<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 {
   void BFS(vector<vector<int>>& graph, vector<int>& lens)
   {
        int n=graph.size();
        vector<bool> visited(n,false);
        
        queue<int> q;
        q.push(0);
        lens[0]=0;
        visited[0]=true;

        while(!q.empty())
        {
            int v=q.front();
            q.pop();

            

            for(auto& neighbour:graph[v])
            {
                if(!visited[neighbour])
                {
                    visited[neighbour]=true;
                    lens[neighbour]=lens[v]+1;
                    q.push(neighbour);
                }

                if(neighbour==n*n-1)
                    return;
            }
        }
   }
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n=grid.size();

        if(grid[0][0]==1 || grid[n-1][n-1]==1)
            return -1;

        vector<vector<int>> graph(n*n);

        for(int i=0;i<n;++i)
        {
            for(int j=0;j<n;++j)
            {
                if( grid[i][j]==1)
                    continue;

                int v=i*n+j;

                if (i<n - 1 && grid[i + 1][j] == 0)
                    graph[v].push_back((i + 1)*n+j);

                if (i > 0 && grid[i - 1][j] == 0 )
                    graph[v].push_back((i - 1)*n+j);

                if (i < n - 1 && j<n - 1 && grid[i + 1][j + 1] == 0)
                    graph[v].push_back((i + 1)*n+j+1);

                if (i < n - 1 && j>0 && grid[i + 1][j - 1] == 0 )
                    graph[v].push_back((i + 1)*n+j-1);



                if (i > 0 && j<n - 1 && grid[i - 1][j + 1] == 0)
                    graph[v].push_back((i - 1)*n+j+1);


                if (i > 0 && j > 0 && grid[i - 1][j - 1] == 0)
                    graph[v].push_back((i - 1)*n+j-1);

                if (j<n - 1 && grid[i][j + 1] == 0)
                    graph[v].push_back((i)*n+j+1);

                if (j > 0 && grid[i][j - 1] == 0)
                    graph[v].push_back((i)*n+j-1);
            }
        }

        vector<int> lens(n*n, -2);

        BFS(graph, lens);

        return lens[n*n-1] + 1;

    }
};
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