Shortest Path in a Grid with Obstacles Elimination
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1293.Shortest%20Path%20in%20a%20Grid%20with%20Obstacles%20Elimination/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1293.Shortest%20Path%20in%20a%20Grid%20with%20Obstacles%20Elimination/images/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>
Breadth-First Search; Array; Matrix
func shortestPath(grid [][]int, k int) int {
	m, n := len(grid), len(grid[0])
	if k >= m+n-3 {
		return m + n - 2
	}
	q := [][]int{[]int{0, 0, k}}
	vis := make([][][]bool, m)
	for i := range vis {
		vis[i] = make([][]bool, n)
		for j := range vis[i] {
			vis[i][j] = make([]bool, k+1)
		}
	}
	vis[0][0][k] = true
	dirs := []int{-1, 0, 1, 0, -1}
	ans := 0
	for len(q) > 0 {
		ans++
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			k = p[2]
			for j := 0; j < 4; j++ {
				x, y := p[0]+dirs[j], p[1]+dirs[j+1]
				if x >= 0 && x < m && y >= 0 && y < n {
					if x == m-1 && y == n-1 {
						return ans
					}
					if grid[x][y] == 0 && !vis[x][y][k] {
						q = append(q, []int{x, y, k})
						vis[x][y][k] = true
					} else if grid[x][y] == 1 && k > 0 && !vis[x][y][k-1] {
						q = append(q, []int{x, y, k - 1})
						vis[x][y][k-1] = true
					}
				}
			}
		}
	}
	return -1
}