<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:

        rows, cols = len(grid), len(grid[0])
        target = (rows - 1, cols - 1)

        def manhattan_distance(row, col):
            return target[0] - row + target[1] - col

        # (row, col, remaining_elimination)
        state = (0, 0, k)

        # (estimation, steps, state)
        # h(n) = manhattan distance,  g(n) = 0
        queue = [(manhattan_distance(0, 0), 0, state)]
        seen = set([state])

        while queue:
            estimation, steps, (row, col, remain_eliminations) = heapq.heappop(queue)

            # we can reach the target in the shortest path (manhattan distance),
            #   even if the remaining steps are all obstacles
            remain_min_distance = estimation - steps
            if remain_min_distance <= remain_eliminations:
                return estimation

            # explore the four directions in the next step
            for new_row, new_col in [(row, col + 1), (row + 1, col), (row, col - 1), (row - 1, col)]:
                # if (new_row, new_col) is within the grid boundaries
                if (0 <= new_row < rows) and (0 <= new_col < cols):
                    new_eliminations = remain_eliminations - grid[new_row][new_col]
                    new_state = (new_row, new_col, new_eliminations)

                    # if the next direction is worth exploring
                    if new_eliminations >= 0 and new_state not in seen:
                        seen.add(new_state)
                        new_estimation = manhattan_distance(new_row, new_col) + steps + 1
                        heapq.heappush(queue, (new_estimation, steps + 1, new_state))

        # did not reach the target
        return -1
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