<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) {
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), 0));
        queue<pair<pair<int, int>, int>> q;
        vector<pair<int, int>> dir{{-1,0},{-1,-1},{-1,1},{1,0},{1,-1},{1,1},{0,1},{0,-1}};
        q.push({{0,0}, 1});
        while(!q.empty()){
            int i = q.front().first.first;
            int j = q.front().first.second;
            int count = q.front().second;
            q.pop();
            if(i < 0 || i > grid.size() - 1 || j < 0 || j > grid[0].size() - 1 || visited[i][j] || grid[i][j] != 0) continue;
            if(i == grid.size() - 1 && j == grid[0].size() - 1) return count;
            visited[i][j] = true;
            for(auto& d : dir){
                q.push({{i + d.first, j + d.second}, count+1});
            }
        }
        return -1;
    }
    /*
    0 0 0
    1 1 0
    1 1 1
    */
};
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