<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:
               
        
        # Approach 1: BFS
        # Time Complexity: O(N * K)
        # Space Complexity: O(N * K)
       
        max_breaks = k 
        R, C = len(grid), len(grid[0])

        def neighbors(r, c, b):
            for nr, nc in [(r-1, c), (r+1, c), (r, c-1), (r, c+1)]:
                if 0 <= nr < R and 0 <= nc < C:
                    nb = b + grid[nr][nc]
                    if nb <= max_breaks:
                        yield nr, nc, nb


        pq = [(0, 0, 0, 0)] # dist, r, c, breaks
        dist = {(0, 0, 0): 0}
        INF = float("inf")
        while pq:
            d, r, c, b = heapq.heappop(pq)
            if dist.get((r, c, b), INF) < d:
                continue

            if r == R - 1 and c == C - 1:
                return d

            for nei in neighbors(r, c, b):
                if dist.get(nei, INF) > d + 1:
                    dist[nei] = d + 1
                    heapq.heappush(pq, (d + 1,) + nei)
        return -1
        

        # Approach 2: Optimize - A* (A Star) Algorithm
        # Time Complexity: O(N * K * \log(N * K))
        # Space Complexity: O(N * K)       

        A = grid
        max_breaks = k
        R, C = len(A), len(A[0])
        
        def neighbors(r, c, b):
            for nr, nc in ((r-1, c), (r, c-1), (r+1, c), (r, c+1)):
                if 0 <= nr < R and 0 <= nc < C:
                    nb = b + A[nr][nc]
                    if nb <= max_breaks:
                        yield nr, nc, nb
    
        pq = [(0, 0, 0, 0)]  # dist, r, c, breaks
        dist = {(0,0,0) : 0}
        INF = float('inf')
        while pq:
            d, r, c, b = heapq.heappop(pq)            
            if dist.get((r, c, b), INF) < d:
                continue                
            if r == R - 1 and c == C - 1:
                return d
            
            for nei in neighbors(r, c, b):
                if dist.get(nei, INF) > d + 1:
                    dist[nei] = d+1
                    heapq.heappush(pq, (d + 1,) + nei)
        return -1

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