<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:

    vector<int> dx = {-1,-1,-1,0,0,1,1,1};
    vector<int> dy = {-1,0,1,-1,1,-1,0,1};

    bool isValid(int row,int col,vector<vector<int>> &grid)
    {
        if(row<0 || col<0 || row>= n || col>=n || grid[row][col] == 1)return false;

        return true;
    }

    vector<pair<int,int>> find(int row,int col,vector<vector<int>> &grid)
    {
        //cout<<row<<" "<<col<<"-> ";
        vector<pair<int,int>> nbr;
        for(int i = 0;i<8;i++)
        {
            if(isValid(row+dx[i],col+dy[i],grid))
            {
                nbr.push_back(make_pair(row+dx[i],col+dy[i]));
            }
        }

        return nbr;
    }

    int n;

    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        n = grid.size();
        vector<vector<bool>> vis(n,vector<bool> (n,false));

        //cout<<n<<" "<<n<<endl;

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

        queue<pair<int,int>> q;
        q.push(make_pair(0,0));
        vis[0][0] = true;
        q.push(make_pair(-1,-1));
        int ans = 1;

        while(!q.empty())
        {
            pair<int,int> front = q.front();
            q.pop();

            if(front.first == n-1 && front.second == n-1)
            return ans;

            if(front == make_pair(-1,-1))
            {
                if(!q.empty())
                {
                    q.push(make_pair(-1,-1));
                    ans++;
                }
            }
            else
            {
                int x = front.first;
                int y = front.second;

                vector<pair<int,int>> neghbour = find(x,y,grid);
                cout<<endl;
                for(auto nbr: neghbour)
                {
                    if(!vis[nbr.first][nbr.second])
                    {
                        q.push(nbr);
                        vis[nbr.first][nbr.second] = true;
                    }

                }
            }
        }

        return -1;
    }
};
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