<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>>& adj_vec) {
        const int N = adj_vec.size();
        int all_visited_state = (1 << N) - 1;
        queue<vector<int>> q;
        unordered_set<string> seen;
        for(int i = 0; i < N; i++) {
            q.push({i, (1 << i)});
            string start_n_visited = to_string(i) + ":" + to_string((1 << i));
            seen.insert(start_n_visited);
        }
        
        int ans = 0;
        while(!q.empty()) {
            int temp_size = q.size();
            for(int i = 0; i < temp_size; i++) {
                vector<int> start_n_visited = q.front();
                q.pop();
                if(start_n_visited[1] == all_visited_state) return ans;
                for(auto& adj_node : adj_vec[start_n_visited[0]]) {
                    int new_state = start_n_visited[1] | (1 << adj_node);
                    string new_start_n_visited = to_string(adj_node) + ":" + to_string(new_state);
                    if(seen.count(new_start_n_visited)) continue;
                    seen.insert(new_start_n_visited);
                    q.push({adj_node, new_state});
                }
            }
            ans++;
        }
        return ans;
    }
};
37619