<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:
//go to anode and look for any node in adj of the nodes that are currently visited 
// if there are no such nodes possible then it might mean that all nodes have been visited already because the graph is already connected 
    
    int solve(int curr,int vis,vector<vector<int>>& graph,vector<vector<int>>& dp){
        int n=graph.size();
        // vis[curr]=1;
        vis=vis|(1<<curr);
        int cnt=0;
        for(int i=0;i<n;i++) if(!(vis&(1<<i))) cnt++;
        if(cnt==0) return 0;
        if(dp[curr][vis]!=-1) return dp[curr][vis];
        
        //calculate the distance of each of these nodes from the current node using only already visited nodes 
        // can we do it with dfs or bfs simple?
        // should we use djikastra here?
        vector<int> dist(n,100);
        set<pair<int,int>> st;
        st.insert({0,curr});
        dist[curr]=0;
        while(!st.empty()){
            auto p=*st.begin();
            st.erase(p);
            int cost=p.first;
            int nod=p.second;
            for(auto x: graph[nod]){
                if(vis&(1<<x) && dist[x]>cost+1){
                    st.erase({dist[x],x});
                    dist[x]=cost+1;
                    st.insert({dist[x],x});
                }
            }
        }
        // now go to all nodes, if it is visited , go to all its adjacency which are not visited yet and then go to them
        int ans=1e9;
        for(int i=0;i<n;i++){
            if(vis&(1<<i)){
                for(auto x: graph[i]){
                    if(!(vis&(1<<x))) ans=min(ans,dist[i]+1+solve(x,vis,graph,dp));
                }
            }
        }
        return dp[curr][vis]=ans;
        

    }
    int shortestPathLength(vector<vector<int>>& graph) {
        // return solve(0,0,graph);
        int n=graph.size();
        int ans=1e9;
        vector<vector<int>> dp(n,vector<int>(pow(2,n),-1));
        for(int i=0;i<n;i++)
        {
            ans=min(ans,solve(i,0,graph,dp));
        }
        return ans;
    }
};
54918