Introduction

In this blog post, we will explore how to solve the LeetCode problem "1348" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 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.

Explanation

• Data Structure: Use a hash map to store tweets. The keys are tweet names, and the values are sorted lists of timestamps for each tweet. This allows for efficient retrieval and counting of tweets within time ranges.
  • Frequency Calculation: Based on the given frequency (minute, hour, day), calculate the time chunk size. Iterate through the time range [startTime, endTime] in steps of the chunk size, counting the number of tweets within each chunk using binary search.
  • Binary Search Optimization: Employ binary search to efficiently count tweets within each time chunk. Since the timestamps are stored in sorted order, binary search reduces the time complexity of counting tweets within a range.

• Runtime Complexity: recordTweet: O(log n), getTweetCountsPerFrequency: O(m log n), where n is the number of tweets for a given tweetName and m is the number of chunks. Storage Complexity: O(N), where N is the total number of tweets recorded.

Code

    from bisect import bisect_left, bisect_right

class TweetCounts:

    def __init__(self):
        self.tweet_data = {}  # tweetName: [timestamps]

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

    def getTweetCountsPerFrequency(self, freq: str, tweetName: str, startTime: int, endTime: int) -> list[int]:
        if tweetName not in self.tweet_data:
            return []

        timestamps = self.tweet_data[tweetName]
        result = []

        if freq == "minute":
            chunk_size = 60
        elif freq == "hour":
            chunk_size = 3600
        else:  # freq == "day"
            chunk_size = 86400

        current_time = startTime
        while current_time <= endTime:
            next_time = min(current_time + chunk_size, endTime + 1)
            count = bisect_right(timestamps, next_time - 1) - bisect_left(timestamps, current_time)
            result.append(count)
            current_time += chunk_size

        return result