<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>

namespace {
    inline void check(vector<vector<int>>& grid, int i, int j, vector<pair<int, int>>& q) {
        if (i < 0 || j < 0 || i == grid.size() || j == grid.size() || grid[i][j]) return;
        q.push_back({i, j});
        grid[i][j] = 1;
    }
}
class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        auto n = grid.size();
        vector<pair<int, int>> q;
        vector<pair<int, int>> q2;
        int step = 1;
        if(grid[0][0] == 0) q.push_back({0, 0});
        while (!q.empty()) {
            for (auto k = 0; k < q.size(); ++k) {
                auto [i, j] = q[k];
                if (i == n - 1 && j == n - 1) return step;
                check(grid, i, j - 1, q2);
                check(grid, i, j + 1, q2);
                check(grid, i - 1, j - 1, q2);
                check(grid, i - 1, j, q2);
                check(grid, i - 1, j + 1, q2);
                check(grid, i + 1, j - 1, q2);
                check(grid, i + 1, j, q2);
                check(grid, i + 1, j + 1, q2);
            }
            ++step;
            q.clear();
            q.swap(q2);
        }
        return - 1;
    }
};
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