Tweet Counts Per Frequency

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A social media company is trying to monitor activity on their site by analyzing the number of tweets that occur in select periods of time. These periods can be partitioned into smaller time chunks based on a certain frequency (every minute, hour, or day).

For example, the period [10, 10000] (in seconds) would be partitioned into the following time chunks with these frequencies:

Every minute (60-second chunks): [10,69], [70,129], [130,189], ..., [9970,10000]
Every hour (3600-second chunks): [10,3609], [3610,7209], [7210,10000]
Every day (86400-second chunks): [10,10000]

Notice that the last chunk may be shorter than the specified frequency's chunk size and will always end with the end time of the period (10000 in the above example).

Design and implement an API to help the company with their analysis.

Implement the TweetCounts class:

TweetCounts() Initializes the TweetCounts object.
void recordTweet(String tweetName, int time) Stores the tweetName at the recorded time (in seconds).
List<Integer> getTweetCountsPerFrequency(String freq, String tweetName, int startTime, int endTime) Returns a list of integers representing the number of tweets with tweetName in each time chunk for the given period of time [startTime, endTime] (in seconds) and frequency freq.
- freq is one of "minute", "hour", or "day" representing a frequency of every minute, hour, or day respectively.

Example:

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for frequency (freq) and what do they represent exactly? Can freq be something other than 'minute', 'hour', or 'day'?
What is the expected range for the timestamp? Are they provided as seconds since the epoch, or some other unit? Are they guaranteed to be non-negative?
If there are no tweets within a specified time range when calling getTweetCountsPerFrequency, what should the corresponding count be in the output? Should I return zero, or exclude that time range altogether?
How should I handle edge cases, such as when the startTime is greater than endTime, or when startTime/endTime fall outside the range of any tweet timestamps?
If multiple tweets occur at the exact same timestamp, should they all be counted within the appropriate frequency interval?

Brute Force Solution

Approach

The brute force strategy for counting tweets involves going through all possible time intervals. We check each tweet individually to see how many times it falls within the specified time range for each request.

Here's how the algorithm would work step-by-step:

For each query, take the start and end time and the frequency (either per minute, hour, or day).
If the frequency is per minute, divide the time range into one-minute intervals. If it is per hour, divide it into one-hour intervals. If it is per day, divide it into one-day intervals.
For each interval, go through the list of all tweets.
For each tweet, check if the time the tweet was created falls within that interval's start and end time.
If it does, increase the count for that interval by one.
After checking all the tweets for an interval, record the count.
Repeat this for every interval within the query's time range.
Return the list of counts for each interval.

Code Implementation

class TweetCounts:

    def __init__(self):
        self.tweet_records = {}

    def recordTweet(self, tweet_name: str, time: int) -> None:
        if tweet_name not in self.tweet_records:
            self.tweet_records[tweet_name] = []
        self.tweet_records[tweet_name].append(time)

    def getTweetCountsPerFrequency(self, frequency: str, tweet_name: str, start_time: int, end_time: int) -> list[int]:
        time_interval_counts = []
        
        if frequency == "minute":
            interval_length = 60
        elif frequency == "hour":
            interval_length = 3600
        else:
            interval_length = 86400
        
        number_of_intervals = (end_time - start_time) // interval_length + 1

        # Iterate over each interval of time
        for interval_number in range(number_of_intervals):
            interval_start_time = start_time + interval_number * interval_length
            interval_end_time = min(start_time + (interval_number + 1) * interval_length - 1, end_time)
            tweet_count = 0

            # Ensure the given tweet name has been recorded
            if tweet_name in self.tweet_records:

                # Go through all of the tweets for a given name
                for tweet_time in self.tweet_records[tweet_name]:

                    # Determine if tweet falls within time range
                    if interval_start_time <= tweet_time <= interval_end_time:
                        tweet_count += 1

            time_interval_counts.append(tweet_count)

        return time_interval_counts

Big(O) Analysis

Time Complexity

O(N * R) – Let N be the total number of tweets stored, and let R be the number of intervals we iterate through for a given query. For each query, we iterate through R intervals (based on the frequency and time range). For each interval, we iterate through all N tweets to determine if they fall within that interval. Thus, for a single query, the time complexity is O(N * R). Since in a real-world scenario the number of intervals R can approach the magnitude of N (e.g., when the time difference is very large and the frequency is minute), the time complexity can approximate O(N*N), but in general we represent it as O(N*R).

Space Complexity

O(1) – The brute force approach iterates through tweets and intervals, but it doesn't store a significant amount of auxiliary data. It primarily uses counters for each interval and reuses them. The number of intervals depends on the query's time range and frequency, but it is processed one at a time, not stored all at once. Hence, the auxiliary space used is constant, regardless of the number of tweets or the size of the time range.

Optimal Solution

Approach

The goal is to count how many tweets happen within specific time windows. We need to efficiently store tweet timestamps and quickly retrieve counts within those windows. The clever approach is to divide the timeline into smaller, manageable chunks to speed up the counting process.

