<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) {
        unordered_set<string> seen;
        // node, visited nodes, dist
        using qtuple = tuple<int,int,int>;
        int maxNodes = graph.size();
        queue<qtuple> q;
      
        for(int i=0;i<graph.size();i++)
            q.push({0,(1<<i),i});

        while(!q.empty()){
            auto [dist, currVisited, node]  = q.front();
            q.pop();

           // cout<<node<<" "<<dist<<" "<<currVisited<<endl;

            if(currVisited == ((1<<maxNodes)-1))
                return dist;
            
            if(seen.count(to_string(node) + " " + to_string(currVisited))>0)
                continue;
            
            seen.insert(to_string(node) + " " + to_string(currVisited));

            for(auto x:graph[node]){
                // if not seen, mark x node as visited
                // push to the queue with increased dist

                if(seen.count(to_string(x) + " " + to_string(currVisited)))
                    continue;

                int newVisited = (currVisited | (1<<x));
             //   cout<<"new push "<<x<<" "<<newVisited<<endl;
                q.push({dist+1,newVisited,x});                
            }

        }

        return -1;
    }

};
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