Analyze User Website Visit Pattern

lcpy gen -n 1152   # by problem number
lcpy gen -s analyze_user_website_visit_pattern   # by problem name

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','home',1],['joe','about',2],['joe','career',3],['james','home',4],['james','cart',5],['james','maps',6],['james','home',7],['mary','home',8],['mary','about',9], and ['mary','career',10].
The pattern ('home', 'about', 'career') has score 2 (joe and mary).
The pattern ('home', 'cart', 'maps') has score 1 (james).
The pattern ('home', 'cart', 'home') has score 1 (james).
The pattern ('home', 'maps', 'home') has score 1 (james).
The pattern ('cart', 'maps', 'home') has score 1 (james).
The pattern ('home', 'home', 'home') has score 0 (no user visited home 3 times).

Input: username = ['ua','ua','ua','ub','ub','ub']
timestamp = [1,2,3,4,5,6]
website = ['a','b','a','a','b','c']
Output: ['a','b','a']

from collections import Counter, defaultdict


class Solution:
    # Time: O(n^3) per user in the worst case, n = visits per user
    # Space: O(n^3) for the distinct triplets
    def most_visited_pattern(
        self, username: list[str], timestamp: list[int], website: list[str]
    ) -> list[str]:
        visits: dict[str, list[tuple[int, str]]] = defaultdict(list)
        for user, ts, site in zip(username, timestamp, website, strict=True):
            visits[user].append((ts, site))

        scores: Counter[tuple[str, str, str]] = Counter()
        for entries in visits.values():
            sites = [site for _, site in sorted(entries)]
            count = len(sites)
            patterns = {
                (sites[i], sites[j], sites[k])
                for i in range(count - 2)
                for j in range(i + 1, count - 1)
                for k in range(j + 1, count)
            }
            scores.update(patterns)

        best = min(scores.items(), key=lambda item: (-item[1], item[0]))[0]
        return list(best)