<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:
        # 2d graph, Djikstra

        import heapq

        h = []
        heapq.heapify(h)
        heapq.heappush(h, (0, (0,0)))

        m = len(grid)
        n = len(grid[0])

        D = [[None]*n for _ in range(m)]

        DIR = [0, 1, 0, -1, 0]
        visited = set()
         
        while h:
            cost, rc = heapq.heappop(h)
            while rc in visited and h:
                cost, rc = heapq.heappop(h)                
            if rc in visited and not h:
                break
            row0, col0 = rc
            visited.add((row0,col0))           
            # explore the connected neighbors.
            for i in range(4):
                row, col = row0 + DIR[i], col0 + DIR[i+1]
                if 0 <= row < m and 0 <= col < n and (row,col) not in visited:
                    val = cost + (1 if grid[row][col] else 0)
                    if D[row][col] is None:
                        D[row][col] = val
                    else:
                        D[row][col] = min(D[row][col], val)
                    
                    heapq.heappush(h, (D[row][col], (row, col)))
                       
        return D[m-1][n-1]



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