<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:
        rows, cols = len(grid), len(grid[0])
        q = deque()
        distances = [[-1 for _ in range(cols)] for _ in range(rows)]
        visited = set()
        #Intially add to queue and set land distances to 0
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 1:
                    q.append((r,c))
                    visited.add((r,c))
                    distances[r][c] = 0
        
        while q:
            row, col = q.popleft()
            parent_distance = distances[row][col]
            directions = [[-1,0],[1,0],[0,-1],[0,1]]
            for direction in directions:
                n_row = row + direction[0]
                n_col = col + direction[1]

                if (n_row in range(rows) and
                    n_col in range(cols) and
                    (n_row, n_col) not in visited and
                    grid[n_row][n_col] == 0):

                    # print(n_row, n_col)
                    # print(distances[n_row][n_col])

                    child_distance = abs(row - n_row) + abs(col - n_col)
                    distances[n_row][n_col] = parent_distance + child_distance
                    visited.add((n_row, n_col))
                    q.append((n_row, n_col))

        #Find maximum from distances array
        max_distance = float('-inf')
        for r in range(rows):
            for c in range(cols):
                if distances[r][c] > max_distance:
                    max_distance = distances[r][c]
        
        if max_distance <= 0:
            return -1
        
        return max_distance
        
618