<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();
        
        // Queue use kar rahe hain BFS traversal ke liye
        queue<pair<int, pair<int, int>>> q; // {visited_nodes, {current_node, distance}}
        
        // Har node se shuru karne ke liye initial state queue mein dal rahe hain
        for (int i = 0; i < n; ++i) {
            // Starting state: Sirf current node visit hui hai (1 << i)
            q.push({1 << i, {i, 0}});
        }
        
        // Har visited state ko track karne ke liye set use kar rahe hain
        unordered_set<string> visited;
        for (int i = 0; i < n; ++i) {
            visited.insert(to_string(1 << i) + " " + to_string(i));
        }
        
        // BFS traversal
        while (!q.empty()) {
            auto [visited_nodes, p] = q.front();
            q.pop();
            int current_node = p.first, distance = p.second;
            
            // Agar saare nodes visit ho chuke hain
            if (visited_nodes == (1 << n) - 1) return distance;
            
            // Current node ke saare neighbors ko check karte hain
            for (int neighbor : graph[current_node]) {
                // Naya state banate hain by visiting this neighbor
                int new_visited_nodes = visited_nodes | (1 << neighbor);
                string state = to_string(new_visited_nodes) + " " + to_string(neighbor);
                
                // Agar yeh state pehle visit nahi hui, toh queue mein daalte hain
                if (visited.find(state) == visited.end()) {
                    q.push({new_visited_nodes, {neighbor, distance + 1}});
                    visited.insert(state);
                }
            }
        }
        
        return -1; // Yeh theoretically kabhi nahi hoga agar graph connected hai
    }
};

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