<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:
    bool chk (string s)
    {
        int i =0;
        int n = s.size();
        while(i<n)
        {
            if(s[i]=='0')return 0;
            i++;
        }
        return 1;
    }



    int shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
        queue<pair<int,string>>q;
        string s = "";
        int k = 0;
        while(k<n)
        {
            s+='0';
            k++;
        }
        map<pair<int,string>,int>mp;
        for(int i=0;i<n;i++)
        {
            s[i] = '1';
            if(mp.find({i,s})==mp.end())q.push({i,s});
            mp[{i,s}] = 1;
            s[i] = '0';
        }
        int ans = 0;
        k = 3;
        while(1)
        {
            int cnt = q.size();
            // cout << cnt << " 1000 " << endl;
            while(cnt--)
            {
                auto tp = q.front();
                q.pop();
                int src = tp.first;
                string s = tp.second;
                // cout << src << " " << ans << " " << s << endl;
                if(chk(s))return ans;
                for(auto i:graph[src])
                {
                    char temp = s[i];
                    s[i] = '1';
                    if(mp.find({i,s})==mp.end())q.push({i,s});
                    mp[{i,s}] = 1;
                    s[i] = temp;
                }
            }
            ans++;
        }
        // return 5;
    }
};
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