<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>

import heapq

class Solution:
    def minimumObstacles(self, grid: List[List[int]]) -> int:
        
        m, n = len(grid), len(grid[0])
        
        d = defaultdict(lambda: float("inf"))
        
        pq = [(int(grid[0][0] == 1), (0, 0))]
        
        seen = set()
        
        deltas = [[0, 1], [1, 0], [0, -1], [-1, 0]]
        def iwb(i, j):
            return 0 <= i < m and 0 <= j < n
        
        while pq:
            current_cost, cp = heapq.heappop(pq)
            
            if d[cp] < current_cost:
                continue
                
            d[cp] = current_cost
            
            for dd in deltas:
                np = (cp[0] + dd[0], cp[1] + dd[1])
                if iwb(*np):
                
                    move_cost = 0
                    if grid[np[0]][np[1]]:
                        move_cost = 1

                    if d[np] > current_cost + move_cost:
                        heapq.heappush(pq, (current_cost + move_cost, np))
                        d[np] = current_cost + move_cost
                    
        return d[(len(grid)-1, len(grid[0])-1)]
                
                
        
    
3790