- Difficulty: Medium
- Tags: LeetCode, Medium, Design, Hash Table, Binary Search, Ordered Set, Sorting, leetcode-1348, O(n)
Problem
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 theTweetCounts
object.void recordTweet(String tweetName, int time)
Stores thetweetName
at the recordedtime
(in seconds).List<Integer> getTweetCountsPerFrequency(String freq, String tweetName, int startTime, int endTime)
Returns a list of integers representing the number of tweets withtweetName
in each time chunk for the given period of time[startTime, endTime]
(in seconds) and frequencyfreq
.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 torecordTweet
andgetTweetCountsPerFrequency
.