<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:
    vector<vector<int>>dp,min_dist;
    int n;
    int slve(int prev_node,int vis_memo)
    {
        if(vis_memo==0)
            return 0;
        if(dp[prev_node][vis_memo]!=-1)
            return dp[prev_node][vis_memo];
        int ans,i;
        ans=696969;
        for(i=0;i<n;i++)
            if((1<<i)&vis_memo)
            ans=min(ans,min_dist[prev_node][i]+slve(i,vis_memo-(1<<i)));
        return dp[prev_node][vis_memo]=ans;
    }
    void floyd_warshall()
    {
        int i,j,k;
        for(k=0;k<n;k++)
            for(i=0;i<n;i++)
                for(j=0;j<n;j++)
                    min_dist[i][j]=min(min_dist[i][k]+min_dist[k][j],min_dist[i][j]);
    }
    int shortestPathLength(vector<vector<int>>& g) {
        n=g.size();
        min_dist.assign(n+1,vector<int>(n+1,696969));
        for(int i=0;i<n;i++)
        {
            for(auto &x:g[i])
                min_dist[i][x]=min_dist[x][i]=1;
            min_dist[i][i]=0;
            min_dist[n][i]=0;
        }
        floyd_warshall();
        dp.assign(n+1,vector<int>((1<<n),-1));
        return slve(n,(1<<n)-1);
    }
};
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