<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 unionFind:
    def __init__(self, size: int):
        self.root = [i for i in range(size)]
        self.rank = [1 for _ in  range(size)]
        
    def find(self, x: int):
        if self.root[x] == x:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    
    def union(self, x: int, y: int):
        rootX = self.find(x)
        rootY = self.find(y)
        
        if rootX != rootY:
            if self.rank[rootX] < self.rank[rootY]:
                self.root[rootX] = rootY
                self.rank[rootY] += self.rank[rootX]
            else:
                self.root[rootY] = rootX
                self.rank[rootX] += self.rank[rootY]
            return True
        return False
    
    def connected(self, x: int, y: int):
        return self.find(x) == self.find(y)

import heapq
    
class Solution:
    def maximumSafenessFactor(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        
        deltas = [[0,1], [1,0], [-1, 0], [0, -1]]

        def manhattan(p1, p2):
            return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])
                
        def iwb(i, j):
            return 0 <= i < m and 0 <= j < n
        
        def bfs(i, j):
            
            if grid[i][j] == 1:
                return 0
            
            stack = deque([(i,j)])
            seen = set([(i,j)])
            
            while stack:
                ci, cj = stack.popleft()
                
                for delta in deltas:
                    ni = ci + delta[0]
                    nj = cj + delta[1]
                    if iwb(ni, nj):
                        if grid[ni][nj]:
                            return manhattan((i, j), (ni, nj))
                        if (ni, nj) not in seen:
                            stack.append((ni, nj))
                            seen.add((ni, nj))
                            
            return float("inf")

        new_grid = [[float("inf")]*n for _ in range(m)]
        
        stack = deque()
        
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    stack.append(((i,j), (i,j)))
                    new_grid[i][j] = 0
                    
        while stack:
            (ci, cj), (si, sj) = stack.popleft()
            
            for delta in deltas:
                ni = ci + delta[0]
                nj = cj + delta[1]
                if iwb(ni, nj) and manhattan((ni, nj), (si,sj)) < new_grid[ni][nj]:
                    new_grid[ni][nj] = manhattan((ni, nj), (si,sj))
                    stack.append(((ni, nj), (si,sj)))
                    
        
        # import numpy as np 
        # print(np.array(new_grid))
        
        
        pq = []
        
        for i in range(m):
            for j in range(n):
                heapq.heappush(pq, (-new_grid[i][j], (i, j)))
                
        uf = unionFind(m*n)        
        
        safety = float("inf")
        
        while pq:
            mcur_safety, p = heapq.heappop(pq)
            
            mcur_safety = -mcur_safety
            
            safety = min(safety, mcur_safety)   

            for delta in deltas:
                ni = p[0] + delta[0]
                nj = p[1] + delta[1]
                if iwb(ni, nj) and new_grid[ni][nj] >= mcur_safety:
                    uf.union(p[0]*n+p[1], ni*n+nj)
            
            if uf.connected(0, m*n-1):
                break

        return safety
6655