<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 {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size();
        if (n == 0)
            return 0;
        queue<int> bfs;
        bfs.push(0);
        if (grid[0][0] || grid[n - 1][n - 1]) {
            return -1;
        }
        vector<vector<int>> dir = {{1, 1}, {1, -1}, {-1, 1}, {-1, -1},
                                   {0, 1}, {1, 0},  {0, -1}, {-1, 0}};
        vector<int> sp(n * n, INT_MAX);
        sp[0] = 1;
        grid[0][0] = -1;
        while (!bfs.empty()) {
            int cell = bfs.front();
            if (cell == n*n-1) return sp[cell];
            bfs.pop();
            int i = cell / n, j = cell % n;
            for (auto d : dir) {
                int x = i + d[0], y = j + d[1];
                if (x < n && x >= 0 && y >= 0 && y < n) {
                    int k = n * x + y;
                    if (grid[x][y] == 0) {
                        if (sp[k] > sp[cell] + 1){
                            sp[k] = sp[cell] + 1;
                            bfs.push(k);
                        }
                        grid[x][y] = -1;
                        if (k == n*n - 1) return sp[k];
                    }
                }
            }
        }
        return sp[n * n - 1] == INT_MAX ? -1 : sp[n * n - 1];
    }
};
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