<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 mn = INT_MAX , n = graph.size();

        for(int i = 0 ; i < n ; i++){
            int curValue = 0;

            vector<vector<bool>> dp(n, vector<bool> (1 << n, false));
            queue<pair<int,pair<int,int>>> q;
            q.push({0,{i,curValue | (1 << i)}});
            dp[i][curValue | (1 << i)] = true;

            while(1) {
                pair<int,pair<int,int>> p = q.front();
                int wt = p.first , node = p.second.first , curValue = p.second.second;
                q.pop();

                dp[node][curValue] = true;

                if(curValue == (1 << n) - 1){
                    mn = min(mn , wt);
                    break;
                }

                for(auto &nbr : graph[node]) {
                    if(!dp[nbr][curValue | (1 << nbr)]){
                        dp[nbr][curValue | (1 << nbr)] = 1;
                        q.push({wt + 1, {nbr , curValue | (1 << nbr)}});
                    }
                }
            }
        }


        return mn;
    }
};
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