<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:
        stack = [(0, 0, k, 0)]
        visited = set()
        while stack:
            curr_i, curr_j, curr_step, step_counter = stack.pop(0)
            if curr_i == len(grid) - 1 and curr_j == len(grid[0]) - 1:
                return step_counter
            next_steps = self.getNeighbor(grid, curr_i, curr_j)
            for step_i, step_j in next_steps:
                if (step_i, step_j, curr_step) not in visited:
                    if grid[step_i][step_j] == 1:
                        if curr_step > 0:
                            visited.add((step_i, step_j, curr_step))
                            stack.append((step_i, step_j, curr_step - 1, step_counter + 1))
                    else:
                        visited.add((step_i, step_j, curr_step))
                        stack.append((step_i, step_j, curr_step, step_counter + 1))
        return -1
                        

    def getNeighbor(self, grid, i, j):
        res = []
        if i - 1 >= 0:
            res.append((i-1, j))
        if j - 1 >= 0:
            res.append((i, j-1))
        if j + 1 < len(grid[0]):
            res.append((i, j+1))
        if i + 1 < len(grid):
            res.append((i+1, j))
        return res
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