<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 typing import Optional


class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        land_tiles = []
        for y, row in enumerate(grid):
            for x, v in enumerate(row):
                if v == 1:
                    land_tiles.append((x, y))

        best_distance = -1
        visited = set(land_tiles)
        q = deque([(x, y, 0) for x, y in land_tiles])
        while q:
            tx, ty, depth = q.popleft()
            v = grid[ty][tx]
            if v == 0 and depth > best_distance:
                best_distance = depth
            if tx != 0 and (tx - 1, ty) not in visited:
                visited.add((tx - 1, ty))
                q.append((tx - 1, ty, depth + 1))
            if tx != len(grid[0]) - 1 and (tx + 1, ty) not in visited:
                visited.add((tx + 1, ty))
                q.append((tx + 1, ty, depth + 1))
            if ty != 0 and (tx, ty - 1) not in visited:
                visited.add((tx, ty - 1))
                q.append((tx, ty - 1, depth + 1))
            if ty != len(grid) - 1 and (tx, ty + 1) not in visited:
                visited.add((tx, ty + 1))
                q.append((tx, ty + 1, depth + 1))
        return best_distance

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