<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:
        num_rows = len(grid)
        num_cols = len(grid[0])

        def get_neighbours(r,c):
            neighbour_lst = []
            if 0 <= r - 1 < num_rows:
                neighbour_lst.append((r - 1, c))
            if 0 <= r + 1 < num_rows:
                neighbour_lst.append((r + 1, c))
            if 0 <= c - 1 < num_cols:
                neighbour_lst.append((r, c - 1))
            if 0 <= c + 1 < num_cols:
                neighbour_lst.append((r, c + 1))
            return neighbour_lst
        
        def distance_calc(neighbour_row, neighbour_col, current_row, current_col):
            return abs(neighbour_row - current_row) + abs(neighbour_col - current_col)

        q = []
        visited = set()
        for row in range(num_rows):
            for col in range(num_cols):
                if grid[row][col] == 1:
                    q.append((row,col))
                    visited.add((row,col))
                
        curr_max = -1
        if len(q) == 0 or len(q) == num_rows * num_cols:
            return -1

        while len(q) > 0:
            current_size = len(q)

            curr_max += 1

            for _ in range(current_size):
                current_node = q.pop(0)
                current_row = current_node[0]
                current_col = current_node[1]

                for neighbour in get_neighbours(current_row,current_col):
                    neighbour_row = neighbour[0]
                    neighbour_col = neighbour[1]

                    if neighbour not in visited:
                        if grid[neighbour_row][neighbour_col] == 0:
                            distance = distance_calc(neighbour_row, neighbour_col, current_row, current_col)
                            grid[neighbour_row][neighbour_col] = distance

                            q.append(neighbour)
                            visited.add(neighbour)


        return curr_max

548