LeetCode 1152. Analyze User Website Visit Pattern

原题链接在这里：https://leetcode.com/problems/analyze-user-website-visit-pattern/

题目：

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.

题解：

Try to get the website sequence visted by most different users. sequence here means chronological order.

First sort the data by timestamp.

And put user and corresponding websites into HashMap<String, List<String>>.

For each user, find all its combination of 3 websites.

And each sequence within this combination, check its count of different user. If count is higher or lexicographically smaller, update the max sequence.

The max sequence is the result.

Time Complexity: O(n ^ 3). n = username.length. sort takes O(nlogn). getCom takes O(n ^ 3).

Space: O(n ^ 3).

AC Java:

class Solution {
    public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
        int n = username.length;
        List<Pair> datas = new ArrayList<>();
        for(int i = 0; i < n; i++){
            datas.add(new Pair(username[i], timestamp[i], website[i]));
        }
        
        Collections.sort(datas, (a, b) -> a.time - b.time);
        HashMap<String, List<String>> userToWebs = new HashMap<>();
        for(Pair data : datas){
            userToWebs.putIfAbsent(data.user, new ArrayList<String>());
            userToWebs.get(data.user).add(data.web);
        }
        
        HashMap<String, Integer> seqToCount = new HashMap<>();
        
        int maxCount = 0;
        String maxSeq = "";
        for(Map.Entry<String, List<String>> entry : userToWebs.entrySet()){
            Set<String> seqCom = getCom(entry.getValue());
            for(String seq : seqCom){
                seqToCount.put(seq, seqToCount.getOrDefault(seq, 0) + 1);
                if(seqToCount.get(seq) > maxCount){
                    maxCount = seqToCount.get(seq);
                    maxSeq = seq;
                }else if(seqToCount.get(seq) == maxCount && seq.compareTo(maxSeq) < 0){
                    maxSeq = seq;
                }    
            }
        }
        
        List<String> res = new ArrayList<>();
        String [] webs = maxSeq.split(",");
        for(String w : webs){
            res.add(w);
        }
        
        return res;
    }
    
    private HashSet<String> getCom(List<String> webs){
        HashSet<String> res = new HashSet<>();
        int n = webs.size();
        for(int i = 0; i < n - 2; i++){
            for(int j = i + 1; j < n - 1; j++){
                for(int k = j + 1; k < n; k++){
                    res.add(webs.get(i) + "," + webs.get(j) + "," + webs.get(k));
                }
            }
        }
        
        return res;
    }
}

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