<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;
        q.push({{0,0},1});
        
        if(grid[0][0]==1)return -1;
        
        if(grid[0][0]==0 && grid.size()==1 && grid[0].size()==1)return 1;
        
        vector<vector<bool>>visited(grid.size(),vector<bool>(grid.size(),false));
        visited[0][0]=true;
        while(!q.empty())
        {
            pair<int,int>p = q.front().first; 
            int x = p.first; 
            int y= p.second; 
            int lengthOfPath = q.front().second; 
            q.pop();
            
            vector<pair<int,int>>neighbours = {{0,1}, {0,-1}, {1,0}, {-1,0},
                                               {1,1}, {-1,-1}, {1,-1}, {-1,1}};
            
            for(pair<int,int>neighbour: neighbours)
            {
                int newx = neighbour.first + x;
                int newy = neighbour.second+ y;
                
         if(newx>=0 && newy>=0 && newx<grid.size() && newy<grid[0].size() && grid[newx][newy]==0 && !visited[newx][newy]){

                    q.push({{newx,newy},lengthOfPath+1});
                    visited[newx][newy] = true;
                    
                    if(newx==grid.size()-1 && newy==grid[0].size()-1)
                     return lengthOfPath+1;
           
                }      
            }
        }
        
        return -1;
        
    }
};
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