<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 {
    bool isValid(int r,int c, int n){
        if(r>=0 && r<n && c>=0 && c<n) return true;
         return false;
    }
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        // node= r*cols+c

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

        int n=grid.size();
        int drow[]={-1,-1,-1,0,1,1,1,0};
        int dcol[]={-1,0,1,1,1,0,-1,-1};
        vector<int>adj[n*n];

        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                int nd=i*n+j;
                if(grid[i][j]==0){
                    for(int k=0;k<8;k++){
                        int r=i+drow[k];
                        int c=j+dcol[k];
                        if(isValid(r,c,n)){
                            if(grid[r][c]==0){
                                int node=r*n+c;
                                adj[node].push_back(nd);
                                adj[nd].push_back(node);
                            }
                        }
                    }
                }
            }
        }

        priority_queue< pair<int,int> ,vector<pair<int,int>> , greater<pair<int,int>>> q;
        
        vector<int>dist(n*n,1e9);
        dist[0]=0;
        
        q.push({0,0});
        
        while(!q.empty()){
            int d=q.top().first;
            int u=q.top().second;
            q.pop();
            
            for(auto it:adj[u]){
                int v=it;
                
                if(d+1< dist[v]){
                    dist[v]=d+1;
                    q.push({dist[v],v});
                }
            }
        }
        if(dist[n*n -1]==1e9 ) return -1;
        return dist[n*n -1]+1;
    }
};
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