<p>You have an undirected, connected graph of <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given an array <code>graph</code> where <code>graph[i]</code> is a list of all the nodes connected with node <code>i</code> by an edge.</p>

<p>Return <em>the length of the shortest path that visits every node</em>. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest1-graph.jpg" style="width: 222px; height: 183px;" />
<pre>
<strong>Input:</strong> graph = [[1,2,3],[0],[0],[0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [1,0,2,0,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest2-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> graph = [[1],[0,2,4],[1,3,4],[2],[1,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [0,1,4,2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == graph.length</code></li>
	<li><code>1 &lt;= n &lt;= 12</code></li>
	<li><code>0 &lt;= graph[i].length &lt;&nbsp;n</code></li>
	<li><code>graph[i]</code> does not contain <code>i</code>.</li>
	<li>If <code>graph[a]</code> contains <code>b</code>, then <code>graph[b]</code> contains <code>a</code>.</li>
	<li>The input graph is always connected.</li>
</ul>

class Solution {
public:
    int shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<int>adj[n];
        int ans=0;
        for(int i=0;i<n;i++)
        {
            int from = i;
            for(auto to:graph[i])
            {
                adj[i].push_back(to);
            }
        }

        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>>pq;
        unordered_map<int,unordered_map<int,bool>>vis;
        for(int i=0;i<n;i++)
        {
            pq.push({0,i,1<<i});
            vis[i][1<<i]=true;
        }
        int req_mask = (1<<n)-1;
        cout<<req_mask<<endl;
        while(!pq.empty())
        {
            auto it =  pq.top();
            pq.pop();

            int node = it[1];
            int dist = it[0];
            int mask = it[2];

            for(auto ele:adj[node])
            {
                int new_mask = mask | 1<<ele;

                if(new_mask==req_mask)
                {
                    return dist+1;
                }
                if(vis[ele][new_mask])
                    continue;
                vis[ele][new_mask]=true;
                pq.push({dist+1,ele,new_mask});
            }
        }
        return ans;
    }
};
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