<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:
        '''
        Precompute safeness(i,j) of each cell first with BFS
        Time: O(row*cols), Space:O(row*cols)

        Let f(i,j) be the max safeness factor on grid[i][j]
            f(i,j) = min(
                max(
                   f(i-1,j)        # coming from above
                   f(i,j-1)        # coming from below  
                ),
                safeness[row][0]  
            )
            f(0,0) = safeness(0,0)
            f(0,x) = min(f(0,x-1), safeness(0,x))
            f(x,0) = min(f(x-1,0), safeness(x,0))

            Pure Recursion: Time:O(2*(rows+cols)), Space: O(row+cols)
            Topdown Memoization; Time:O(row*cols),  Space:O(row*cols)
            DP tabulation: O(row,cols), Space: O(rows)

        Overall Complexity: 
        Time: O(row*cols), Space:  O(row*cols)  
        '''
        ROWS = len(grid)
        COLS = len(grid[0])

        # Precompute safeness
        safeness = [[-1] * COLS for _ in range(ROWS)]

        def precomputeSafeness():
            q = deque([])
            # BFS starting from thieves position
            for row in range(ROWS):
                for col in range(COLS):
                    if grid[row][col] == 1:
                        safeness[row][col] = 0
                        q.append((row,col,0))
            while q:
                row,col,safeVal = q.popleft()
                for neighRow, neighCol in [(row-1,col),(row+1,col),(row,col-1),(row,col+1)]:
                    if 0<=neighRow<ROWS and 0<=neighCol<COLS:
                        if safeness[neighRow][neighCol] == -1:
                            safeness[neighRow][neighCol] = safeVal+1
                            q.append((neighRow,neighCol,safeVal+1))
        

        def canReachEnd(safenessVal: int) -> bool:
            q = deque([])
            visited = set()
            if safeness[0][0] >= safenessVal:
                q.append((0,0))
                visited.add((0,0))
            while q:
                row,col = q.popleft()
                # Last cell reached
                if row==ROWS-1 and col==COLS-1:
                    return True
                for neighRow, neighCol in [(row-1,col),(row+1,col),(row,col-1),(row,col+1)]:
                    if 0<=neighRow<ROWS and 0<=neighCol<COLS: 
                        if (neighRow,neighCol) not in visited:
                            if safeness[neighRow][neighCol] >= safenessVal:
                                visited.add((neighRow,neighCol))
                                q.append((neighRow,neighCol))
            # unreachable
            return False
                
        precomputeSafeness()
        # Binanry search max safeness value
        start, end = 0, (ROWS+COLS)
        while start<=end:
            mid = start + (end-start)//2
            if canReachEnd(mid):
                start = mid+1
            else:
                end = mid-1
        return end


         

        
        
        

        
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