CPMagical CP Library

Problem solution · Java

Tweet Counts Per Frequency

Tweet Counts Per Frequency: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
23 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Tweet Counts Per Frequency, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 23 lines of Java from the credited upstream file 1348.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 1 loop block detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeTweet Counts Per Frequency · JavaJava
Use this to learn the idea, then write your own version.
class TweetCounts {  public void recordTweet(String tweetName, int time) {    tweetNameToTimeCount.putIfAbsent(tweetName, new TreeMap<>());    tweetNameToTimeCount.get(tweetName).merge(time, 1, Integer::sum);  }   public List<Integer> getTweetCountsPerFrequency(String freq, String tweetName, int startTime,                                                  int endTime) {    final int chunkSize = freq.equals("minute") ? 60 : freq.equals("hour") ? 3600 : 86400;    int[] counts = new int[(endTime - startTime) / chunkSize + 1];    TreeMap<Integer, Integer> timeCount = tweetNameToTimeCount.get(tweetName);     for (Map.Entry<Integer, Integer> entry : timeCount.subMap(startTime, endTime + 1).entrySet()) {      final int index = (entry.getKey() - startTime) / chunkSize;      counts[index] += entry.getValue();    }     return Arrays.stream(counts).boxed().collect(Collectors.toList());  }   private Map<String, TreeMap<Integer, Integer>> tweetNameToTimeCount = new HashMap<>();}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