<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:
        if k >= len(grid) + len(grid[0]) - 2:
            return len(grid) + len(grid[0]) - 2
        start = deque((((0, 0), k), ))
        visited = set(((0, 0), k))
        step = 0
        while start:
            for _ in range(len(start)):
                curr, n = start.popleft()
                visited.add(((curr, n), ))
                if curr == (len(grid) - 1, len(grid[0]) - 1):
                    return step
                for direction in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                    i, j = curr[0] + direction[0], curr[1] + direction[1]
                    if 0 <= i < len(grid) and 0 <= j < len(grid[0]):
                        if grid[i][j] == 0 and ((i, j), n) not in visited:
                            visited.add(((i, j), n))
                            start.append(((i, j), n))
                        else:
                            if n > 0 and ((i, j), n - 1) not in visited:
                                visited.add(((i, j), n - 1))
                                start.append(((i, j), n - 1))
            step += 1
        return -1

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