<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();
         vector<vector<int>> dist(n,vector<int>(n,1e9));  
          vector<vector<pair<int,int>>> par(n, vector<pair<int,int>>(n, {0,0}));

        queue<vector<int>> q;
        if(!grid[0][0]){q.push({0,0,0,0});
        dist[0][0] = 0;  }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++) par[i][j]={i,j};
        }
        vector<int> xx={0,0,1,-1,1,-1,1,-1};
        vector<int> yy={1,-1,0,0,1,-1,-1,1};
        while (!q.empty()) {
           int row=q.front()[0];
           int col=q.front()[1];
           int d=q.front()[2];
            q.pop();
            for (int i=0;i<8;i++) {
                int r=xx[i]+row;
                int c=yy[i]+col;
                if(r<n && r>-1 && c>-1 && c<n && grid[r][c]==0){
                      if(dist[r][c]>d+1){
                        dist[r][c]=d+1;
                        q.push({r,c,d+1});
                        par[r][c]={row,col};
                      }
                }
            }
        }
        
        if(dist[n-1][n-1]==1e9) return -1;
        else return dist[n-1][n-1]+1;


    }
};
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