<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) {
        if (grid[0][0] == 1) return -1;
        int m = grid.size();
        int n = grid[0].size();
        if (m == 1 && n == 1 && grid[0][0] == 0) return 1;
        queue<vector<int>> q;
        q.push({0,0,1});
        set<pair<int, int>> visited;
        while (!q.empty()) {
            auto cur = q.front();
            q.pop();
            int row = cur[0], col = cur[1], len = cur[2];
            for (auto  [i, j] : std::vector<std::pair<int, int>>{{-1,-1}, {-1,0}, {-1,1}, {0,-1}, {0,1}, {1,-1}, {1,0}, {1,1}}) {
                int nr = row+i;
                int nc = col+j;
                if (nr >= 0 && nc >= 0 && nr < m && nc < n && grid[nr][nc] == 0 && visited.find({nr, nc}) == visited.end()) {
                    visited.insert({nr, nc});
                    q.push({nr, nc, len+1});
                    if (nr == m - 1 && nc == n - 1) {
                        return len+1;
                    }
                }              
            }
        }
        return -1;
    }
};
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