<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:
    
    typedef pair<int,int> P;
    int n,m;
    bool isValid(int i,int j,vector<vector<int>>& grid){
        return i>=0 && j>=0 && i<n && j<m && grid[i][j]==0;
    }

    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0]==1) return -1;
        n = grid.size();
        m = grid[0].size();
        vector <P> dirs = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
        vector<vector<int>> dist(n,vector<int>(m,INT_MAX));
        vector<vector<P>> par(n,vector<P>(m));
        priority_queue <pair<int,P>,vector<pair<int,P>>,greater<pair<int,P>>> pq;
        dist[0][0]=0;
        pq.push({0,{0,0}});
        int ans = 1;
        par[0][0]={0,0};
        while(!pq.empty()){
            int cx = pq.top().second.first;
            int cy = pq.top().second.second;
            pq.pop();
            for(auto &it : dirs){
                int nx = cx + it.first;
                int ny = cy + it.second;
                int w = 1;
                if(isValid(nx,ny,grid) && dist[cx][cy]+w<dist[nx][ny]){
                    par[nx][ny] = {cx,cy};
                    dist[nx][ny] = dist[cx][cy]+w;
                    pq.push({dist[nx][ny],{nx,ny}});
                }
            }
        }
        if(dist[n-1][m-1]==INT_MAX) return -1;
        int cnt = 1;
        int i = n-1;
        int j = m-1;
        while(i!=0 || j!=0){
            pair p1 = par[i][j];
            i = p1.first;
            j = p1.second;
            cnt++;
        }
        return cnt;
    }
};
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