<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>

int dix(vector<vector<int>> &grid){
    int n = grid.size();
    vector<vector<int>> dist(n,vector<int> (n,INT_MAX));
    if(grid[0][0] == 1) return -1;

   set<vector<int>> s;
    s.insert({0,0,1});
    dist[0][0] = 1;
    vector<int> drow = {-1,1,0,0,-1,-1,1,1};
    vector<int> dcol = {0,0,-1,1,-1,1,-1,1};

    while(!s.empty()){
        auto it = *s.begin();
        int d = it[2];
        int xcord = it[0];
        int ycord = it[1];
        s.erase(it);

        for(int i=0;i<8;i++){
            int newr = xcord+drow[i];
            int newc = ycord+dcol[i];

            if(newr>=0 && newc>=0 && newr<n && newc<n &&grid[newr][newc] == 0){
                if(dist[newr][newc] > 1+d){
                   s.insert({newr,newc,1+d});
                    dist[newr][newc] = 1+d;               
                    }
            }
        }
    }
    if(dist[n-1][n-1] == INT_MAX){
            return -1;
        }
    else return dist[n-1][n-1];
}

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        return dix(grid);
    }
};


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