<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 get_positions(self, i, j, m, n):
        if (i, j) in self.pos_dp:
            return self.pos_dp[(i, j)]
        
        positions = [(i-1, j), (i, j+1), (i+1, j), (i, j-1)]
        positions = [pos for pos in positions if pos[0]>=0 and pos[0]<m and pos[1]>=0 and pos[1]<n]
        self.pos_dp[(i, j)] = positions 
        return positions

    # def dfs(self, i, j, k, m, n):
    #     if ((i, j), k) in self.dp:
    #         return self.dp[((i, j), k)]
        
    #     neighbours = self.get_positions(i, j, m, n)



    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        # self.grid = grid
        queue = [(0, (0, 0), k)]
        dp = {}
        self.pos_dp = {}

        while queue:
            # print(len(self.pos_dp))
            dist, pos, num_obs = queue.pop(0)
            if pos == (m-1, n-1):
                return dist
            
            
            neighbours = self.get_positions(pos[0], pos[1], m, n)
            for nb in neighbours:
                if (not grid[nb[0]][nb[1]]) and (not (nb, num_obs) in dp):
                    dp[(nb, num_obs)] = 0
                    queue.append([dist+1, nb, num_obs])
                elif grid[nb[0]][nb[1]] and num_obs >0 and not ((nb, num_obs) in dp):
                    dp[(nb, num_obs)] = 0
                    queue.append((dist+1, nb, num_obs-1))
        return -1


        

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