<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 {
    vector<vector<int>>adj;
    int n;
public:

    int solve(int src){
        queue<pair<int,pair<int,int>>>q;
        vector<vector<int>>vis(n,vector<int>(1<<n,0));
        int mask = 1<<src;
        q.push({0,{src,mask}});
        vis[src][mask]=1;

        while(!q.empty()){
            int s = q.front().first;
            int node = q.front().second.first;
            int m = q.front().second.second;
            q.pop();
            if(m==(1<<n)-1){
                return s;
            }
            for(auto it:adj[node]){
                if(vis[it][m|(1<<it)]==0){
                    q.push({s+1,{it,m|(1<<it)}});
                    vis[it][m|(1<<it)]=1;
                }
            }
        }
        return -1;

    }

    int shortestPathLength(vector<vector<int>>& graph) {
        // queue<pair<int,pair<int,int>>q;
        n=graph.size();
        adj=graph;
        int ans=1e9;
        for(int i=0 ; i<n ; i++){
            ans=min(ans,solve(i));
        }
        return ans;
    }
};
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