<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 setMask(int mask , int num){
        return mask|(1 << num);
    }
    int shortestPathLength(vector<vector<int>>& graph) {
        unordered_map<int,vector<int>> adjList;
        int n = graph.size();
        for(int i=0;i<graph.size();i++){
            int size = graph[i].size();
            for(int j=0;j<size;j++){
                adjList[i].push_back(graph[i][j]);
            }
        }
        int row = (1<<n);
        int col = n;
        vector<vector<int>> distance(row , vector<int>(col , -1));
        queue<vector<int>> q;
        for(int i=0;i<n;i++){
            int mask = setMask(0 , i);
            distance[mask][i] = 0;
            q.push({i , mask});
        }
        while(q.size() > 0){
            int size = q.size();
            for(int i=0;i<size;i++){
                auto path = q.front();
                q.pop();
                int currNode = path[0];
                int currMask = path[1];
                if(currMask == row - 1)
                    return distance[currMask][currNode];
                if(adjList.find(currNode) != adjList.end()){
                    for(auto child : adjList[currNode]){
                        int adjNode = child;
                        int newMask = setMask(currMask , adjNode);
                        if(distance[newMask][adjNode] != -1)
                            continue;
                        distance[newMask][adjNode] = distance[currMask][currNode] + 1;
                        q.push({adjNode , newMask});
                    }
                }
            }
        }
        return -1;
    }
};
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