<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 {
    private:
    struct pair_hash {
    template <class T1, class T2>
    size_t operator() (const pair<T1, T2>& pair) const {
        return hash<T1>()(pair.first) ^ hash<T2>()(pair.second);
    }
};


public:
    int shortestPathLength(vector<vector<int>>& graph) {
        int  n = graph.size();
        if(n == 1){
            return 0;

        }
        queue<tuple<int,int,int>>q;
        unordered_set<pair<int, int>, pair_hash> vis;

        for(int i =1 ; i < n  ;i++){
            int mask = (1 << i);
            q.push({i , mask , 0});
            vis.insert({i , mask});

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

            if(mask == (1 << n) - 1){
                return dist;
            }
            for(int neighbour : graph[node]){
                int next_mask = mask | (1 << neighbour);
               if (!vis.count({neighbour, next_mask})) {  
                vis.insert({neighbour, next_mask});
                q.push({neighbour, next_mask, dist + 1});
            }
            }
            
        }
        return -1;
        
    }
};
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