Leetcode #355: Design Twitter
In this guide, we solve Leetcode #355 Design Twitter 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
Design a simplified version of Twitter where users can post tweets, follow/unfollow another user, and is able to see the 10 most recent tweets in the user's news feed. Implement the Twitter class: Twitter() Initializes your twitter object.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Hash Table, Linked List, Heap (Priority Queue)
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
["Twitter", "postTweet", "getNewsFeed", "follow", "postTweet", "getNewsFeed", "unfollow", "getNewsFeed"]
[[], [1, 5], [1], [1, 2], [2, 6], [1], [1, 2], [1]]
Output
[null, null, [5], null, null, [6, 5], null, [5]]
Explanation
Twitter twitter = new Twitter();
twitter.postTweet(1, 5); // User 1 posts a new tweet (id = 5).
twitter.getNewsFeed(1); // User 1's news feed should return a list with 1 tweet id -> [5]. return [5]
twitter.follow(1, 2); // User 1 follows user 2.
twitter.postTweet(2, 6); // User 2 posts a new tweet (id = 6).
twitter.getNewsFeed(1); // User 1's news feed should return a list with 2 tweet ids -> [6, 5]. Tweet id 6 should precede tweet id 5 because it is posted after tweet id 5.
twitter.unfollow(1, 2); // User 1 unfollows user 2.
twitter.getNewsFeed(1); // User 1's news feed should return a list with 1 tweet id -> [5], since user 1 is no longer following user 2.
Python Solution
class Twitter:
def __init__(self):
"""
Initialize your data structure here.
"""
self.user_tweets = defaultdict(list)
self.user_following = defaultdict(set)
self.tweets = defaultdict()
self.time = 0
def postTweet(self, userId: int, tweetId: int) -> None:
"""
Compose a new tweet.
"""
self.time += 1
self.user_tweets[userId].append(tweetId)
self.tweets[tweetId] = self.time
def getNewsFeed(self, userId: int) -> List[int]:
"""
Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user herself. Tweets must be ordered from most recent to least recent.
"""
following = self.user_following[userId]
users = set(following)
users.add(userId)
tweets = [self.user_tweets[u][::-1][:10] for u in users]
tweets = sum(tweets, [])
return nlargest(10, tweets, key=lambda tweet: self.tweets[tweet])
def follow(self, followerId: int, followeeId: int) -> None:
"""
Follower follows a followee. If the operation is invalid, it should be a no-op.
"""
self.user_following[followerId].add(followeeId)
def unfollow(self, followerId: int, followeeId: int) -> None:
"""
Follower unfollows a followee. If the operation is invalid, it should be a no-op.
"""
following = self.user_following[followerId]
if followeeId in following:
following.remove(followeeId)
# Your Twitter object will be instantiated and called as such:
# obj = Twitter()
# obj.postTweet(userId,tweetId)
# param_2 = obj.getNewsFeed(userId)
# obj.follow(followerId,followeeId)
# obj.unfollow(followerId,followeeId)
Complexity
The time complexity is O(n). The space complexity is O(n).
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.