<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>

class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:
        queue = deque()
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == 1:
                    queue.append((i, j))

        if len(queue) == len(grid) * len(grid):
            return -1
            
        if queue:
            visited = {}
            count = 0
            while queue:
                length = len(queue)
                for _ in range(length):
                    r, c = queue.popleft()
                    if r >= 0 and r < len(grid) and c >= 0 and c < len(grid[r]) and (r, c) not in visited:
                        visited[(r, c)] = count
                        for dr, dc in [[0, 1], [1, 0], [0, -1], [-1, 0]]:
                            new_r, new_c = r + dr, c + dc
                            if new_r >= 0 and new_r < len(grid) and new_c >= 0 and new_c < len(grid[new_r]) and (new_r, new_c) not in visited and grid[new_r][new_c] == 0:
                                queue.append((r + dr, c + dc))
                count += 1
            return max(visited.values())
        else:
            return -1
534