8 August 2020
LeetCode(1152) -- Analyze User Website Visit Pattern
by Jerry Zhang
Problem
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.
Discuss Solution
class Solution {
class Pair {
int time;
String web;
public Pair(int time, String web) {
this.time = time;
this.web = web;
}
}
public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
Map<String, List<Pair>> map = new HashMap<>();
Map<String, Integer> count = new HashMap<>();
int length = username.length;
for (int i = 0; i < length; i++) {
map.putIfAbsent(username[i], new ArrayList<>());
map.get(username[i]).add(new Pair(timestamp[i], website[i]));
}
String ans = "";
for (String user : map.keySet()) {
List<Pair> list = map.get(user);
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 str = list.get(i).web + " " + list.get(j).web + " " + list.get(k).web;
if (!set.contains(str)) {
count.put(str, count.getOrDefault(str, 0) + 1);
set.add(str);
}
if (ans.equals("") || count.get(str) > count.get(ans) || count.get(str) == count.get(ans) && str.compareTo(ans) < 0) {
ans = str;
}
}
}
}
}
return Arrays.asList(ans.split(" "));
}
}