Here's how the algorithm would work step-by-step:

When a tweet is recorded, store its timestamp in a way that makes it easy to find later. Think of this like putting it in a sorted list.
When someone asks for the tweet count for a specific time window (either every minute, hour, or day), break that time window into smaller, equal-sized sections.
For each section, count the number of tweets that fall within it. Because we stored the timestamps in order, we can quickly find the first and last tweet in each section and count everything in between without looking at every single tweet.
Add up the tweet counts from each section to get the total tweet count for the requested time window.
If a request spans multiple days, hours, or minutes, repeat the counting process for each period and add them all together. This prevents looking at too many unrelated tweets.
By splitting time into small chunks, we process only the necessary tweets, avoiding a slow, comprehensive search every time.

Code Implementation

class TweetCounts:

    def __init__(self):
        self.tweet_times = {}

    def recordTweet(self, tweetName: str, time: int) -> None:
        if tweetName not in self.tweet_times:
            self.tweet_times[tweetName] = []
        self.tweet_times[tweetName].append(time)
        self.tweet_times[tweetName].sort()

    def getTweetCountsPerFrequency(self, freq: str, tweetName: str, startTime: int, endTime: int) -> list[int]:
        time_intervals = []
        if freq == "minute":
            interval = 60
        elif freq == "hour":
            interval = 3600
        else:
            interval = 86400

        #Determine the time ranges that we will use to count tweets
        current_time = startTime
        while current_time <= endTime:
            time_intervals.append(current_time)
            current_time += interval

        result = [0] * len(time_intervals)

        #Iterate through each tweet to count occurences
        for tweet_time in self.tweet_times[tweetName]:
            if startTime <= tweet_time <= endTime:
                for i in range(len(time_intervals)):

                    #Tweets are counted within each time interval
                    if i == len(time_intervals) - 1 or (time_intervals[i] <= tweet_time < time_intervals[i + 1]):
                        result[i] += 1
                        break

        return result

Big(O) Analysis

Time Complexity

O(n + klog(n)) – Adding a tweet involves inserting a timestamp into a sorted list of existing tweets which takes O(log n) time using binary search, where n is the number of tweets. Therefore, adding m tweets takes O(m log n) time which we can consider O(n) if m = n. Retrieving tweet counts involves iterating through k time windows (minutes, hours, or days) and for each window, performing a binary search to find the start and end indices of tweets within that window, which takes O(log n) time per window. Thus, the overall time complexity for getting the counts is O(klog(n)), and the total time complexity is O(n + klog(n)).

Space Complexity

O(N) – The solution stores tweet timestamps in a sorted manner, implying a data structure like a list or an array. The size of this data structure will grow linearly with the number of tweets recorded. Therefore, if N is the number of tweets, the auxiliary space used to store these timestamps is proportional to N. The additional space for section counts during retrieval is considered negligible compared to the storage of the tweet timestamps. Thus the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty content name.	Return an empty list or throw an IllegalArgumentException based on the API's intended behavior.
Invalid frequency (not 'minute', 'hour', 'day').	Throw an IllegalArgumentException or return an empty list to signal an error.
startTime greater than endTime.	Return an empty list or swap the start and end times after validation since order shouldn't matter for aggregation purposes.
Tweet time outside of the allowed range (before the beginning or after the end of the frequency time range).	Ensure all tweets outside the requested timerange are discarded before bucketing.
Large number of tweets for a given content name causing memory issues.	Consider using a more memory-efficient data structure, like a streaming algorithm, to reduce peak memory usage or use database aggregation.
Large time range causing integer overflow when calculating bucket indexes.	Use long type to store timestamps and calculate bucket indices, preventing potential integer overflow.
Content name already exists, and adding a new tweet should update counts in existing buckets.	Ensure the implementation correctly merges new tweet times into the existing bucket structure.
Edge case when startTime and endTime are very close together, creating tiny time ranges (e.g., startTime=0, endTime=1 with frequency=minute)	Handle near-zero time ranges correctly, and ensure a bucket is created for each valid point within the range.

Null or empty content name.

How to Handle:

Return an empty list or throw an IllegalArgumentException based on the API's intended behavior.

Invalid frequency (not 'minute', 'hour', 'day').

How to Handle:

Throw an IllegalArgumentException or return an empty list to signal an error.

startTime greater than endTime.

How to Handle:

Return an empty list or swap the start and end times after validation since order shouldn't matter for aggregation purposes.

Tweet time outside of the allowed range (before the beginning or after the end of the frequency time range).

How to Handle:

Ensure all tweets outside the requested timerange are discarded before bucketing.

Large number of tweets for a given content name causing memory issues.

How to Handle:

Consider using a more memory-efficient data structure, like a streaming algorithm, to reduce peak memory usage or use database aggregation.

Large time range causing integer overflow when calculating bucket indexes.

How to Handle:

Use long type to store timestamps and calculate bucket indices, preventing potential integer overflow.

Content name already exists, and adding a new tweet should update counts in existing buckets.

How to Handle:

Ensure the implementation correctly merges new tweet times into the existing bucket structure.

Edge case when startTime and endTime are very close together, creating tiny time ranges (e.g., startTime=0, endTime=1 with frequency=minute)

How to Handle:

Handle near-zero time ranges correctly, and ensure a bucket is created for each valid point within the range.