<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:
		ymax, xmax = len(grid), len(grid[-1])

		A = set()
		for ay, ax in product(range(ymax), range(xmax)):
			if grid[ay][ax] == 1:
				grid[ay][ax] = 0 # water
				A.add((ay, ax))
			else:
				grid[ay][ax] = -1 # terra

		res = -1
		while A:
			# for row in grid:
			# 	print(' '.join('.' if r == -1 else str(r) for r in row))
			# print()

			B = set() # using B --> *B*FS, using A.pop() --> *D*FS
			for ay, ax in A:
				val_ayax = grid[ay][ax] + 1
				for y, x in ((ay - 1, ax), (ay + 1, ax), (ay, ax + 1), (ay, ax - 1)):
					if ymax > y >= 0 and xmax > x >= 0 and grid[y][x] == -1:
						grid[y][x] = val_ayax
						if val_ayax > res:
							res = val_ayax
						B.add((y, x))
			A = B
		return res

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