<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])
        
        # Initialize a 2D array to store the minimum number of obstacles removed to reach each cell
        removed_obstacles = [[float('inf')] * n for _ in range(m)]
        removed_obstacles[0][0] = 0  # Start with 0 removed obstacles
        
        # Define the directions (up, down, left, right)
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        
        # Priority queue to process cells with the minimum number of removed obstacles first
        pq = [(0, 0, 0)]  # (removed_obstacles, row, col)
        
        while pq:
            removed, r, c = heapq.heappop(pq)
            
            # If we reach the bottom-right corner, return the number of removed obstacles
            if r == m - 1 and c == n - 1:
                return removed
            
            # Process neighbors
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                
                # Check if the neighbor is within bounds
                if 0 <= nr < m and 0 <= nc < n:
                    new_removed = removed + grid[nr][nc]
                    
                    # If we can reach the neighbor with fewer removed obstacles, update and add to the queue
                    if new_removed < removed_obstacles[nr][nc]:
                        removed_obstacles[nr][nc] = new_removed
                        heapq.heappush(pq, (new_removed, nr, nc))
        
        return -1  # This should never be reached with valid inputs

# Example usage:
solution = Solution()
grid = [[0,1,1],[1,1,0],[1,1,0]]
print(solution.minimumObstacles(grid))  # Output: 2

grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
print(solution.minimumObstacles(grid))  # Output: 0
41060