<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>

"""
- m x n grid
- 0 empty; 1 obstacle
- 1 step in each direction (up, down, left, right)
- minimum steps to go from upper left to lower right
    - can eliminate at most k obstacles


Algorithm
- we start with the initial cell
- recursively explore all directions; keep track of k, visited nodes
    - if no obstacle; we can move freely
    - if we encounter an obstacle, there are two possible outcomes:
        - pass through obstacle and reduce k if k > 0
        - dont pursue this path and retain current value of k
- continue traversing till we exhaust the grid

Time: O(mnk)
Space: O(mnk)
"""

from collections import deque

class Solution:

    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        if not grid or not grid[0] or (grid[0][0] == 1 and k == 0):
            return -1
        
        rows, cols = len(grid), len(grid[0])
        queue = deque([(0, 0, 0, k)])
        visited = set()

        while queue:
            i, j, dist, rem_k = queue.popleft()
            if (i, j) == (rows - 1, cols - 1):
                return dist
            if (i, j, rem_k) in visited:
                continue
            visited.add((i, j, rem_k))
            for r, c in [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]:
                if not (0 <= r < rows and 0 <= c < cols and (r, c, rem_k) not in visited):
                    continue
                if grid[r][c] == 1:
                    if rem_k > 0:
                        queue.append((r, c, dist + 1, rem_k - 1))
                else:
                    queue.append((r, c, dist + 1, rem_k))
        
        return -1
                

        
        
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