<p>Given an <code>n x n</code> <code>grid</code>&nbsp;containing only values <code>0</code> and <code>1</code>, where&nbsp;<code>0</code> represents water&nbsp;and <code>1</code> represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.&nbsp;If no land or water exists in the grid, return <code>-1</code>.</p>

<p>The distance used in this problem is the Manhattan distance:&nbsp;the distance between two cells <code>(x0, y0)</code> and <code>(x1, y1)</code> is <code>|x0 - x1| + |y0 - y1|</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex1.JPG" style="width: 185px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,1],[0,0,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The cell (1, 1) is as far as possible from all the land with distance 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex2.JPG" style="width: 184px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The cell (2, 2) is as far as possible from all the land with distance 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>grid[i][j]</code>&nbsp;is <code>0</code> or <code>1</code></li>
</ul>

class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        N = len(grid)
        seen = set()

        # multi-start BFS
        q = [] #(r,c, d)
        for r in range(N):
            for c in range(N):
                if grid[r][c] == 1:
                    q.append((r,c,0))
                    seen.add((r,c))
        if len(q) == N*N or len(q) == 0:
            return -1
        
        while q:
            next_q = []
            for r,c,d in q:
                if r+1 < N and (r+1,c) not in seen:
                    next_q.append((r+1,c,d+1))
                    seen.add((r+1,c))
                if r-1 >= 0 and (r-1,c) not in seen:
                    next_q.append((r-1,c,d+1))
                    seen.add((r-1,c))
                if c+1 < N and (r,c+1) not in seen:
                    next_q.append((r,c+1,d+1))
                    seen.add((r,c+1))
                if c-1 >= 0 and (r,c-1) not in seen:
                    next_q.append((r,c-1,d+1))
                    seen.add((r,c-1))

            q = next_q
        return d
18700