<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) {

        int n= grid.size();
        int m=grid[0].size();

        vector<vector<int>> dis(n,vector<int> (m,1e9));
        priority_queue< vector<int> , vector<vector<int>> ,greater<vector<int>> >q;
        // queue<vector<int>> q;

        // pq.push({0,0,0});:
        if(grid[0][0]==1) return -1;
        q.push({0,0,1});
        dis[0][0]=1;

        while(!q.empty()){

            auto temp=q.top();
            q.pop();
            int i=temp[0];
            int j=temp[1];
            int d=temp[2];

            cout<<i<<" "<<j<<" "<<d<<endl;

            int row[8]={-1,1,0,0,-1,-1,+1,+1};
            int col[8]={0,0,-1,1,-1,+1,-1,+1};

            for(int k=0;k<8;k++)
            {

                int p= i+row[k];
                int r= j+col[k];

                if(p<n && r<m && p>=0&& r>=0 && grid[p][r]==0  && dis[p][r] > d+1){
                    q.push({p,r,d+1});
                    dis[p][r]=d+1;
                } 
            }
        }
        if(dis[n-1][m-1]==1e9) return -1;
        return dis[n-1][m-1];
        
    }
};
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