<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;
    int done[12][(1 << 12) + 10];
    int rec(int node, int mask, vector<vector<int>>& g, vector<vector<int>>& dis) {
        if (mask == ((1 << n) - 1)) {
            return 0;
        }
        if (done[node][mask] != -1) {
            return done[node][mask];
        }
        int ans = 1e9;
        for (int i=0;i<n;i++) {
            if ((mask & (1 << i)) == 0) {
                ans = min(ans, dis[node][i] + rec(i, mask | (1 << i), g, dis));
            }
        }
        return done[node][mask] = ans;
    }
    int shortestPathLength(vector<vector<int>>& g) {
        n = g.size();
        int ans = 1e9;
        vector<vector<int>> dis(n, vector<int>(n, 1e9));
        for (int i = 0; i < n; i++) {
            dis[i][i] = 0;
        }
        for (int i = 0; i < n; i++) {
            for (auto j : g[i]) {
                // This was the error, we need to consider the edges as undirected
                dis[i][j] = 1;
                dis[j][i] = 1;
            }
        }
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (dis[i][j] > dis[i][k] + dis[k][j]) {
                        dis[i][j] = dis[i][k] + dis[k][j];
                    }
                }
            }
        }

        memset(done, -1, sizeof(done));
        for (int i = 0; i < n; i++) {
            ans = min(ans, rec(i, 1 << i, g, dis));
        }

        return ans;
    }
};
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