<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();
        if(grid[0][0]==1)
        return -1;
        vector<int>adj[n*n];
        for(int i=0;i<n;i++)
        {
            int y=i*n;
            for(int j=0;j<n;j++)
            {
                if(((y+j)%n)>0 && grid[i][j-1]==0)
                adj[y+j].push_back(y+j-1);
                if(((y+j)%n)!=n-1 && grid[i][j+1]==0)
                adj[y+j].push_back(y+j+1);
                if((y+j)>n-1)
                {
                    if(grid[i-1][j]==0)
                    adj[y+j].push_back(y+j-n);
                    if(((y+j)%n)>0 && grid[i-1][j-1]==0)
                    adj[y+j].push_back(y+j-n-1);
                    if(((y+j)%n)!=n-1 && grid[i-1][j+1]==0)
                    adj[y+j].push_back(y+j-n+1);
                }
                if((y+j)<(n-1)*n)
                {
                    if(grid[i+1][j]==0)
                    adj[y+j].push_back(y+j+n);
                    if(((y+j)%n)>0 && grid[i+1][j-1]==0)
                    adj[y+j].push_back(y+j+n-1);
                    if(((y+j)%n)!=n-1 && grid[i+1][j+1]==0)
                    adj[y+j].push_back(y+j+n+1);
                }
            }
        }
        vector<int>dis(n*n,INT_MAX);
        dis[0]=1;
        queue<int>q;
        q.push(0);
        while(!q.empty())
        {
            int p=q.front();
            q.pop();
            for(auto it : adj[p])
            {
                if(dis[it]>dis[p]+1)
                {
                    dis[it]=dis[p]+1;
                    q.push(it);
                }
            }
        }
        return dis[n*n-1]==INT_MAX ? -1:dis[n*n-1];
        
        
    }
};
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