<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:
    """
        we're going to use BFS
        in addition to queueing the row and coliumn name, we're also going to track the obstacleBalance

        let's start off by writing a BFS approach, and then we'll modify it to utilize obstacles

        [   i ->. n
          j [0,0,0],
          | [1,1,0],
          m [0,0,0],
            [0,1,1],
            [0,0,0]]
    """
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        start = (0, 0)

        # queue = deque(([start], 0)) # (position, steps, k) ((x, y), steps, k)
        queue = deque([(start, 0, k)])
        cols = len(grid[0])
        rows = len(grid)

        if k >= rows + cols - 2:
            return rows + cols - 2

        visited = set([(start, k)])

        while queue:
            (x, y), steps, k = queue.popleft()

            if (x, y) == (rows-1, cols - 1):
                return steps

            # Explore the four possible directions
            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                nx, ny = x + dx, y + dy

                if 0 <= nx < rows and 0 <= ny < cols:
                    new_k = k - grid[nx][ny]

                    # Only consider this state if we have enough k left and haven't visited it before
                    if new_k >= 0 and ((nx, ny), new_k) not in visited:
                        visited.add(((nx, ny), new_k))
                        queue.append(((nx, ny), steps + 1, new_k))

        
        return -1 

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