<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) {
        queue<vector<int>> q;
        q.push({0,0});
        int n=grid.size();
        map<int,map<int,bool> >dp;
        if(grid[0][0]){
            return -1;
        }
        dp[0][0]=true;
        vector<vector<int>>vis(n,vector<int>(n,INT_MAX/10));
        vis[0][0]=0;
        while(q.size()){
            int  nn=q.size();
          
                auto it=q.front();
                q.pop();
                dp[it[0]][it[1]]=true;
                vector<int> dx{0,1,0,-1,-1,1,1,-1};
                vector<int> dy{1,0,-1,0,1,-1,1,-1};

                for(int i=0;i<8;i++){
                    int nx=dx[i]+it[0];
                    int ny=dy[i]+it[1];
                    if(nx<0 or ny<0 or nx>=n or ny>=n or grid[nx][ny] or  dp[nx][ny] ){
                        continue;
                    }
                    if(vis[it[0]][it[1]]+1<vis[nx][ny]){
                        vis[nx][ny]=vis[it[0]][it[1]]+1;
                        q.push({nx,ny});
                    }

                
            }
        }
        return (vis[n-1][n-1]==(INT_MAX/10))?-1:1+vis[n-1][n-1];
    }
};
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