<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 dx[] = {1,-1,0,0,1,-1,-1,1};
        int dy[] = {0,0,1,-1,1,-1,1,-1};
        int n = grid.size();
        //dikstra's algorithm question
        vector<vector<int>>dist(n,vector<int>(n,INT_MAX));
        set<vector<int>> s;
        dist[0][0] = 1;
        s.insert({0,0,1});
        while(!s.empty()){
            auto it = s.begin();
            auto arr = *it;
            int cx = arr[0];
            int cy = arr[1];
            int distTillNow = arr[2];
            s.erase(it);
            for(int i=0;i<8;i++){
                int nx = cx + dx[i];
                int ny = cy + dy[i];
                if(nx>=0 && nx<n && ny>=0 && ny<n){
                    if(grid[nx][ny]==0){
                        if(distTillNow + 1 < dist[nx][ny]){
                            if(s.find({nx,ny,dist[nx][ny]})!=s.end()){
                                s.erase({nx,ny,dist[nx][ny]});
                            }
                            dist[nx][ny] = distTillNow + 1;
                            s.insert({nx,ny,dist[nx][ny]});
                        }
                    }
                }
            }
        }
        if(dist[n-1][n-1]!=INT_MAX) return dist[n-1][n-1];
        return -1; 
    }
};
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