<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 bfs(int node,int mask,int n,vector<int>adj[])
    {
        int cnt=0;
        mask|=1<<node;
        queue<pair<int,int>>q;
        q.push({node,mask});
        vector<vector<int>>vis(n,vector<int>((1<<n),0));
        vis[node][mask]=1;
        while(!q.empty())
        {
            int sz=q.size();
            for(int i=0;i<sz;i++)
            {
               auto it=q.front();
               q.pop();
               int u=it.first;
               int bitmask=it.second;
               if(bitmask==((1<<n)-1)) return cnt;
               for(auto v:adj[u])
               {
                  int newBitMask=bitmask|(1<<v);
                  if(vis[v][newBitMask]==1) continue;
                  q.push({v,newBitMask});
                  vis[v][newBitMask]=1;
               }  
            }
            cnt++;
        }
        return -1;
    }
    int shortestPathLength(vector<vector<int>>& graph) {
        int n=graph.size();
        vector<int>adj[n];
        for(int i=0;i<n;i++)
        {
            for(auto it:graph[i]) adj[i].push_back(it);
        }
        int mini=INT_MAX;
        for(int i=0;i<n;i++)
        {
            mini=min(mini,bfs(i,0,n,adj));
        }
        return mini;
    }
};
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