<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 UF:
    def __init__(self, n):
        self.parent = list(range(n*n))
        self.cnt = [1]*(n*n)
    
    def find(self, x):
        if self.parent[x] == x:
            return x
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px == py:
            return False

        if self.cnt[px] > self.cnt[py]:
            self.parent[py] = px
            self.cnt[px] += self.cnt[py]
        else:
            self.parent[px] = py
            self.cnt[py] += self.cnt[px]
        return True


class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        # grid[r][c] = 1 means there is a thief
        # 0,0 ). min manhattane distance from any cell in the path to any thief in the grid

        # max safeness factr of all paths
        # min manhatten distance from any cell to any thief grid
        # run bfs from all thieve cells to mark safenes factor, it's going to be min (cur+1, path)
        n = len(grid)

        # 1 thief, 0 no thief
        # safeness factor is the minimum manhattanc distance from any path to any thief
        # max factor to the path leading to (n-1, n-1
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return 0

        # we can do dfs from each at distance
        # n*n

        # min among all max
        # for each ones mark all distance to 0
        # do dfs picking max distances

        # let's start with bfs
        dp = [[-1]*n for _ in range(n)]

        queue = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    queue.append((i, j))
                    dp[i][j] = 0

        while queue:
            i, j = queue.popleft()
            for di, dj in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
                if not (0<= i + di < n and 0 <= j + dj < n):
                    continue
                if dp[i+di][j+dj] >= 0:
                    continue
                dp[i+di][j+dj] = dp[i][j] + 1
                queue.append((i+di, j+dj))

        
        # we wil bin search on the minimum possible distance
        # all distances must be >= mid

        def can_connect(dist):
            uf = UF(n)

            for i in range(n):
                for j in range(n):                   
                    if dp[i][j] >= dist:
                        if i < n-1 and dp[i+1][j] >= dist:
                            uf.union(i*n+j, (i+1)*n + j)
                        if j < n-1 and dp[i][j+1] >= dist:
                            uf.union(i*n+j, i*n + j+1)

            return uf.find(0) == uf.find(n*n-1)

        low, high = 0, 2*n

        while low <= high:
            mid = (low + high) >> 1
            if can_connect(mid):
                low = mid + 1
            else:
                high = mid - 1

        return high

        

5590