<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) {
    const int n = graph.size();
    const int goal = (1 << n) - 1;

    int ans = 0;
    queue<pair<int, int>> q;  // (u, state)
    vector<vector<bool>> seen(n, vector<bool>(1 << n));

    for (int i = 0; i < n; ++i)
      q.emplace(i, 1 << i);

    while (!q.empty()) {
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [u, state] = q.front();
        q.pop();
        if (state == goal)
          return ans;
        if (seen[u][state])
          continue;
        seen[u][state] = true;
        for (const int v : graph[u])
          q.emplace(v, state | (1 << v));
      }
      ++ans;
    }

    return -1;
  }
};
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