<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])
        queue = deque()
        visited = set()
        for r,c in product(range(ROWS), range(COLS)): # prepopulate BFS queue with sources
            if grid[r][c]: 
                queue.append((r,c, 0))
                visited.add((r,c, 0))


        if not queue or ROWS * COLS == len(queue): # no land or all land
            return -1
        
        
        # level = 0
        # while queue: # count number of BFS layers
        #     for _ in range(len(queue)):
        #         r,c = queue.popleft()
        #         for dr,dc in (0,1),(1,0),(0,-1),(-1,0):
        #             nr,nc = r+dr,c+dc
        #             if ROWS > nr >= 0 <= nc < COLS and not grid[nr][nc] and (nr,nc) not in visited:
        #                 visited.add((nr,nc))
        #                 queue.append((nr,nc))
        #     level += 1
        # return level - 1
        ans = 0
        while(queue):
            row, col, dist = queue.popleft()
            for dr,dc in (0,1),(1,0),(0,-1),(-1,0):
                nr = row + dr
                nc = col + dc
                if ROWS > nr >= 0 <= nc < COLS and not grid[nr][nc] and (nr,nc) not in visited:
                    visited.add((nr, nc))
                    queue.append((nr, nc, dist + 1))
                    ans = max(ans, dist+1)
        return ans

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