#1348

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Tweet Counts Per Frequency

Design an efficient solution to track and retrieve tweet counts over different frequencies using binary search and hash tables.

hash table binary search design sorting ordered set

Problem Statement

A social media platform needs to analyze tweet counts across various time intervals, partitioning the time period based on a given frequency. Each tweet is recorded at a specific timestamp, and you are tasked with efficiently counting how many tweets occur in specific time intervals, which are divided based on the given frequency (e.g., minute, hour, day).

For example, given a time period [10, 10000], you can partition it into chunks with specified frequencies. The last chunk may be shorter than the others and always ends with the time boundary. You need to implement methods to record tweets and get the count of tweets in a frequency-defined period for a given time range.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["TweetCounts","recordTweet","recordTweet","recordTweet","getTweetCountsPerFrequency","getTweetCountsPerFrequency","recordTweet","getTweetCountsPerFrequency"] [[],["tweet3",0],["tweet3",60],["tweet3",10],["minute","tweet3",0,59],["minute","tweet3",0,60],["tweet3",120],["hour","tweet3",0,210]]

Output [null,null,null,null,[2],[2,1],null,[4]]

Explanation TweetCounts tweetCounts = new TweetCounts(); tweetCounts.recordTweet("tweet3", 0); // New tweet "tweet3" at time 0 tweetCounts.recordTweet("tweet3", 60); // New tweet "tweet3" at time 60 tweetCounts.recordTweet("tweet3", 10); // New tweet "tweet3" at time 10 tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 59); // return [2]; chunk [0,59] had 2 tweets tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); // return [2,1]; chunk [0,59] had 2 tweets, chunk [60,60] had 1 tweet tweetCounts.recordTweet("tweet3", 120); // New tweet "tweet3" at time 120 tweetCounts.getTweetCountsPerFrequency("hour", "tweet3", 0, 210); // return [4]; chunk [0,210] had 4 tweets

Constraints

0 <= time, startTime, endTime <= 109
0 <= endTime - startTime <= 104
There will be at most 104 calls in total to recordTweet and getTweetCountsPerFrequency.

Solution Approach

Binary Search Over Valid Answer Space

The key to solving this problem is performing binary search over the valid answer space to efficiently determine the number of tweets in each chunk. By leveraging binary search, we minimize the need to scan through each time chunk sequentially, which improves the time complexity significantly.

Use of Hash Tables for Efficient Storage

Tweets are stored in hash tables, where the tweet name is the key, and the value is an ordered list of timestamps when that tweet was recorded. This structure allows quick lookups for each tweet's timestamp, enabling efficient counting of tweets within each time chunk.

Handling Time Chunks Based on Frequency

The partitioning of time intervals into chunks based on the frequency (minute, hour, day) is handled efficiently through pre-calculated ranges. By calculating the start and end of each time chunk dynamically, the solution ensures optimal performance while retrieving tweet counts for specific periods.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach but is primarily influenced by the binary search over the valid range of time chunks and the hash table operations. With a well-designed hash table and binary search strategy, the solution can achieve efficient time and space complexities.

What Interviewers Usually Probe

Understanding of binary search over a range of values and its application to time intervals.
Ability to design a system with hash tables for efficient data storage and retrieval.
Clear grasp of how to partition time periods into chunks and handle edge cases (e.g., last chunk being smaller).

Common Pitfalls or Variants

Common pitfalls

Failure to properly partition time ranges based on the given frequency, leading to incorrect counts.
Inefficient handling of large data sets or too many function calls, resulting in high time complexity.
Incorrect or inefficient implementation of binary search for counting tweets in a given time period.

Follow-up variants

Different time frequencies (minute, hour, day) can change the complexity of the chunking and binary search logic.
Handling extremely large numbers of timestamps or a high number of function calls may require additional optimizations.
Designing the system for a dynamic range of time periods rather than fixed ones, which may require extra handling of edge cases.

FAQ

How does binary search help in solving the Tweet Counts Per Frequency problem?

Binary search helps by narrowing down the possible answer space for tweet counts within a specific time period, minimizing unnecessary checks.

What is the best way to store tweet timestamps for efficient retrieval?

The best way is to use a hash table, where tweet names are keys, and their corresponding timestamps are stored in ordered lists for fast access.

How do time intervals impact the solution's complexity?

The partitioning of time intervals into smaller chunks based on frequency can increase the complexity if not handled efficiently, requiring careful attention to how binary search and data structures are used.

What are the main challenges when implementing this problem?

The main challenges include correctly handling time chunking based on the frequency, efficiently counting tweets using binary search, and managing high numbers of tweets and function calls.

How does GhostInterview assist with solving this problem?

GhostInterview provides a structured approach, breaking down the binary search, hash table usage, and time chunk handling, helping you focus on optimal solutions.

terminal

Solution

Solution 1

#### Python3

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class TweetCounts:
    def __init__(self):
        self.d = {"minute": 60, "hour": 3600, "day": 86400}
        self.data = defaultdict(SortedList)

    def recordTweet(self, tweetName: str, time: int) -> None:
        self.data[tweetName].add(time)

    def getTweetCountsPerFrequency(
        self, freq: str, tweetName: str, startTime: int, endTime: int
    ) -> List[int]:
        f = self.d[freq]
        tweets = self.data[tweetName]
        t = startTime
        ans = []
        while t <= endTime:
            l = tweets.bisect_left(t)
            r = tweets.bisect_left(min(t + f, endTime + 1))
            ans.append(r - l)
            t += f
        return ans


# Your TweetCounts object will be instantiated and called as such:
# obj = TweetCounts()
# obj.recordTweet(tweetName,time)
# param_2 = obj.getTweetCountsPerFrequency(freq,tweetName,startTime,endTime)

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