<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] != 0) return -1;
        int n = grid.size();
        
        vector<vector<int>> vis(n,vector<int>(n,0));
        queue<pair<int,pair<int,int>>> q;

        q.push({1,{0,0}});
        vis[0][0] = 1;
        while(!q.empty()){
           auto it = q.front();
           q.pop();

           int dis = it.first;
           int x = it.second.first;
           int y = it.second.second;
           if(x == n-1 && y==n-1) return dis;
           vector<int> del={1,0,-1};

           for(int i=0; i<3; i++){
            for(int j=0; j<3; j++){
                int nx = x+del[i];
                int ny = y+del[j];
               
                if(nx < 0 || ny < 0 || nx == n || ny == n || vis[nx][ny]!=0 || grid[nx][ny]!=0){
                    continue;
                }
                else{
                    vis[nx][ny] = 1;
                    if(nx == n-1 && ny == n-1){return dis+1;}
                    else q.push({dis+1,{nx,ny}});

                    vis[nx][ny] = 1;
                }
            }
           }
        }
        return -1;
    }
};
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