<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 shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
    if (n == 1) return 0;  // If there is only one node, the shortest path is 0.

    // A queue that stores tuples of (current node, visited bitmask, path length).
    queue<tuple<int, int, int>> q;
    // A set to track visited states: (node, bitmask)
    unordered_set<string> visited;

    // Initialize the queue with each node as a starting point.
    for (int i = 0; i < n; ++i) {
        int bitmask = 1 << i;  // Mark the ith node as visited.
        q.push({i, bitmask, 0});
        visited.insert(to_string(i) + "_" + to_string(bitmask));  // Mark initial state as visited.
    }

    // BFS traversal
    while (!q.empty()) {
        auto [node, bitmask, dist] = q.front();
        q.pop();

        // If all nodes have been visited, return the current path length.
        if (bitmask == (1 << n) - 1) {
            return dist;
        }

        // Explore the neighbors
        for (int neighbor : graph[node]) {
            int new_mask = bitmask | (1 << neighbor);  // Mark the neighbor as visited.
            string state = to_string(neighbor) + "_" + to_string(new_mask);
            if (visited.find(state) == visited.end()) {
                q.push({neighbor, new_mask, dist + 1});
                visited.insert(state);  // Mark this state as visited.
            }
        }
    }

    return -1;
    }
};
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