<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 get_moves(self,point,grid,n,m):
        i,j = point
        moves = [(i+1,j),(i+1,j-1),(i+1,j+1),(i,j-1),(i,j+1),(i-1,j-1),(i-1,j),(i-1,j+1)]
        valid_moves = []
        for x,y in moves:
            if 0 <= x < n and 0 <= y < m and grid[x][y] == 0:
                valid_moves.append((x,y))
                
        return valid_moves
    
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        
        n,m = len(grid), len(grid[0])
        
        if grid[0][0] == 1 or grid[-1][-1] == 1:
            return -1
        
        queue = deque([((0,0),1)])
        seen = set()
        max_depth = 1
        
        while queue:
            point,depth = queue.popleft()
            seen.add(point)
            max_depth = max(max_depth,depth)
            
            if point == (n-1,m-1):
                return max_depth
            else:
                for move in self.get_moves(point,grid,n,m):
                    if move not in seen:
                        seen.add(move)
                        queue.append((move,depth+1))
                
        return -1
        
18200