<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();
        if(grid[0][0]!=0 || grid[n-1][m-1]!=0){
            return -1;
        }
        vector<int>dx={0,0,-1,1,-1,1,-1,1};
        vector<int>dy={-1,1,0,0,-1,1,1,-1};
        vector<vector<int>>vis(n,vector<int>(m,-1));
        queue<pair<int,int>>q;
        q.push({0,0});
        int ans=0;
        while(!q.empty()){
            int s=q.size();
            ans++;
            for(int i=0;i<s;i++){
                pair<int,int>p=q.front();
                q.pop();
                int x=p.first;
                int y=p.second;
                if(vis[x][y]!=-1){
                    continue;
                }
                // cout<<x<<" "<<y<<endl;
                if(x==n-1 && y==m-1){
                    return ans;
                }
                vis[x][y]=1;
                for(int i=0;i<8;i++){
                    int nx=x+dx[i];
                    int ny=y+dy[i];
                    if(nx>=0 && nx<n && ny>=0 && ny<m){
                        if(vis[nx][ny]==-1 && grid[nx][ny]==0){
                            q.push({nx,ny});
                        }
                    }
                }
            }
        }
        return -1;
    }
};
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