<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:
        def bfs(grid, r, c):
            ROWS, COLS = len(grid), len(grid[0])
            queue = deque()
            visited = set()
            length = 1
            queue.append((r, c))

            while queue:
                for _ in range(len(queue)):
                    r, c = queue.popleft()
                    if min(r, c) < 0 or r == ROWS or c == COLS or (r, c) in visited or grid[r][c] == 1:
                        continue
                    else:
                        if r == ROWS - 1 and c == COLS - 1:
                            return length

                        visited.add((r, c))
                        dirs = [[0, 1], [0, -1], [1, 0], [-1, 0], [1, 1], [1, -1], [-1, 1], [-1, -1]]                      
                        for dr, dc in dirs:
                            queue.append((r + dr, c + dc))
                length += 1
            else:
                return -1
        return bfs(grid, 0, 0)
        


18300