<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;

        queue<vector<int>> q;  // To store {cost, node, mask}
        set<pair<int, int>> visited;  // To store {node, mask}

        // Initialize the queue with all nodes as starting points
        for (int i = 0; i < n; i++) {
            int mask = (1 << i);  // Initial mask for the node
            q.push({0, i, mask}); // {cost, node, mask}
            visited.insert({i, mask}); // {node, mask}
        }

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

            int cost = vec[0];
            int node = vec[1];
            int mask = vec[2];

            // If we have visited all nodes
            if (mask == (1 << n) - 1) {
                return cost;  // Return the minimum cost to visit all nodes
            }

            // Explore the neighbors
            for (int neighbor : graph[node]) {
                int newMask = mask | (1 << neighbor); // Update the mask
                if (visited.find({neighbor, newMask}) == visited.end()) {
                    visited.insert({neighbor, newMask}); // Mark as visited
                    q.push({cost + 1, neighbor, newMask}); // Push next state
                }
            }
        }

        return -1;  // Should never reach here
    }
};

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