<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();
        vector<int> v[n];
        for(int i = 0; i < graph.size(); i ++) {
            for(int j = 0; j < graph[i].size(); j ++) {
                v[i].push_back(graph[i][j]);
                v[graph[i][j]].push_back(i);
            }
        }
        
        vector<vector<int>> d(n, vector<int> (n, 2000));
        
        for(int i = 0; i < n; i ++) {
            queue<int> q;   q.push(i);
            vector<bool> vis(n, false);
            vis[i] = 1;  
            
            d[i][i] = 0;
            while(!q.empty()) {
                int x = q.front();
                for(auto it : v[x]) {
                    if(vis[it])     continue;
                    vis[it] = true;
                    
                    d[i][it] = d[i][x] + 1;
                    q.push(it);
                }
                q.pop();
            }
        }
        
        vector<vector<int>> dp(1<<n, vector<int> (n, 1e5));
        for(int i = 0; i < (1 << n); i++) {
            int bit = __builtin_popcount(i);
            if(bit == 1) {
                for(int j = 0; j < n; j ++) {
                    if(i & (1 << j)) {
                        dp[i][j] = 0;
                    }
                }
            }
        }
        
        vector<int> ss[n+1];
        for(int i = 0; i < (1 << n); i++) {
            int bit = __builtin_popcount(i);
            ss[bit].push_back(i);
        }
        
        for(int bit = 2; bit <= n; bit ++) {
            for(auto i : ss[bit]) {
                for(int j = 0; j < n; j ++) {
                    if(i & (1 << j)) {
                        for(int k = 0; k < n; k ++) {
                            if(j == k)      continue;
                            if(i & (1 << k)) {
                                dp[i][j] = min(dp[i][j], dp[i ^ (1 << j)][k] + d[j][k]);   
                            }
                        }
                    }
                }
            }
        }

        int ans = INT_MAX;
        for(int i = 0; i < n; i ++) {
            ans = min(ans, dp[(1 << n) - 1][i]);
        }
        return ans;
        
    }
};
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