<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>

import numpy as np
class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        def setbit(mask, i):
            return mask | (1<<i)
        
        hm = {}
        n = len(graph)

        for i in range(n):
            if i not in hm:
                hm[i] = []
            m = len(graph[i])
            for j in range(m):
                hm[i].append(graph[i][j])

        # dist 2d array
        # row: bitmask --> visited node set bits are 1
        # column: leading node

        row = pow(2, n)
        col = n
        dist = []
        for i in range(row):
            dist.append([-1] * col)

        # queue: [{leading node 1, mask}, {leading node 2, mask}]
        q = []
        for i in range(n):
            lead = i
            mask = setbit(0, i)
            q.append([lead, mask])
            dist[mask][lead] = 0
        
        # applying simultaneous BFS
        while len(q) > 0:
            size = len(q)
            for i in range(size):
                path = q.pop(0)
                lead = path[0]
                mask = path[1]

                if mask == row-1:
                    return dist[mask][lead]

                # iterate over neighbors of lead
                if lead in hm:
                    for child in hm[lead]:
                        newlead = child
                        newmask = setbit(mask, newlead)

                        # avoid cycle: intelligent bfs: check if this set is already visited
                        # set: lead, mask(visited nodes)
                        if dist[newmask][newlead] != -1:
                            continue
                        
                        dist[newmask][newlead] = dist[mask][lead] + 1
                        q.append([newlead, newmask])

        return 1221

30000