<p>You are given a <strong>0-indexed</strong> 2D integer array <code>grid</code> of size <code>m x n</code>. Each cell has one of two values:</p>

<ul>
	<li><code>0</code> represents an <strong>empty</strong> cell,</li>
	<li><code>1</code> represents an <strong>obstacle</strong> that may be removed.</li>
</ul>

<p>You can move up, down, left, or right from and to an empty cell.</p>

<p>Return <em>the <strong>minimum</strong> number of <strong>obstacles</strong> to <strong>remove</strong> so you can move from the upper left corner </em><code>(0, 0)</code><em> to the lower right corner </em><code>(m - 1, n - 1)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/04/06/example1drawio-1.png" style="width: 605px; height: 246px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1],[1,1,0],[1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can remove the obstacles at (0, 1) and (0, 2) to create a path from (0, 0) to (2, 2).
It can be shown that we need to remove at least 2 obstacles, so we return 2.
Note that there may be other ways to remove 2 obstacles to create a path.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/04/06/example1drawio.png" style="width: 405px; height: 246px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> We can move from (0, 0) to (2, 4) without removing any obstacles, so we return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> <strong>or</strong> <code>1</code>.</li>
	<li><code>grid[0][0] == grid[m - 1][n - 1] == 0</code></li>
</ul>

class Solution:
    def minimumObstacles(self, grid: List[List[int]]) -> int:
        n = len(grid)      # number of rows
        m = len(grid[0])   # number of columns
        visited = set()
        def get_unvisited_neighbours(i, j):
            ret = []
            for ii in range(max(i - 1, 0), min(i + 2, n)):
                val = (ii, j, grid[ii][j])
                if val not in visited:
                    ret.append(val)
                    visited.add(val)
            for jj in range(max(j - 1, 0), min(j + 2, m)):
                val = (i, jj, grid[i][jj])
                if val not in visited:
                    ret.append(val)
                    visited.add(val)
            return ret
        assert grid[0][0] == 0
        assert grid[-1][-1] == 0
        empty_stack = [(0, 0, grid[0][0])]
        obstacle_stack = []
        obstacle_2nd_stack = []
        
        ret = 0
        while True:
            # print(ret)
            # print(empty_stack)
            # print(obstacle_stack)
            # print(obstacle_2nd_stack)
            while empty_stack:
                current = empty_stack.pop()
                visited.add(current)
                if current[:2] == (n - 1, m - 1):       # it means we have reached the exit cell
                    return ret
                
                i, j, v = current               # v means value :)
                neighbours = get_unvisited_neighbours(i, j)
                empty_stack.extend(n for n in neighbours if n[-1] == 0)
                obstacle_stack.extend(n for n in neighbours if n[-1] != 0)
            ret += 1                            # we have traversed all available empty cells
            while obstacle_stack:
                current = obstacle_stack.pop()
                visited.add(current)
                i, j, v = current
                neighbours = get_unvisited_neighbours(i, j)
                empty_stack.extend(n for n in neighbours if n[-1] == 0)
                obstacle_2nd_stack.extend(n for n in neighbours if n[-1] != 0)
            obstacle_stack = obstacle_2nd_stack
            obstacle_2nd_stack = []
        # unreachable
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