<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:
        '''
        traverse graph with BFS + min-heap
        calculate how many ways to remove obstacles while traversing
        find out the min obstacles to remove!
        '''

        import heapq

        dist = {} # key: (r, c), value: rm_cnt
        dist[(0,0)] = grid[0][0]

        rows, cols = len(grid), len(grid[0])
        for r in range(rows):
            for c in range(cols):
                dist[(r, c)] = float('inf')

        DIR = [(0,1), (0,-1), (1,0), (-1,0)]
        heap = [(grid[0][0], 0, 0)] # rm_cnt, r, c
        while heap:
            rm_cnt, r, c = heappop(heap)
            if r == rows-1 and c == cols-1:
                return dist[(r, c)]
            
            for dr, dc in DIR:
                nr, nc = dr+r, dc+c
                if (0 <= nr < rows and 0 <= nc < cols and rm_cnt+grid[nr][nc] < dist[(nr, nc)]):
                    dist[(nr, nc)] = rm_cnt + grid[nr][nc]
                    heappush(heap, (dist[(nr, nc)], nr, nc))

        return dist[(rows-1, cols-1)]
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