<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();
        int mask = (1<<n) - 1;
        queue<pair<int,pair<int,int>>> q;

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

        for (int node = 0; node < n; ++node) {
            int origMask = (1 << node);
            q.push({node, {origMask, 1}});
            visited[origMask][node] = true;
        }

        
        while (!q.empty()) {
            auto curr = q.front();
            q.pop();

            int currNode = curr.first;
            int currMask = curr.second.first;
            int currLength = curr.second.second;

            if (currMask == mask)
                return currLength - 1;

            for (int i = 0; i < graph[currNode].size(); ++i) {
                int neighbor = graph[currNode][i];
                int newMask = currMask | (1 << neighbor);

                if (visited[newMask][neighbor])
                    continue;

                q.push({neighbor, {newMask, currLength + 1}});
                visited[newMask][neighbor] = true;
            }
        }
        return -1;  
    }
};
18673