<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 find_land(self, grid):
        n = len(grid)
        result = []
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    result.append((i, j))
        return result
    def maxDistance(self, grid: List[List[int]]) -> int:
        n = len(grid)
        max_distance = [[float('inf')]*n for _ in range(n)]
        direction = [(-1, 0), (1, 0), (0, 1), (0, -1)]
        land = self.find_land(grid)
        def bfs(index):
            deque = collections.deque()
            visited = [[False]*n for _ in range(n)]
            x, y = index
            deque.append((x, y ,0))
            visited[x][y] = True

            while deque:
                x, y, dist = deque.popleft() 

                for dx, dy in direction:
                    nx = x + dx
                    ny = y + dy
                    if nx >= 0 and nx < n and ny >= 0 and ny < n and visited[nx][ny] == False:
                        if grid[nx][ny] == 0:
                            visited[nx][ny] = True
                            max_distance[nx][ny] = min(max_distance[nx][ny], dist + 1)
                            deque.append((nx, ny, dist + 1))
                        else:
                            visited[nx][ny] = True

        for i in range(len(land)):
            bfs(land[i])
        result = 0
        #print(max_distance)
        for i in range(n):
            for j in range(n):
                if max_distance[i][j] > result and max_distance[i][j]<=2*n:
                    result = max_distance[i][j]
        if result > 0:
            return result
        else:
            return - 1

        
17400