<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

WATER = 0
LAND = 1

class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        n = len(grid)
        water = []
        land = []
        for x, row in enumerate(grid):
            for y, val in enumerate(row):
                if val == WATER:
                    water.append((x, y))
                else:
                    land.append((x, y))
        if not water or not land:
            return -1

        visited = set()
        q = deque(land)
        last_water = None
        while q:
            x, y = q.popleft()
            if (x, y) in visited:
                continue
            visited.add((x, y))
            if grid[x][y] == WATER:
                last_water = (x, y)
            cands = [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]
            for cx, cy in cands:
                if 0 <= cx < n and 0 <= cy < n and (cx, cy) not in visited:
                    q.append((cx, cy))

        lx, ly = last_water
        return min(abs(x - lx) + abs(y - ly) for x, y in land)

        


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