<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:     
        q = deque()
        visit = set()
        res = 0
        temp = len(grid)
        temp2 = len(grid[0])
        def bfs(q):
            nonlocal res
            direction = [(-1,0),(1,0),(0,-1),(0,1)]
            while q:
                flag = 0
                for i in range(len(q)):
                    x,y = q.popleft()
                    for dx,dy in direction:
                        row = x+dx
                        col = y+dy
                        if row in range(len(grid)) and col in range(len(grid[0])) and (row,col) not in visit:
                            q.append((row,col))
                            visit.add((row,col))
                            flag = 1

                if flag==1:
                    res = res+1

        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1:
                    q.append((i,j))
                    visit.add((i,j)) 
        if grid==None or len(visit)==0 or len(visit)== temp*temp2:
            return -1
        bfs(q)
        print(q)
        return res
581