<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:
        n = len(grid)
        if 1 in (grid[0][0], grid[n-1][n-1]):
            return -1
        
        directions = [
            (0, 1),
            (0, -1),
            (1, 0),
            (-1, 0),
            (1, 1),
            (1, -1),
            (-1, 1),
            (-1, -1)
        ]

        def get_neighbors(x, y):
            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                if 0 <= nx < n and 0 <= ny < n and not grid[nx][ny] and not visited[nx][ny]:
                    yield nx, ny
        
        q = deque([(1, 0, 0)])
        visited = [[False for _ in range(n)] for _ in range(n)]
        visited[0][0] = True

        while q:
            length, x, y = q.popleft()
            visited[x][y] = True
            if (x, y) == (n - 1, n - 1):
                return length
            for nx, ny in get_neighbors(x, y):
                visited[nx][ny]
                grid[nx][ny] = 1
                q.append((length + 1, nx, ny))
        return -1

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