How do you solve Tweet Counts Per Frequency on LeetCode?

Tweet Counts Per Frequency (LeetCode #1348) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient storage and retrieval of tweet times.

What is the time complexity of Tweet Counts Per Frequency?

The optimal solution for Tweet Counts Per Frequency runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only loop through the tweets once and count them based on their respective intervals.

What is the brute force solution for Tweet Counts Per Frequency?

The brute force approach for Tweet Counts Per Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through the entire time range and count the tweets for each time chunk. This method is straightforward but inefficient as it checks each tweet for every time chunk. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Tweet Counts Per Frequency?

Always clarify the problem statement and constraints before diving into coding. Think about how you can optimize your solution from the start. Practice explaining your thought process as you code.

What are common mistakes when solving Tweet Counts Per Frequency?

Not considering edge cases like tweets falling exactly on the boundary of intervals. Forgetting to initialize the tweet list in the hash map.

#1348

Tweet Counts Per Frequency

Medium

Hash TableStringBinary SearchDesignSortingOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal solution, we will use a hash map to store the tweet times and then calculate the counts based on the frequency without nested loops. This significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Store each tweet's time in a hash map where the key is the tweet name and the value is a list of times.
2Step 2: Calculate the interval based on the frequency (minute, hour, day).
3Step 3: Iterate through the time range and count the tweets that fall into each interval using a single loop.

solution.py22 lines

1# Full working Python code
2class TweetCounts:
3    def __init__(self):
4        self.tweets = {}
5
6    def recordTweet(self, tweetName, time):
7        if tweetName not in self.tweets:
8            self.tweets[tweetName] = []
9        self.tweets[tweetName].append(time)
10
11    def getTweetCountsPerFrequency(self, freq, tweetName, startTime, endTime):
12        if tweetName not in self.tweets:
13            return []
14        counts = []
15        interval = 60 if freq == "minute" else 3600 if freq == "hour" else 86400
16        num_chunks = (endTime - startTime) // interval + 1
17        counts = [0] * num_chunks
18        for time in self.tweets[tweetName]:
19            if startTime <= time <= endTime:
20                index = (time - startTime) // interval
21                counts[index] += 1
22        return counts

ℹ

Complexity note: The time complexity is O(n) because we only loop through the tweets once and count them based on their respective intervals.

1Using a hash map allows for efficient storage and retrieval of tweet times.
2Understanding how to calculate intervals is key to optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.