<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) {
        set<pair<int,pair<int,int>>> s;
        int n=grid.size();
        if(grid[0][0]==1)return -1;
        int ans=1e9;
        s.insert({1,{0,0}});
        
        while(!s.empty()){
            int i=(*s.begin()).second.first;
            int j=(*s.begin()).second.second;
            int d=(*s.begin()).first;
            //cout<<i<<j<<d;
            s.erase(s.begin());
            grid[i][j]=1;
            if(i==n-1 && j==n-1)ans=min(ans,d);
            if(i>0 && grid[i-1][j]==0){
                s.insert({d+1,{i-1,j}});
            }
            if(i>0 && j>0 && grid[i-1][j-1]==0){
                s.insert({d+1,{i-1,j-1}});
            }
            if(i>0 && j<n-1 && grid[max(i-1,0)][j+1]==0){
                s.insert({d+1,{i-1,j+1}});
            }
            if(j>0 && grid[i][j-1]==0){
                s.insert({d+1,{i,j-1}});
            }
            if(i<n-1 && j>0 && grid[i+1][j-1]==0){
                s.insert({d+1,{i+1,j-1}});
            }
            if(i<n-1 && j<n-1 && grid[i+1][j+1]==0){
                s.insert({d+1,{i+1,j+1}});
            }
            if(i<n-1 && grid[i+1][j]==0){
                s.insert({d+1,{i+1,j}});
            }
            if(j<n-1 && grid[i][j+1]==0){
                s.insert({d+1,{i,j+1}});
            }
        }
        if(ans==1e9)return -1;
        return ans;

    }
};
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