<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) {
        queue<pair<pair<int,int>,int>>q;
          int n=grid.size();
        vector<vector<bool>> vis(n, vector<bool>(n, false));
        vector<int> r = {1, 1, 0, -1, 0, -1, -1, 1};
        vector<int> c = {1, 0, 1, 1, -1, -1, 0, -1};
       if(grid[0][0]==0&&n==1) return 1;
        if(grid[0][0]==1||grid[n-1][n-1]==1) return -1;
        q.push({{0,0},0});
        vis[0][0]=true;
        while(!q.empty()){
            auto p=q.front();
            q.pop();
            int row=p.first.first;
            int col=p.first.second;
            int dis=p.second;
             
        for(int i = 0; i < 8; i++)
        {
            int R = row + r[i];
            int C = col+ c[i];
            if(R < n && R >= 0 && C < n && C >= 0 && grid[R][C] == 0 && !vis[R][C])
            {  vis[R][C]=true;
               if(R==n-1&&C==n-1) return dis+2;
                q.push({{R,C},dis+1});
            }
        }

        }
        return -1;

        }
    
};
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