<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:
    ''' 

        input => list[list[int]]
        output => int

        [[1,0,1],
         [0,0,0],
         [1,0,1]]

        visited = set()
        stack = []

         loop through the grid
            find all cell with 1
            stack.add (0, (i, j))
            visited.add((i,j))

    '''
    def maxDistance(self, grid: List[List[int]]) -> int:
        
        visited = set()
        stack = []
        max_distance = -1
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if(grid[i][j] == 1):
                    stack.append((0, (i, j)))
                    visited.add((i,j))
        
        direction = [(0, -1), (0, 1), (1, 0), (-1, 0)]
        while(stack):
            distance, coord = heapq.heappop(stack)
            x, y = coord

            for dir_x, dir_y in direction:
                new_x = x + dir_x
                new_y = y + dir_y

                if(new_x > -1 and new_y > -1 and new_x < len(grid) and new_y < len(grid[i]) and (new_x, new_y) not in visited):
                    max_distance = max(distance + 1, max_distance)
                    heapq.heappush(stack, (distance + 1, (new_x, new_y)))
                    visited.add((new_x, new_y))
                

        return max_distance
590