<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 target = (1 << n) -1;
        
        vector<int> ans;
        for(int i = 0 ; i < n ; i++){
            queue<pair<int, int>> q;
            vector<vector<int>> dp(n, vector<int>(target+1, 0));

            q.push({i, (1 << i)});
            dp[i][1 << i] = true;


            int steps = 0;

            while(!q.empty()){
                for(int j = q.size() ; j > 0 ; j--){
                    int node = q.front().first;
                    int val = q.front().second;
                    q.pop();

                    if(val == target){
                        ans.push_back(steps);
                        break;
                    }

                    for(auto it: graph[node]){
                        int newVal = val | (1 << it);
                        if(dp[it][newVal]) continue;
                        dp[it][newVal] = true;
                        q.push({it, newVal});
                    }
                }

                steps++;
            }
        }

        return *min_element(ans.begin(), ans.end());

    }
};
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