<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<vector<pair<int,int>>> congraph(n);
        vector<vector<int>> arr(n,vector<int>(n,1e9));
        for(int i=0;i<n;i++){
            for(auto child: graph[i]){
                arr[i][child] = 1;
                arr[i][child] = 1;
            }
        }
        for(int k=0;k<n;k++){
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    arr[i][j] = min(arr[i][j],arr[i][k]+arr[k][j]);
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                congraph[i].push_back({j,arr[i][j]});
                congraph[j].push_back({i,arr[j][i]});
            }
        }
        map<pair<int,int>,int> dp;
        function<int(int,int)> tsp = [&](int u, int mask){
            if(mask == (1<<n)-1){
                return 0;
            }
            if(dp.find({u,mask}) != dp.end()){
                return dp[{u,mask}];
            }
            int mx = 1e9;
            for(auto child: congraph[u]){
                int v = child.first;
                int d = child.second;
                if(!((1<<v)&mask)){
                    int res = d+tsp(v,mask|(1<<v));
                    mx = min(res,mx);
                }
            }
            dp[{u,mask}] = mx;
            return mx;
        };
        int ans = 1e9;
        for(int i=0;i<n;i++){
            ans = min(ans,tsp(i,1<<i));
        }
        return ans;
    }
};
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