<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>

from collections import deque
from itertools import product, starmap
class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        cols = len(grid[0])

        is_valid = lambda rc: (lambda r,c: 0 <= r < rows and 0 <= c < cols and grid[r][c] == 0)(*rc)
        directions = [[0,1], [0,-1], [1,0], [-1,0]]
        get_valid = lambda r,c: filter(is_valid, starmap(lambda a, b: [a[0] + b[0], a[1]+b[1]], product(directions, [[r,c]])))

        lands = list(filter(lambda x: grid[x[0]][x[1]] == 1, product(range(rows), range(cols))))
        if len(lands) == rows*cols:
            return -1
        q = deque()
        q.extend(lands)
        dist = -1
        while q:
            qlen = len(q)
            dist += 1
            for _ in range(qlen):
                 r,c = q.popleft()
                 for row, col in get_valid(r,c):
                    grid[row][col] = 1
                    q.append((row,col))
            
        return dist

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