<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 ax[8]={0,0,1,1,-1,-1,1,-1};
    int ay[8]={1,-1,1,-1,1,-1,0,0};
    int n,m;
    vector<vector<int>>gr;
    int check(int x,int y)
    {
        return (x>=0&&x<n&&y>=0&&y<m&&gr[x][y]==0);
    }
    
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        gr=grid;
        n=grid.size(),m=grid[0].size();
        if(grid[0][0]==1)return -1;
        queue<pair<int,int>>q;
        q.push({0,0});
        vector<vector<int>>dist(n,vector<int>(n,1e9));
        dist[0][0]=1;
        while(!q.empty())
        {
            int x=q.front().first,y=q.front().second;q.pop();
            for(int i=0;i<8;i++)
            {
                if(check(x+ax[i],y+ay[i])&&dist[x+ax[i]][y+ay[i]]>1+dist[x][y]){
                    dist[x+ax[i]][y+ay[i]]=1+dist[x][y];
                    q.push({x+ax[i],y+ay[i]});
                }
            }
        }
        
        return (dist[n-1][n-1]==1e9 ? -1 : dist[n-1][n-1]);
    }
};
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