<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:
        def aux(cell1, cell2):
            x1,y1=cell1
            x2,y2=cell2
            return abs(x1-x2)+abs(y1-y2)

        n=len(grid)
        dist={}
        q=deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j]==1:
                    q.append((i,j))
                    dist[i,j]=0
        if len(dist) in [0, n*n]:
            return -1
        
        ans=0
        while q:
            x,y=q.pop()
            if grid[x][y]==0:
                ans=max(ans, dist[x,y])
            for u,v in [(x+1,y), (x-1,y), (x,y+1), (x,y-1)]:
                if min(u,v)>=0 and max(u,v)<n and (u,v) not in dist and grid[u][v]==0:
                    dist[u,v]=dist[x,y]+1
                    q.appendleft((u,v))

        return ans
        
571