[LC] 1152. Analyze User Website Visit Pattern

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.

Note:

1. 3 <= N = username.length = timestamp.length = website.length <= 50
2. 1 <= username[i].length <= 10
3. 0 <= timestamp[i] <= 10^9
4. 1 <= website[i].length <= 10
5. Both username[i] and website[i] contain only lowercase characters.
6. It is guaranteed that there is at least one user who visited at least 3 websites.
7. No user visits two websites at the same time.

class Solution {
    public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
        Map<String, List<Pair>> map = new HashMap<>();
        int len = username.length;
        for (int i = 0; i < len; i++) {
            if (map.get(username[i]) == null) {
                map.put(username[i], new ArrayList<>());
            }
            map.get(username[i]).add(new Pair(timestamp[i], website[i]));
        }
        
        Map<String, Integer> count = new HashMap<>();
        String res = "";
        for (String str: map.keySet()) {
            List<Pair> list = map.get(str);
            // avoid the same user visit 3-seq
            Set<String> set = new HashSet<>();
            Collections.sort(list, (a, b) -> a.time - b.time);
            for (int i = 0; i < list.size(); i++) {
                for (int j = i + 1; j < list.size(); j++) {
                    for (int k = j + 1; k < list.size(); k++) {
                        String cur = list.get(i).web + " " + list.get(j).web + " " + list.get(k).web;
                        if (!set.contains(cur)) {
                            count.put(cur, count.getOrDefault(cur, 0) + 1);
                            set.add(cur);
                        }
                        if (res.equals("") || count.get(cur) > count.get(res) || count.get(cur) == count.get(res) && cur.compareTo(res) < 0) {
                            res = cur;
                        }
                    }
                }
            }
        }
        
        String[] strArr = res.split(" ");
        List<String> resList = new ArrayList<>();
        for (String str: strArr) {
            resList.add(str);
        }
        return resList;
    }
}

class Pair {
    String web;
    int time;
    public Pair(Integer time, String web) {
        this.web = web;
        this.time = time;
    }
}