<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) {
        if(grid[0][0]==1){
            return -1;
        }
        int n = grid.size();
        if(grid[0][0]==0 && n==1){
            return 1;
        }
        vector<vector<int>> vis(n, vector<int> (n, 0));
        vector<vector<int>> dis(n, vector<int> (n, 1e9));
        queue<pair<int, int>> q;
        vis[0][0] = 1;
        dis[0][0] = 1;
        q.push({0, 0});
        int d1[8] = {0, 0, 1, -1, -1, 1, -1, 1};
        int d2[8] = {1, -1, 0, 0, -1, 1, 1, -1};
        while(!q.empty()){
            pair<int, int> p = q.front();
            q.pop();
            for(int i=0; i<8; i++){
                int nx = p.first+d1[i];
                int ny = p.second+d2[i];
                if(nx==n-1 && ny==n-1 && grid[nx][ny]==0){
                    return dis[p.first][p.second]+1;
                }
                if(nx>=0 && ny>=0 && nx<n && ny<n && grid[nx][ny]==0 && vis[nx][ny]==0){
                    q.push({nx, ny});
                    vis[nx][ny] = 1;
                    dis[nx][ny] = dis[p.first][p.second]+1;
                }
            }
        }
        return -1;
    }
};
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