<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:
        # concise version:
        n = len(grid)
        if grid[0][0] or grid[n-1][n-1]:
            return -1
        q = [(0,0,1)]
        grid[0][0] = 1
        for i,j,d in q:
            if i == n-1 and j == n-1: return d
            for x,y in ((i-1,j-1),(i-1,j),(i-1,j+1),(i,j-1),(i,j+1),(i+1,j-1),(i+1,j),(i+1,j+1)):
                if 0 <= x < n and 0 <= y < n and not grid[x][y]:
                    grid[x][y] = 1
                    q.append((x,y,d+1))
        return -1

        # bfs
        if grid[0][0] == 1:
            return -1
        n = len(grid)
        dq = deque([(0,0)])
        dis = defaultdict(int)
        seen = set()
        dis[(0,0)] = 1
        dirs = [(-1,-1),(-1,0),(-1,1),
                (1,0), (1,1), (0,1),
                (1,-1),(0,-1)]
        while dq:
            cur = dq.popleft()
            if cur == (n-1, n-1):
                return dis[cur]
            for d in dirs:
                ne = (cur[0] + d[0], cur[1] + d[1])
                if (ne not in seen and 
                    0 <= ne[0] < n and 0 <= ne[1] < n and
                    grid[ne[0]][ne[1]] == 0):
                    dq.append(ne)
                    dis[ne] = dis[cur] + 1
                    seen.add(ne)
        return -1

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