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.

vector<string> mostVisitedPattern(vector<string>& username, vector<int>& timestamp, vector<string>& website) {
    unordered_map<string, map<int, string> > m; // <user -> <time -> website> >
    for (size_t i = 0; i < username.size(); ++i) m[username[i]][timestamp[i]] = website[i];
    unordered_map<string, int> cnt;
    unordered_set<string> st;
    for (auto user_timeline : m) {
        st.clear();
        for (auto it = begin(user_timeline.second); it != end(user_timeline.second); ++it) {
            for (auto it1 = next(it); it1 != end(user_timeline.second); ++it1) {
                for (auto it2 = next(it1); it2 != end(user_timeline.second); ++it2) {
                    st.insert(it->second + "$" + it1->second + "#" + it2->second);
                }
            }
        }
        for (auto s : st) ++cnt[s];
    }
    int count = 0;
    string res;
    for (auto t : cnt) {
        if (t.second >= count) {
            res = (res.empty() || t.second > count) ? t.first : min(res, t.first); 
            count = t.second;
        }
    }
    int pos1 = res.find("$"), pos2 = res.find("#");
    return {res.substr(0, pos1), res.substr(pos1 + 1, pos2 - pos1 - 1), res.substr(pos2 + 1)};
}