<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:
        rows, cols = len(grid), len(grid[0])
        queue = deque()
        directions = [(0, -1), (0, 1), (1, 0), (-1, 0)]

        # Initialize the queue with all land cells (1s)
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    queue.append((i, j))

        # If there are no water cells or all are water, return -1
        if len(queue) == 0 or len(queue) == rows * cols:
            return -1

        max_distance = -1

        # BFS from all land cells
        while queue:
            x, y = queue.popleft()

            for d in directions:
                nx, ny = x + d[0], y + d[1]

                # Check if the new position is within bounds and is a water cell
                if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] == 0:
                    # Update the max distance for the water cell
                    grid[nx][ny] = grid[x][y] + 1  # Mark visited with the distance
                    max_distance = max(max_distance, grid[nx][ny] - 1)  # Store the max distance
                    queue.append((nx, ny))

        return max_distance
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