<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();
        string mask = "";
        string eq = "";
        for(int i=0; i<n; i++){
            mask += '0';
            eq += '1';
        }
        queue<pair<int,string>>q;
        set<pair<int,string>>s;
        for(int i=0; i<n; i++){
            string temp = mask;
            temp[i] = '1';
            q.push({i,temp});
            s.insert({i,temp});
        }
        int c = 0;
        while(!q.empty()){
            int size = q.size();
            for(int i=0; i<size; i++){
                auto top = q.front();
                q.pop();
                if(top.second == eq) return c;
                for(auto p: graph[top.first]){
                    string temp1 = top.second;
                    temp1[p] = '1';
                    if(s.count({p,temp1}) == 0){
                        q.push({p,temp1});
                        s.insert({p,temp1});
                    }
                }
            }
            c++;
        }
        return -1;
    }
};
 // int n = graph.size(),res = 0;
 //    queue<tuple<int,int,int>> q;
 //    vector<vector<int>> seen(n,vector<int>(1<<n));
 //    for(int i=0;i<n;i++){
 //        q.push(tuple<int,int,int>(i,1<<i,0));
 //        seen[i][1<<i] = true;
 //    }
 //    while (!q.empty())
 //    {
 //        auto [idx,mask,dist] = q.front();
 //        q.pop();
 //        if(mask==(1<<n)-1){
 //            res = dist;
 //            break;
 //        }
 //        for(auto v:graph[idx]){
 //            int mask_v = mask|1<<v;
 //            if(!seen[v][mask_v]){
 //                q.push(tuple<int,int,int>(v,mask_v,dist+1));
 //                seen[v][mask_v] = true;
 //            }
 //        }
 //    }
 //    return res;
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