<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>

from queue import deque
class Solution:
    def maxDistance(self, grid: List[List[int]]) -> int:

        max_distance = 0

        def bfs(q):
            n = len(grid)
            vis = set()
            direc= [(0,1),(1,0),(0,-1),(-1,0)]

            max_distance = 0

            while(q):
                r,c,count = q.popleft()
                max_distance = max(max_distance,count)
                
                for k in range(4):
                    nr  = direc[k][0] + r
                    nc  = direc[k][1] + c
                    if 0 <= nr < n and 0 <= nc  <n and (nr,nc) not in vis and grid[nr][nc] == 0:
                        vis.add((nr, nc))
                        q.append((nr,nc,count + 1))
            return max_distance
        
        n = len(grid)
        q  = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    q.append((i,j,0))
   
        if not q or len(q) == n * n:
            return -1
        
        return bfs(q)




        

18400