<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:

        self.ROWS, self.COLS = len(grid), len(grid[0])

        self.q = deque([])  # (i, j, distance)
        self.grid = grid 

        # for g in self.grid: 
        #     print(g)

        self.visited = set() 

        # put all land in the queue
        for i in range(self.ROWS): 
            for j in range(self.COLS): 
                if self.grid[i][j] == 1: 
                    self.visited.add((i, j))
                    self.q.append((i, j, 0))

        if len(self.q) == 0 or len(self.q) == self.ROWS * self.COLS: 
            return -1 

        self.max_distance = -1

        self.moves = [(-1, 0), (0, -1), (0, 1), (1, 0)]

        # BFS from all the land pieces 
        while self.q: 
            current_i, current_j, current_distance = self.q.popleft() 
            self.max_distance = max(self.max_distance, current_distance)

            for move in self.moves: 
                new_i, new_j = current_i + move[0], current_j + move[1]

                if new_i < 0 or new_j < 0 or new_i >= self.ROWS or new_j >= self.COLS or (new_i, new_j) in self.visited: 
                    continue 
                
                new_distance = abs(new_i - current_i) + abs(new_j - current_j)
                self.visited.add((new_i, new_j))
                self.q.append((new_i, new_j, current_distance + new_distance))

        return self.max_distance                

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