<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 dp[n][4100] ; 
        for(int i = 0 ;i<n;i++){
            for(int j =0;j<4100;j++){
                dp[i][j] = 1e9 ; 
            }
        }
        queue<vector<int> >Q ; 
        for(int i = 0 ;i<n;i++){
            dp[i][(1<<i)] = 0 ; 
            Q.push({0,i,(1<<i)}) ;
        }
        while(!Q.empty()){
            auto it = Q.front() ; 
            Q.pop() ; 
            int ind = it[1] ; 
            int mask = it[2] ; 
            int val = it[0] ;
            if(dp[ind][mask]<val) continue ; 
            if((mask+1)==(1<<n)) return val ; 
            for(auto v:graph[ind]){
                int nmask = (mask)|(1<<v) ;
                if(dp[v][nmask]>val+1){
                    dp[v][nmask] = val+1 ; 
                    Q.push({val+1,v,nmask}) ; 
                }
            }
        }
        return 2 ; 
    }
};
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