<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 List

class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        
        n = len(grid)
        q = deque()

        # Initialize the queue with land cells and distance 0
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append((i, j, 0))

        # If there is no land or no water, return -1
        if len(q) == 0 or len(q) == n**2:
            return -1
        
        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        visited = set()
        d = -1

        # BFS to calculate the maximum distance
        while q:
            k = len(q)
            for _ in range(k):
                i, j, d = q.popleft()

                if (i, j) in visited:
                    continue

                visited.add((i, j))

                # Explore neighbors
                for di, dj in directions:
                    ni = i + di
                    nj = j + dj

                    # Add unvisited water cells to the queue
                    if 0 <= ni < n and 0 <= nj < n and grid[ni][nj] == 0:
                        q.append((ni, nj, d + 1))
                        grid[ni][nj] = 1  # Mark the water cell as visited by turning it into land
        
        return d  # Return the maximum distance

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