<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:
        def within_grid(r, c):
            return 0 <= r < rows and 0 <= c < cols
        
        # since we store by cost, if we visit a cell and it already has a lower cost to get to (min_steps), dont visit, otherwise
        # update
        rows, cols = len(grid), len(grid[0])

        cost = [[float("inf")] * cols for _ in range(rows)]
        cost[0][0] = 0

        h = [(0, (0, 0))] # num_obstacles_removed, (r, c)

        heapq.heapify(h)

        while h:
            num_removals, (r, c) = heapq.heappop(h)
            if (r, c) == (rows - 1, cols - 1):
                return num_removals

            # if cost[r][c] < num_removals:
            #     continue

            for direction in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                dy, dx = direction
                new_r, new_c = r + dy, c + dx
                if within_grid(new_r, new_c):
                    if grid[new_r][new_c] == 1 and num_removals + 1 < cost[new_r][new_c]:
                        cost[new_r][new_c] = num_removals + 1
                        heapq.heappush(h, (num_removals + 1, (new_r, new_c)))
                    elif grid[new_r][new_c] == 0 and num_removals < cost[new_r][new_c]:
                        cost[new_r][new_c] = num_removals
                        heapq.heappush(h, (num_removals, (new_r, new_c)))
        return -1
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