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1152.Analyze-User-Website-Visit-Pattern

1152. Analyze User Website Visit Pattern

题目地址

题目描述

We are given some website visits: the user with name username[i] visited the website website[i] at time timestamp[i].
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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.)
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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.
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Example 1:
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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.
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Note:
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3 <= N = username.length = timestamp.length = website.length <= 50
1 <= username[i].length <= 10
0 <= timestamp[i] <= 10^9
1 <= website[i].length <= 10
Both username[i] and website[i] contain only lowercase characters.
It is guaranteed that there is at least one user who visited at least 3 websites.
No user visits two websites at the same time.

代码

Approach #1

class Solution {
public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
int n = timestamp.length;
// 1. sort session list by time, list[ list[name, timestap, web], ...]
List<List<String>> sessions = new ArrayList<>();
for (int i = 0; i < n; i++) {
sessions.add(new ArrayList<>());
sessions.get(i).add(username[i]);
sessions.get(i).add("" + timestamp[i]);
sessions.get(i).add(website[i]);
}
sessions.sort((a, b) -> Integer.parseInt(a.get(1)) - Integer.parseInt(b.get(1)));
​
// 2. add each person visited list: {name => list(web)}
Map<String, List<String>> visited = new HashMap<>();
for (int i = 0; i < n; i++) {
visited.putIfAbsent(sessions.get(i).get(0), new ArrayList<>());
visited.get(sessions.get(i).get(0)).add(sessions.get(i).get(2));
}
​
// 3. find each user list and build all 3-sequences and count by map
// {3seq => count}
Map<String, Integer> sequenceMap = new HashMap<>();
int maxCount = 0;
String maxseq = "";
for (String name: visited.keySet()) {
List<String> list = visited.get(name);
if (list.size() < 3) continue;
Set<String> subsequences = getSubsequences(list);
for (String seq: subsequences) {
int count = sequenceMap.getOrDefault(seq, 0) + 1;
sequenceMap.put(seq, count);
if (count > maxCount) {
maxCount = count;
maxseq = seq;
} else if (count == maxCount && seq.compareTo(maxseq) < 0){
maxseq = seq;
}
}
}
​
String[] strs = maxseq.split(",");
List<String> res = new ArrayList<>();
for (String s: strs) {
res.add(s);
}
​
return res;
}
​
public Set<String> getSubsequences(List<String> list) {
int n = list.size();
Set<String> res = new HashSet<>();
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(list.get(i) + "," + list.get(j) + "," + list.get(k));
}
}
}
return res;
}
}