<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):
        n, cans = len(grid), []
        for y, row in enumerate(grid):
            for x, val in enumerate(row):
                if val:
                    cans.append((x, y))

        if len(cans) == 0 or len(cans) == n*n:
            return -1
        if len(cans) == n*n-1:
            return 1

        ctr = 0
        while cans:
            _cans = set()
            for x, y in cans:
                if x > 0 and not grid[y][x-1]:
                    _cans.add((x-1, y))
                    grid[y][x-1] = 1
                if x+1 < n and not grid[y][x+1]:
                    _cans.add((x+1, y))
                    grid[y][x+1] = 1
                if y > 0 and not grid[y-1][x]:
                    _cans.add((x, y-1))
                    grid[y-1][x] = 1
                if y+1 < n and not grid[y+1][x]:
                    _cans.add((x, y+1))
                    grid[y+1][x] = 1
            cans = _cans
            ctr += 1
        return ctr-1
318