<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:
        # First BFS
        n = len(grid)
        distance = [[1000 for c in range(n)] for r in range(n)]
        thieves = []
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    distance[i][j] = 0
                    thieves.append((i,j))

        queue = deque([(i, j, 0) for i, j in thieves])
        while queue:
            i, j, d = queue.popleft()
            for p, q in getNeighbors(i,j,n,n):
                if distance[p][q] == 1000:
                    distance[p][q] = min(d+1, distance[p][q])
                    queue.append((p,q,d+1))

        # # 1. Then backtracking
        # safeness = 0
        # visited = [[0 for c in range(n)] for r in range(n)]
        # return backtracking(0, 0, distance[0][0], visited, distance)

        # # 2. Then binary search
        # if not meet_safeness(distance, 1):
        #     return 0

        # left = 1
        # right = n
        # while left < (right - 1):
        #     mid = left + (right-left) // 2
        #     if meet_safeness(distance, mid):
        #         left = mid
        #     else:
        #         right = mid
        # return left

        # # 2. Then union-find
        byDist = defaultdict(list)
        for i in range(n):
            for j in range(n):
                byDist[distance[i][j]].append((i,j))
        
        uf = UnionFind(n)
        # print(uf.find(0,0))
        # print(uf.union(0,1,1,0))

        for v in sorted(byDist.keys(), reverse=True):
            for i, j in byDist[v]:
                for p, q in getNeighbors(i,j,n,n):
                    if distance[p][q] >= v:
                        # print(v, (i,j),(p,q))
                        uf.union(i,j,p,q)
            
            # print(v, byDist[v], uf.find(0,0), uf.find(n-1,n-1), uf.nodes, distance)

            if uf.find(0,0) == uf.find(n-1,n-1):
                return v
        return 0
    
class UnionFind:
    def __init__(self, n):
        self.nodes = [[(r, c) for c in range(n)] for r in range(n)]
    
    def find(self, r, c):
        # with path compression
        p, q = self.nodes[r][c]
        if (p,q) == (r,c):
            return (r,c)
        i,j = self.find(p,q)
        self.nodes[r][c]=(i,j)
        return (i,j)
    
    def union(self, r1, c1, r2, c2):
        i1, j1 = self.find(r1, c1)
        i2, j2 = self.find(r2, c2)
        if (i1, j1) < (i2, j2):
            self.nodes[i2][j2] = (i1,j1)
        else:
            self.nodes[i1][j1] = (i2,j2)


def meet_safeness(distance, safeness):
    if distance[0][0] < safeness:
        return False

    n = len(distance)
    visited = [[0 for c in range(n)] for r in range(n)]

    queue = deque([(0,0)])
    visited[0][0] = 1

    while queue:

        i, j = queue.popleft()
        if i == (n-1) and j == (n-1):
            return True
        
        for p, q in getNeighbors(i,j,n,n):
            if distance[p][q] >= safeness and not visited[p][q]:
                queue.append((p,q))
                visited[p][q] = 1

    return False


def backtracking(r, c, s, visited, distance):
    s = min(s, distance[r][c])
    if s == 0:
        return 0

    n = len(visited)
    if r == (n-1) and c == (n-1):
        return s

    visited[r][c] = 1

    safenest = 0
    for p, q in getNeighbors(r,c,n,n):
        if not visited[p][q]:
            safe = backtracking(p, q, s, visited, distance)
            if safe > safenest:
                safenest = safe
    visited[r][c] = 0
    return safenest



def getNeighbors(r, c, m, n):
    if c < (n - 1):
        yield (r, c + 1)
    if c > 0:
        yield (r, c - 1)
    if r < (m - 1):
        yield (r + 1, c)
    if r > 0:
        yield (r - 1, c)

4658