<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(vector<vector<int>>& dist, int start, int to_visit, vector<vector<int>>& mem){
        if(to_visit == 0) return 0;
        if(mem[start][to_visit] == -1){
            int m = INT_MAX;
            for(int i = 0; i < dist.size(); i++){
                if(((to_visit >> i) & 1) == 1){
                    int temp = dp(dist, i, to_visit ^ (1 << i), mem);
                    if(dist[start][i] != INT_MAX && temp != INT_MAX)
                        m = min(m, dist[start][i] + dp(dist, i, to_visit ^ (1 << i), mem));
                }
            }
            mem[start][to_visit] = m;
        }
        return mem[start][to_visit];
    }
    int shortestPathLength(vector<vector<int>>& graph) {
        if(graph[0].size()  == 0) return 0;
        int n = graph.size();
        vector<vector<int>> dist(n, vector<int>(n, INT_MAX));
        for(int i = 0; i < n; i++){
            dist[i][i] = 0;
            for(int& node: graph[i]){
                dist[i][node] = 1;
            }
        }

        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                for(int k = 0; k < n; k++){
                    if(dist[i][k] != INT_MAX && dist[k][j] != INT_MAX){
                        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        }

        int to_visit = ~((~0) << n);
        int m = INT_MAX;
        for(int start = 0; start < n; start++){
            vector<vector<int>> mem(n, vector<int>(pow(2, n), - 1));
            m = min(m, dp(dist, start, to_visit ^ (1 << start), mem));
        }
        return m;
    }
};
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