<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 shortestPathFromNode(int start,int n,int curMask,vector<vector<int>>& graph,unordered_map<int,unordered_map<int,int>> &vis){
             queue<pair<int,int>> q;
             vis[start][curMask]=1;
             q.push({start,curMask});
             int steps=0;
             while(!q.empty()){
                    int sz=q.size();
                    while(sz--){
                        auto src=q.front();
                        q.pop();
                        int node=src.first;
                        int mask=src.second;
                        if(mask==(1<<n)-1) return steps;
                        for(auto adjNode:graph[node]){
                            int nextMask=mask|(1<<adjNode);
                             if(vis[adjNode][nextMask]!=1){
                                 vis[adjNode][nextMask]=1;
                                 q.push({adjNode,nextMask});
                             }
                        }
                    }
                    steps++;
             }
        return -1;
    }
    int shortestPathLength(vector<vector<int>>& graph) {
        // Basically here we can start with any node and need to find the shortest path      covering all nodes and then take the minimum out of all paths.
        // The main catch here is that we can reuse edges and revisit nodes.so it can't be done using simple BFS and maintaining simple visited array for a node
        // we have to maintain visited for a state which is being repetitively used ,i.e , we can maintain a direction state which denotes whether we have reached any node from same direction as i have reached before
        // hence for this for every node , i have to keep track of parent as a state
        // I can use a bitmask to denote the path i have taken to reach a particular node
        //once all nodes have been visited , the bitmask contains all set bits and BFS ensures that it has been achieved through shortest paths
        int shortestPath=INT_MAX;
        int n=graph.size();
        for(int i=0;i<n;i++){
             unordered_map<int,unordered_map<int,int>> mp;
             int curMask = 0|(1<<i);
            //  cout<<"mask"<<curMask<<endl;
             shortestPath = min(shortestPath,shortestPathFromNode(i,n,curMask,graph,mp));
            //  cout<<"path"<<shortestPath<<endl;
        }
        return shortestPath;
    }
};
57389