<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:
        queue = deque()
        m,n = len(grid),len(grid[0])
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        max_dist = 0
        # dist = [[0]*m for _ in range(n)]
        
        for i in range(m):
            for j in range(n):
                if grid[i][j] ==1:
                    queue.append((i,j))
                    grid[i][j] = (i,j)
        while queue:
            x,y = queue.popleft()
            for dx, dy in directions:
                new_x , new_y = dx + x, dy + y
                if 0<=new_x<m and 0<=new_y<n and grid[new_x][new_y] ==0:
                    grid[new_x][new_y] = grid[x][y]
                    # dist[new_x][new_y] = dist[x][y] + 1
                    max_dist = max(max_dist,abs(grid[new_x][new_y][0]-new_x)+
                    abs(grid[new_x][new_y][1]-new_y))
                    
                    queue.append((new_x,new_y))
        # print(grid)
        return max_dist if max_dist > 0 else -1





        
17500