<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) {
        // If there's only one node, no paths are needed
        if (graph.size() == 1) return 0;

        // Create a string to represent all nodes visited ('1' for visited, '0' for not visited)
        string fullPath (graph.size() , '1');

        // Queue to perform BFS and track current state of visited nodes
        queue <pair <int, string>> q;

        string currPath (graph.size() , '0'); // Initially no nodes are visited

        set <pair <int, string>> visited; // Set to track visited nodes with their respective states

        // Initialize BFS from each node individually
        for (int i = 0 ; i < graph.size(); i++) {
            currPath[i] = '1'; // Mark current node as visited
            q.push ({i, currPath}); // Push node index and current state to the queue
            visited.insert ({i, currPath}); // Mark this state as visited
            currPath[i] = '0'; // Reset current node to unvisited for next iteration
        }

        int steps = 0; // Step counter to track distance
        int node; // Current node index
        char tempChar; // Temporary character to store original state of visited node
        
        // Perform BFS
        while (! q.empty()) {
            int size = q.size();

            while (size --) {
                node = q.front().first; // Current node index
                currPath = q.front().second; // Current state of visited nodes
                q.pop();

                // Explore neighbors of current node 'node'
                for (int i = 0 ; i < graph[node].size() ; i++) {
                    tempChar = currPath[graph[node][i]]; // Store original state of the neighbor node
                    currPath[graph[node][i]] = '1'; // Mark the neighbor node as visited

                    // If all nodes are visited, return the current step count
                    if (currPath == fullPath) {
                        return steps + 1;
                    }

                    // If this state hasn't been visited before, enqueue it
                    if (! visited.count({graph[node][i], currPath})) {
                        q.push({graph[node][i], currPath});
                        visited.insert ({graph[node][i], currPath});
                    }

                    currPath[graph[node][i]] = tempChar; // Restore the state of the neighbor node
                }
            }

            steps ++; // Increment step count after processing each level in BFS
        }

        return -1; // Return -1 if no shortest path found        
    }
};
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