<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:
        m, n = len(grid), len(grid[0])
        dir = [0, 1, 0, -1, 0]

        # bfs using deque
        # que = deque()
        # que.append([0, 0, 0, k])
        # visit = set()
        # while que:
        #     r, c, d, p = que.popleft()
        #     if min(r, c) < 0 or r == m or c == n: continue
        #     if r == m - 1 and c == n - 1: return d
        #     key = (r, c, p)
        #     if key in visit: continue
        #     visit.add(key)
        #     p -= grid[r][c]
        #     if p < 0: continue
        #     d += 1
        #     for i in range(4):
        #         x, y = dir[i] + r, dir[i + 1] + c
        #         que.append([x, y, d, p])


        # bfs
        st = [[0, 0, k]]
        visit = set()
        dist = 0
        while st:
            aux = []
            while st:
                r, c, p = st.pop()
                if r == m - 1 and c == n - 1: return dist
                key = (r, c, p)
                if key in visit: continue
                visit.add(key)
                for i in range(4):
                    x, y = dir[i] + r, dir[i + 1] + c
                    if 0 <= x < m and 0 <= y < n:
                        if grid[x][y] == -1 or p < grid[x][y]: continue
                        aux.append([x, y, p - grid[x][y]])
            st.extend(aux)
            aux.clear()
            dist += 1
        

        # dijkestra failed
        # pq = [[0, 0, 0, k]]
        # res = {(0, 0, k): 0}
        # while pq:
        #     d, r, c, p = heappop(pq)
        #     if r == m - 1 and c == n - 1: return d
        #     for i in range(4):
        #         x, y = dir[i] + r, dir[i + 1] + c
        #         if 0 <= x < m and 0 <= y < n:
        #             pp = p - grid[x][y]
        #             if pp < 0: continue
        #             if res.get((x, y, pp), math.inf) < d + 1: continue
        #             heappush(pq, [d + 1, x, y, pp])
        #             res[(x, y, pp)] = d + 1
        return -1
25648