<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>

from collections import deque
class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        n1, n2 = len(grid), len(grid[0])
        visited = [[set() for _ in range(n2)] for _ in range(n1)]
        visited[0][0].add(0)
        
        queue = deque([(0, 0, 0)])
        diffs = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        steps = 0
        while queue:
            for _ in range(len(queue)):
                i, j, cost = queue.popleft()
                print(i, j, cost)
                if i == n1-1 and j == n2-1:
                    return steps
                
                for diff in diffs:
                    new_i, new_j = i + diff[0], j + diff[1]
                    if new_i >= 0 and new_i < n1 and new_j >= 0 and new_j < n2:
                        new_cost = cost + (1 if grid[new_i][new_j] == 1 else 0)
                        if new_cost <= k and new_cost not in visited[new_i][new_j]:
                            queue.append((new_i, new_j, new_cost))
                            visited[new_i][new_j].add(new_cost)
            steps += 1
        return -1
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