Leetcode #1152: Analyze User Website Visit Pattern
In this guide, we solve Leetcode #1152 Analyze User Website Visit Pattern in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two string arrays username and website and an integer array timestamp. All the given arrays are of the same length and the tuple [username[i], website[i], timestamp[i]] indicates that the user username[i] visited the website website[i] at time timestamp[i].
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Hash Table, String, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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).
Python Solution
class Solution:
def mostVisitedPattern(
self, username: List[str], timestamp: List[int], website: List[str]
) -> List[str]:
d = defaultdict(list)
for user, _, site in sorted(
zip(username, timestamp, website), key=lambda x: x[1]
):
d[user].append(site)
cnt = Counter()
for sites in d.values():
m = len(sites)
s = set()
if m > 2:
for i in range(m - 2):
for j in range(i + 1, m - 1):
for k in range(j + 1, m):
s.add((sites[i], sites[j], sites[k]))
for t in s:
cnt[t] += 1
return sorted(cnt.items(), key=lambda x: (-x[1], x[0]))[0][0]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.