<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();
        vector<vector<int>> adj(n);
        for (int i = 0; i < n; i++) {
            for (int x : graph[i]) {
                adj[x].push_back(i);
                adj[i].push_back(x);
            }
        }

        queue<pair<int, pair<int, int>>> q;
        for (int i = 0; i < n; i++) {
            q.push(make_pair(0, make_pair(i, 1 << i)));
        }

        vector<vector<bool>> visited(n, vector<bool>(1 << n, false));

        while (!q.empty()) {
            int level = q.front().first;
            int node = q.front().second.first;
            int state = q.front().second.second;
            q.pop();

            if (visited[node][state]) continue;
            visited[node][state] = true;

            if (state == ((1 << n) - 1)) return level;

            for (auto x : adj[node]) {
                q.push(make_pair(level + 1, make_pair(x, state | 1 << x)));
            }
        }
        
        return -1;
    }
};
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