<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();
        if(grid[0][0] != 0 || grid[n-1][n-1] != 0) return -1;
        
        int drow[] = {-1,0,1,0,-1,1,1,-1};
        int dcol[] = {0,-1,0,1,-1,1,-1,1};
        map<pair<int,int>,int> mp;
        int temp = 0;

        for(int i = 0 ; i < n ; i++)
        {
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == 0)
                {
                    mp[{i,j}] = temp;
                    temp++;
                }
            }
        }

        vector<vector<int>> adj(temp);

        for(int i = 0 ; i < n ; i++)
        {
            for(int j = 0 ; j < n ; j++)
            {
                if(grid[i][j] == 1) continue;
                int v = mp[{i,j}];
                for(int k = 0 ; k < 8 ; k++)
                {
                    int row = i+drow[k];
                    int col = j+dcol[k];
                    if(row >= 0 && col >= 0 && row < n && col < n)
                    {
                        if(grid[row][col] == 0)
                        {
                            adj[v].push_back(mp[{row,col}]);
                        }
                    }
                }
            }
        }

        vector<int> dist(temp,1e9);
        priority_queue<pair<int,int> , vector<pair<int,int>> , greater<pair<int,int>>> pq;
        pq.push({0,0});
        dist[0] = 0;
        while(!pq.empty())
        {
            int dis = pq.top().first;
            int node = pq.top().second;
            pq.pop();
            for(auto it : adj[node])
            {
                if(1+dis < dist[it])
                {
                    dist[it] = 1+dis;
                    pq.push({dist[it] , it});
                }
            }
        }
        if(dist[temp-1] == 1e9) return -1;
        return dist[temp-1]+1;
    }
};
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