mid + 1
            else: # seach on left part
                hi = mid
        
        return lo - 1

63931