<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) {
        queue<pair<int,pair<int,int>>>q;
        int n = graph.size();
        set<pair<int,int>> st;
        if(n==1)
        return 0;
        int maskAll = (1<<n) -1 ;

        for(int i=0;i<n;++i)
        {
            int mask = 1<<i;
            q.push({i,{0,mask}});
            st.insert({i,mask});
        }
        while(!q.empty()){
            auto element = q.front();
            q.pop();
            int node = element.first;
            int dist = element.second.first;
            int mask = element.second.second;
            for(auto neighbour : graph[node]){
                int newMask = (mask | (1<<neighbour));
                if(newMask ==maskAll){
                    return dist +1;
                }
                if(st.count({neighbour,newMask})){
                    continue;
                }
                st.insert({neighbour,newMask});
                q.push({neighbour,{dist+1,newMask}});
            }
        }
        return 0;
    }
};
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