<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 dp[5000][20];

    void computeShortestPaths(vector<vector<int>>& graph, vector<vector<int>>& dist) {
        int n = graph.size();
        for (int i = 0; i < n; ++i) {
            vector<int> d(n, 1e9);
            queue<int> q;
            q.push(i);
            d[i] = 0;
            
            while (!q.empty()) {
                int node = q.front();
                q.pop();
                
                for (int neighbor : graph[node]) {
                    if (d[neighbor] > d[node] + 1) {
                        d[neighbor] = d[node] + 1;
                        q.push(neighbor);
                    }
                }
            }
            dist[i] = d;
        }
    }

    int solve(int mask,int currnode, vector<vector<int>>&adj, vector<vector<int>>&dist){
        int n = adj.size(), res=1e9;
        if(mask == (1<<n)-1){return 0;}

        if(dp[mask][currnode] != -1){return dp[mask][currnode];}

        for(int i=0; i<n; i++){
            if(!(mask&(1<<i))){
                int dis = dist[currnode][i];
                res = min(res, dis + solve(mask|(1<<i),i,adj,dist));
            }
        }
        return dp[mask][currnode] = res;
    }

    int shortestPathLength(vector<vector<int>>& graph) {
        memset(dp,-1,sizeof(dp));
        int n = graph.size();
        vector<vector<int>> adj(n);
        for(int i=0; i<n; i++){
            for(auto it: graph[i]){
                adj[i].push_back(it);
                adj[it].push_back(i);
            }
        }
        vector<vector<int>> dist(n, vector<int>(n));
        computeShortestPaths(graph, dist);
        int ret = 1e9;
        for(int i=0; i<n; i++){
            ret = min(ret, solve((1<<i),i,adj,dist));
        }
        return ret;
    }
};
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