<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 n, mind;
    int final_mask;

    void solve(vector<vector<int>> &g, int u, int &mask, int d, vector<vector<int>> &dis) {
        dis[u][mask] = d;
        if(mask == final_mask) {
            mind = min(d, mind);
            return ;
        }
        int temp = mask;
        for(int i=0; i < g[u].size(); i++) {
            int v = g[u][i];
            mask = mask|(1<<v);

            if(dis[v][mask] > d+1)
                solve(g, v, mask, d+1, dis);

            mask = temp;
        }
    }

    int shortestPathLength(vector<vector<int>>& graph) {
        n = graph.size();
        final_mask = pow(2,n) - 1;
        mind = INT_MAX;
        
        for(int i=0; i<n; i++) {
            vector<int> x(final_mask+1, INT_MAX);
            vector<vector<int>> dis;
            for(int j=0; j<n; j++) dis.push_back(x);
            int mask = (1<<i);
            solve(graph, i, mask, 0, dis);
        }

        return mind;
    }
};
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