<p>Given an <code>n x n</code> binary matrix <code>grid</code>, return <em>the length of the shortest <strong>clear path</strong> in the matrix</em>. If there is no clear path, return <code>-1</code>.</p>

<p>A <strong>clear path</strong> in a binary matrix is a path from the <strong>top-left</strong> cell (i.e., <code>(0, 0)</code>) to the <strong>bottom-right</strong> cell (i.e., <code>(n - 1, n - 1)</code>) such that:</p>

<ul>
	<li>All the visited cells of the path are <code>0</code>.</li>
	<li>All the adjacent cells of the path are <strong>8-directionally</strong> connected (i.e., they are different and they share an edge or a corner).</li>
</ul>

<p>The <strong>length of a clear path</strong> is the number of visited cells of this path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example1_1.png" style="width: 500px; height: 234px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example2_1.png" style="height: 216px; width: 500px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j] is 0 or 1</code></li>
</ul>

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        if grid[0][0] == 1:
            return -1
        if len(grid) == 1:
            return 1
        q = deque()
        q.append(((0,0), 1))
        n = len(grid)

        while q:
            p, steps = q.popleft()
            grid[p[0]][p[1]] = 1
            options = [(0,1), (1,0), (1,1), (-1, 0), (0, -1), (-1,1), (1, -1), (-1,-1)]
            options = [
                (p[0]+opt[0], p[1]+opt[1]) for opt in options
            ]
            options = [opt for opt in options if 0<=opt[0]<n and 0<=opt[1]<n and grid[opt[0]][opt[1]] == 0]
            for opt in options:
                if opt == (n-1, n-1):
                    return steps+1
                grid[opt[0]][opt[1]] = 1
                q.append((opt, steps+1))
        return -1

16500