<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>

struct hash_pair {
    template <class T1, class T2>
    size_t operator() (const pair<T1, T2>& pair) const {
        return hash<T1>()(pair.first) ^ hash<T2>()(pair.second);
    }
};
class Solution {
public:
    int shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
        if (n == 1) return 0;  // If there's only one node, the path length is 0
        
        // Initialize the BFS queue and the visited set
        queue<tuple<int, int, int>> q; // {current_node, bitmask, steps}
        unordered_set<pair<int, int>, hash_pair> visited; // {current_node, bitmask}
        
        // Initialize the queue with each node as a starting point
        for (int i = 0; i < n; ++i) {
            int bitmask = 1 << i; // Set only the i-th bit
            q.push({i, bitmask, 0});
            visited.insert({i, bitmask});
        }
        
        int all_visited_mask = (1 << n) - 1; // The mask when all nodes have been visited (111...111)
        
        // BFS
        while (!q.empty()) {
            auto [node, mask, steps] = q.front();
            q.pop();
            
            // If all nodes have been visited, return the number of steps
            if (mask == all_visited_mask) return steps;
            
            // Explore the neighbors
            for (int neighbor : graph[node]) {
                int next_mask = mask | (1 << neighbor); // Mark the neighbor as visited
                
                // If this state (neighbor, next_mask) hasn't been visited, add it to the queue
                if (!visited.count({neighbor, next_mask})) {
                    q.push({neighbor, next_mask, steps + 1});
                    visited.insert({neighbor, next_mask});
                }
            }
        }
        
        return -1; // This line should never be reached
    }
};



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