<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
import heapq

class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        n = len(grid)
        multi_src_queue = deque()
        # Mark thieves as 0 and empty cells as -1, and push thieves to the queue
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    multi_src_queue.append((i, j))
                    grid[i][j] = 0
                else:
                    grid[i][j] = -1
        # Calculate safeness factor for each cell using BFS
        while multi_src_queue:
            level_len = len(multi_src_queue)
            for _ in range(level_len):
                curr = multi_src_queue.popleft()
                safeness = grid[curr[0]][curr[1]]
                for d in dirs:
                    nxt_i, nxt_j = curr[0] + d[0], curr[1] + d[1]
                    if self.valid_loc(nxt_i, nxt_j, n) and grid[nxt_i][nxt_j] == -1:
                        multi_src_queue.append((nxt_i, nxt_j))
                        grid[nxt_i][nxt_j] = safeness + 1
        res = self.get_max_sf_from_binary_search(grid, dirs)
        # res = self.get_max_sf_from_greedy(grid, dirs)
        return res

    def get_max_sf_from_greedy(self, grid, dirs):
        # Priority queue to prioritize cells with higher safeness factor
        n = len(grid)
        pq = [(-grid[0][0], 0 , 0)]
        grid[0][0] = -1
        while pq:
            max_sf, i, j = heapq.heappop(pq)
            max_sf = -max_sf
            if i == n - 1 and j == n - 1:
                return max_sf
            
            for d in dirs:
                nxt_i, nxt_j = i + d[0], j + d[1]
                if self.valid_loc(nxt_i, nxt_j, n) and grid[nxt_i][nxt_j] != -1:
                    # Each path's safeness factor will be the minimum one to thieves
                    nxt_sf = min(max_sf, grid[nxt_i][nxt_j])
                    heapq.heappush(pq, (-nxt_sf, nxt_i, nxt_j))
                    grid[nxt_i][nxt_j] = -1
        return -1

    def valid_loc(self, i, j, n):
        return 0 <= i < n and 0 <= j < n

    def get_max_sf_from_binary_search(self, grid, dirs):
        n = len(grid)
        
        def is_valid(min_sf):
            # Check if a path exists with given minimum safeness value
            if grid[0][0] < min_sf or grid[-1][-1] < min_sf:
                return False
            q = deque([(0, 0)])
            seen = set((0, 0))
            while q:
                i, j = q.popleft()
                if i == n - 1 and j == n - 1:
                    return True
                for d in dirs:
                    nxt_i, nxt_j = i + d[0], j + d[1]
                    if self.valid_loc(nxt_i, nxt_j, n) and (nxt_i, nxt_j) not in seen and grid[nxt_i][nxt_j] >= min_sf:
                        q.append((nxt_i, nxt_j))
                        seen.add((nxt_i, nxt_j))
            return False

        start, end = 0, 0
        for i in range(n):
            for j in range(n):
                end = max(end, grid[i][j])

        while start + 1 < end:
            mid = (start + end) // 2
            if is_valid(mid):
                start = mid
            else:
                end = mid
        if is_valid(end): return end
        return start if is_valid(start) else -1





5590