We are given some website visits: the user with name username[i] visited the website website[i] at time timestamp[i].

A 3-sequence is a list of websites of length 3 sorted in ascending order by the time of their visits.  (The websites in a 3-sequence are not necessarily distinct.)

Find the 3-sequence visited by the largest number of users. If there is more than one solution, return the lexicographically smallest such 3-sequence.

Example 1:

Input: username = ["joe","joe","joe","james","james","james","james","mary","mary","mary"], timestamp = [1,2,3,4,5,6,7,8,9,10], website = ["home","about","career","home","cart","maps","home","home","about","career"]
Output: ["home","about","career"]
Explanation: 
The tuples in this example are:
["joe", 1, "home"]
["joe", 2, "about"]
["joe", 3, "career"]
["james", 4, "home"]
["james", 5, "cart"]
["james", 6, "maps"]
["james", 7, "home"]
["mary", 8, "home"]
["mary", 9, "about"]
["mary", 10, "career"]
The 3-sequence ("home", "about", "career") was visited at least once by 2 users.
The 3-sequence ("home", "cart", "maps") was visited at least once by 1 user.
The 3-sequence ("home", "cart", "home") was visited at least once by 1 user.
The 3-sequence ("home", "maps", "home") was visited at least once by 1 user.
The 3-sequence ("cart", "maps", "home") was visited at least once by 1 user.

 1 class Pair {
 2     int time;
 3     String web;
 4 
 5     public Pair(int time, String web) {
 6         this.time = time;
 7         this.web = web;
 8     }
 9 }
10 
11 class Solution {
12     public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
13         Map<String, List<Pair>> map = new HashMap<>();
14         int n = username.length;
15         for (int i = 0; i < n; i++) {
16             map.putIfAbsent(username[i], new ArrayList<>());
17             map.get(username[i]).add(new Pair(timestamp[i], website[i]));
18         }
19         // count map to record every 3 combination occuring time for the different user.
20         Map<String, Integer> count = new HashMap<>();
21         String res = "";
22         for (String key : map.keySet()) {
23             Set<String> set = new HashSet<>();
24             List<Pair> list = map.get(key);
25             if (list.size() < 3) {
26                 continue;
27             }
28             Collections.sort(list, (a, b) -> (a.time - b.time)); // sort by time
29             // brutal force O(N ^ 3)
30             for (int i = 0; i < list.size(); i++) {
31                 for (int j = i + 1; j < list.size(); j++) {
32                     for (int k = j + 1; k < list.size(); k++) {
33                         String str = list.get(i).web + " " + list.get(j).web + " " + list.get(k).web;
34                         if (!set.contains(str)) {
35                             count.put(str, count.getOrDefault(str, 0) + 1);
36                             set.add(str);
37                         }
38                         if (res.equals("") || count.get(res) < count.get(str) || (count.get(res) == count.get(str) && res.compareTo(str) > 0)) {
39                             // make sure the right lexi order
40                             res = str;
41                         }
42                     }
43                 }
44             }
45         }
46         return Arrays.asList(res.split(" "));
47     }
48 }