<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])
        
        def is_in_bounds(row,col):
            return row >= 0 and row < m and col >= 0 and col < n

        def get_neighbors(row,col):
            adjs = [(0,1),(1,0),(0,-1),(-1,0)]
            neighbors = []

            for adj in adjs:
                i,j = row + adj[0], col + adj[1]
                if is_in_bounds(i,j):
                    weight = 1 if grid[i][j] == 1 else 0
                    neighbors.append((weight,i,j))
            return neighbors

        distances = [n*[float('inf')] for row in range(m)]
        start = (0,0,0)
        distances[0][0] = 0
        pq = [start]

        while pq:
            distance,i,j = heapq.heappop(pq)

            if distance > distances[i][j]:
                continue
            
            for weight,x,y in get_neighbors(i,j):
                distance = distances[i][j] + weight
                if distance < distances[x][y]:
                    distances[x][y] = distance
                    heapq.heappush(pq, (distance,x,y))
                
        return distances[m-1][n-1]

        
    def dijkstra(self, start, graph):
        distances = {node: float('inf') for node in graph}
        pq = [(0, start)]
        distances[start] = 0

        while pq:
            current_distance, node = heapq.heappop(pq)

            if current_distance > distances[node]:
                continue
            
            for neighbor, weight in graph.items():
                distance = distances[node] + weight

                if distance < distances[neighbor]:
                    distances[neighbor] = distance
                    heapq.heappush(pq, (distance, neighbor))
        return distances


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