<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>

// BFS 

class Solution {
public:
    int shortestPathLength(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<vector<int>> vis(n, vector<int>(1<<n, 0));
        map<int, vector<int>> mpp;

        for(int i = 0 ; i < n ; i++) {
            mpp[graph[i].size()].push_back(i);
        }
        vector<int> temp = mpp.begin()->second;

        queue<vector<int>> q;
        for(int i : temp) {
            q.push({i, 1<<i, 0});
            vis[i][1<<i] = 1;
        }

        while(!q.empty()) {
            vector<int> temp = q.front();
            q.pop();
            int node = temp[0];
            int mask = temp[1];
            int d = temp[2];

            if(mask == (1<<n)-1) return d;

            for(int child : graph[node]) {
                int newMask = (mask | (1<<child));
                if(vis[child][newMask] == 0) {
                    vis[child][newMask] = 1;
                    q.push({child, newMask, d+1});
                }
            }
        }
        return -1;
    }
};



// BFS 

// class Solution {
// public:
//     int shortestPathLength(vector<vector<int>>& graph) {
//         int n = graph.size();
//         int finalMask = (1 << n) - 1;  // All nodes visited.
//         queue<pair<int, int>> q;       // {node, mask}
//         vector<vector<int>> dist(n, vector<int>(1 << n, 1e9));

//         for(int i = 0; i < n; i++) {
//             q.push({i, 1 << i});
//             dist[i][1 << i] = 0;
//         }

//         while(!q.empty()) {
//             auto [node, mask] = q.front();
//             q.pop();
//             cout << "\nVisiting " << node;
//             int d = dist[node][mask];
            
//             for(int neighbor : graph[node]) {
//                 int nextMask = mask | (1 << neighbor);
//                 if(dist[neighbor][nextMask] > d + 1) {
//                     dist[neighbor][nextMask] = d + 1;
//                     q.push({neighbor, nextMask});
//                 }
//             }
//         }

//         int ans = 1e9;
//         for(int i = 0; i < n; i++) {
//             ans = min(ans, dist[i][finalMask]);
//         }
//         return ans;
//     }
// };



// floyd warshall + DFS + DP

// class Solution {
// public:
//     int shortestPathLength(vector<vector<int>>& graph) {
//         int n = graph.size();
//         vector<vector<int>> dp(n, vector<int>(1<<n, -1));
//         vector<vector<int>> dist(n, vector<int>(n, 1e5));

//         for(int i = 0 ; i < n ; i++) {
//             dist[i][i] = 0;
//             for(int j = 0 ; j < graph[i].size() ; j++) {
//                 dist[i][graph[i][j]] = 1;
//             }
//         }

//         for(int via = 0 ; via < n ; via++) {
//             for(int i = 0 ; i < n ; i++) {
//                 for(int j = 0 ; j < n ; j++) {
//                     dist[i][j] = min(dist[i][j], dist[i][via] + dist[via][j]);
//                 }
//             }
//         }

//         auto f = [&] (auto& f, int node, int mask) -> int {
//             if(mask == ((1 << n)-1)) return 0;
//             if(dp[node][mask] != -1) return dp[node][mask];

//             int mn = 1e9;
//             for(int j = 0 ; j < n ; j++) {
//                 if(!(mask & (1<<j))) {
//                     mn = min(mn, dist[node][j] + f(f, j, (mask | (1<<j))));
//                 }
//             }
//             return dp[node][mask] = mn;
//         };

//         int mn = 1e9;
//         for(int start = 0 ; start < n ; start++) {
//             mn = min(mn, f(f, start, (0|(1<<start))));
//         }
//         return mn;
//     }
// };



// DFS TLE

// class Solution {
// public:
//     int shortestPathLength(vector<vector<int>>& graph) {
//         int n = graph.size();
//         vector<vector<int>> vis(n, vector<int>(1<<n, 0));

//         auto f = [&] (auto& f, int node, int mask) -> int {
//             if(mask == ((1 << n)-1)) return 0;

//             int mn = 1e9;
//             for(int child : graph[node]) {
//                 int newMask = (mask | (1<<child));
//                 if(vis[child][newMask] == 0) {
//                     vis[child][newMask] = 1;
//                     mn = min(mn, 1 + f(f, child, newMask));
//                     vis[child][newMask] = 0;
//                 }
//             }
//             return mn;
//         };

//         int mn = 1e9;
//         for(int start = 0 ; start < n ; start++) {
//             mn = min(mn, f(f, start, (0|(1<<start))));
//         }
//         return mn;
//     }
// };
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