<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>

#input = list of list nodes and their given neighbors
#output = shortest path to visit all nodes

#idea:
#start at node neighboring the node with the most edges
#when visiting each node, add each unvisited neighbor to minHeap pq based on the num of edges that a node has
#

class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)
        endingMask = (1 << n) - 1
        dp = [[math.inf for _ in range(endingMask+1)] for _ in range(n)]

        def exploreState(node, mask):
            if dp[node][mask] != math.inf:
                return dp[node][mask]
            #visited all nodes
            elif (mask & (mask - 1)) ==  0:
                return 0

            minPath = math.inf
            dp[node][mask] = 99999999

            #for each neighbor explore case where we are visiting for first time or repeat visit
            for adj in graph[node]:
                if mask & (1 << adj):
                    path = 1 + min(exploreState(adj, mask), exploreState(adj, mask ^ (1 << node)))
                    dp[node][mask] = min(dp[node][mask], path)

            return dp[node][mask]
        
        minSteps = math.inf
        for i in range(n):
            minSteps = min(minSteps, exploreState(i, endingMask))

        return minSteps
21100