<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:
    vector<int>dx = {0,1,1,1,0,-1,-1,-1};
    vector<int>dy = {1,1,0,-1,-1,-1,0,1};
    bool isValid(int x, int y, int n, int m, vector<vector<int>>&vis,vector<vector<int>>& grid){
        if(x<0 || y<0 || x>=n || y>=m || vis[x][y]==1 || grid[x][y]==1)
            return false;
        return true;
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        if(grid[0][0]==1 || grid[n-1][m-1]==1)
            return -1;
        vector<vector<int>>vis(n,vector<int>(m)), dis(n,vector<int>(m,1e9));
        queue<vector<int>>q;
        dis[0][0]=1;
        q.push({1,0,0});
        while(!q.empty()){
            auto vec = q.front();
            if(vec[1]==n-1 && vec[2]==m-1)
                return vec[0];
            vis[vec[1]][vec[2]]=1;
            q.pop();
            for(int i=0; i<8; ++i){
                int newX = vec[1]+dx[i];
                int newY = vec[2]+dy[i];
                if(isValid(newX,newY,n,m,vis,grid) && dis[newX][newY] > 1+vec[0]){
                    dis[newX][newY] = 1+vec[0];
                    vector<int>tmp = {1+vec[0],newX,newY};
                    q.push(tmp);
                }
            }
        }
        return -1;
    }
};
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