<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();
        int mval=(1<<n)-1;
        // cout<<mval<<endl;

        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>>pq;
        // priority_queue<pair<int,pair<int,int>>>pq;


        int ans=INT_MAX;
        for(int i=0; i<n; i++){
            int val=(1<<i);
            pq.push({0,{i,val}});
            map<pair<int,int>,int>mp;
            mp[{i,val}]++;
            while(!pq.empty()){
                int count=pq.top().first;
                int node=pq.top().second.first;
                int val=pq.top().second.second;
                pq.pop();

                if(val==mval){
                    ans=min(ans,count);
                }

                for(auto x:graph[node]){
                    int nval=val | (1<<x);
                    if(mp[{x,nval}]>0) continue;

                    pq.push({count+1,{x,nval}});
                    mp[{x,nval}]++;
                }
            }
        }

        return ans;
    }
};
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