<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:
        r,c=len(grid),len(grid[0])
        dp=[[float('inf') for _ in range(c)] for _ in range(r)]
        for i in range(r):
            for j in range(c):
                if grid[i][j]==1:
                    dp[i][j]=0
        print(dp)
        for i in range(r):
            for j in range(c):
                up,left=float('inf'),float('inf')
                if i-1>=0:
                    up=dp[i-1][j]
                if j-1>=0:
                    left=dp[i][j-1]
                print(up,left)
                dp[i][j]=min(dp[i][j],min(up,left)+1)
        print(dp)
        for i in range(r-1,-1,-1):
            for j in range(c-1,-1,-1):
                up,left=float('inf'),float('inf')
                if i+1<r:
                    up=dp[i+1][j]
                if j+1<c:
                    left=dp[i][j+1]
                dp[i][j]=min(dp[i][j],min(up,left)+1)
        m=max([max(i) for i in dp])
        return m if m!=0 and m!=float('inf') else -1
17800