<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:
        n = len(grid)
        dis = [[0 for __ in range(n)] for __ in range(n)]
        res = -1

        dq = deque([])

        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    dq.append((i, j))

        l = 1

        while dq:
            cnt = len(dq)

            for __ in range(cnt):
                x, y = dq.popleft()

                if x and grid[x - 1][y] == 0 and dis[x - 1][y] == 0:
                    dis[x - 1][y] = l
                    res = max(res, dis[x - 1][y])
                    dq.append((x - 1, y))
                if x <= n - 2 and grid[x + 1][y] == 0 and dis[x + 1][y] == 0:
                    dis[x + 1][y] = l
                    res = max(res, dis[x + 1][y])
                    dq.append((x + 1, y))
                if y and grid[x][y - 1] == 0 and dis[x][y - 1] == 0:
                    dis[x][y - 1] = l
                    res = max(res, dis[x][y - 1])
                    dq.append((x, y - 1))
                if y <= n - 2 and grid[x][y + 1] == 0 and dis[x][y + 1] == 0:
                    dis[x][y + 1] = l
                    res = max(res, dis[x][y + 1])
                    dq.append((x, y + 1))

            l += 1
        
        return res
327