<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:
        n = len(grid)
        # perform a multisource bfs on all 1s in the grid
        # use a q to store all the initial 1s r, c
        q = deque()
        for r in range(n):
            for c in range(n):
                if grid[r][c] == 1:
                    q.append([r, c])
        
        # define directions for bfs
        directions = [[0, -1], [0, 1], [-1, 0], [1, 0]]
        # store the farthest water count in res, this will be the last processes element
        res = -1

        # go through the q processing each element
        while q:
            r, c = q.popleft()
            # update res with the grid element we're at
            res = grid[r][c]
            # check all 4 directions from the current r,c
            for dirR, dirC in directions:
                newR, newC = r + dirR, c + dirC
                # validate the new coordinates
                if (newR < n and newC < n and 
                    newR >= 0 and newC >= 0 and
                    grid[newR][newC] == 0):
                    # add the direction to the queue
                    q.append([newR, newC])
                    # update grid in the new direction to store the distance from land + 1
                    # since we're overwriting the 0 we dont have to worry about revisited spots
                    grid[newR][newC] = grid[r][c] + 1

        return res - 1 if res > 1 else -1
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