<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<vector<int>> vis(n,vector<int>(1<<n,0));
        queue<vector<int>> q;
        // q.push({i,(1<<i),0});
        // vis[i][1<<i]=1;
        for(int i=0;i<n;i++)
        {
            q.push({i,1<<i});
            vis[i][1<<i]=1;
        }
        int lvl=0;
        while(!q.empty())
        {
            int sz=q.size();
            for(int x=0;x<sz;x++)
            {
                auto it=q.front();
                q.pop();
                int node=it[0];
                int mask=it[1];
                
                // int s=it[2];
                if(mask+1==(1<<n))
                return lvl;
                for(auto idx:graph[node])
                {
                    if(!vis[idx][mask|(1<<idx)])
                    {
                        vis[idx][mask|(1<<idx)]=1;
                        q.push({idx,mask|(1<<idx)});
                    }
                }
            }
            lvl++;

        }
        return -1;
        
    }
};
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