<p>You are given an <code>m x n</code> integer matrix <code>grid</code> where each cell is either <code>0</code> (empty) or <code>1</code> (obstacle). You can move up, down, left, or right from and to an empty cell in <strong>one step</strong>.</p>

<p>Return <em>the minimum number of <strong>steps</strong> to walk from the upper left corner </em><code>(0, 0)</code><em> to the lower right corner </em><code>(m - 1, n - 1)</code><em> given that you can eliminate <strong>at most</strong> </em><code>k</code><em> obstacles</em>. If it is not possible to find such walk return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/30/short1-grid.jpg" style="width: 244px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
The shortest path without eliminating any obstacle is 10.
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -&gt; (0,1) -&gt; (0,2) -&gt; (1,2) -&gt; (2,2) -&gt; <strong>(3,2)</strong> -&gt; (4,2).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/09/30/short2-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
<strong>Output:</strong> -1
<strong>Explanation:</strong> We need to eliminate at least two obstacles to find such a walk.
</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;= 40</code></li>
	<li><code>1 &lt;= k &lt;= m * n</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 shortestPath(self, grid: List[List[int]], k: int) -> int:
        R = len(grid)
        C = len(grid[0])
        target = (R - 1, C - 1)
        
        def manhattan_distance(a, b):
            return (target[0] - a) + (target[1] - b)

        steps = 0
        state = (0, 0, k)
        queue = []
        #queue.append((steps, state))
        estimated_distance = manhattan_distance(0,0)
        heapq.heappush(queue, (estimated_distance, steps, state))
        visited = set()
        visited.add(state)

        while queue:
            # steps, (x, y, k) = queue.pop(0)
            estimated_distance, steps, (x, y, k) = heapq.heappop(queue)
            
            if estimated_distance - steps <= k:
                return estimated_distance
            
            for dx, dy in [ (0, 1), (0, -1), (1, 0), (-1, 0)]:
                nx = x + dx
                ny = y + dy
                if 0 <= nx < R and 0 <= ny < C:
                    new_k = k - grid[nx][ny]
                    new_state = (nx, ny, new_k)
                    if new_k >= 0 and new_state not in visited:
                        new_estimated_distance = manhattan_distance(nx, ny) + steps + 1
                        new_steps = steps + 1
                        heapq.heappush(queue, (new_estimated_distance, new_steps, new_state))
                        visited.add(new_state)
                        

        return -1

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