<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 (n-1 == 0 && grid[0][0] == 0)
            return 1;
        if (grid[0][0] == 1 || grid[n-1][n-1] == 1)
            return -1;
        // queue<pair<int,pair<int,int>>>q;
        queue<pair<int,int>>q;
        vector<vector<int>>dis(n,vector<int>(n,1e9));
        dis[0][0]=1;
        q.push({0,0});

        while(q.size()>0){
            vector<int>drow={-1,-1,0,1,1,1,0,-1};
            vector<int>dcol={0,1,1,1,0,-1,-1,-1};
            int row=q.front().first;
            int col=q.front().second;
            q.pop();

            for(int i=0;i<8;i++){
                int nrow=drow[i]+row;
                int ncol=dcol[i]+col;
                if(nrow>=0 && ncol>=0 && nrow<n && ncol<n && grid[nrow][ncol]==0){
                  if(dis[row][col]+1 < dis[nrow][ncol]){
                    dis[nrow][ncol]=dis[row][col]+1;
                    if(nrow==n-1 && ncol==n-1){
                        return dis[n-1][n-1];
                    }
                    q.push({nrow,ncol});
                  }  
                }
            }

        }
        return -1;
    }
};
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