<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:
        """
        Problem
        -------
        Return the minimum number of steps to walk from the upper left corner (0, 0) 
        to the lower right corner (m - 1, n - 1) given that you can eliminate 
        at most k obstacles. If it is not possible to find such walk return -1.
        - Given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle).
        - Can move up, down, left, or right from and to an empty cell in one step.

        Example
        -------
        Input: grid = [[0,0,0],
                       [1,1,0],
                       [0,0,0],
                       [0,1,1*], <-- *removed obstacle
                       [0,0,0]],
                k = 1
        Output: 6
        Explanation: 
        - 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) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).
        """
        rows, cols = len(grid), len(grid[0])
        if k >= rows + cols - 2: return rows + cols - 2

        def get_children(curr):
            ## child options
            # get curr node coords
            row, col = curr[0], curr[1]
            children = []
            for new_row, new_col in [(row, col + 1), (row + 1, col), (row, col - 1), (row - 1, col)]:
                # if (new_row, new_col) is within the grid boundaries
                if (0 <= new_row < rows) and (0 <= new_col < cols):
                    children.append((new_row, new_col))
            return children
                
        ## BFS
        curr, target = (0,0), (len(grid)-1, len(grid[0])-1)
        # ( (row, col), remaining quota to eliminate obstacles) )
        state = (curr, k)
        # (steps, state)
        queue = deque([(0, state)])
        seen = set([state])
        # start node is the target
        if curr == target: return 1
        
        ## Loop to traverse the graph in BFS manner
        while queue:
            # pop queue
            steps, state = queue.popleft()
            node, k = state

            print(f"State: {state}")

            ## Conditional Criteria
            if node == target: 
                return steps
                
            # get neighbors
            children = get_children(node)
            
            # Loop to iterate over the 
            # neighbors of the node
            for child in children:
                new_elims = k - grid[child[0]][child[1]]
                new_state = (child, new_elims)
                if new_elims >= 0 and new_state not in seen:
                    # visit node
                    seen.add(new_state)
                    # enqueue new path
                    queue.append((steps+1, new_state))
    
        # shortest path not found
        return -1
18805