<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>

#define pb push_back
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
#define sz(v) (int)v.size()
#define deb(x) cout<<#x<<"="<<x<<endl;
#define pii pair<int,int>
#define vi vector<int>

class Solution {
public:
    int bfs(vector<vector<int>> graph,int st)
    {
        int n=graph.size();
        int mask=(1LL<<n);
        vector<vector<int>> dp(n+1,vector<int>(mask+1));
        queue<pii> q; 
        q.push({st,(1LL<<st)});
        dp[st][(1LL<<st)]=1;
        while(!q.empty())
        {
            int node=q.front().ff;
            int curr_mask=q.front().ss;
            q.pop();
            for(auto &it:graph[node])
            {
                int new_mask=(curr_mask | (1LL<<it));
                if(dp[it][new_mask]) continue;
                q.push({it,new_mask});
                dp[it][new_mask]=dp[node][curr_mask]+1;
                if(new_mask==mask-1)
                {
                    return dp[it][new_mask];
                }
            }
        }
        return -1;
    }

    int shortestPathLength(vector<vector<int>>& graph) 
    {
        int n=graph.size();
        int masks=(1LL<<n);
        int mn_ans=1e9;
        for(int i=0;i<n;i++)
        {
            int curr_ans=bfs(graph,i);
            if(curr_ans!=-1) mn_ans=min(mn_ans,curr_ans);
        }
        if(mn_ans==1e9) return 0;
        return mn_ans-1;
    }
};
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