LeetCode #1348 — MEDIUM

Tweet Counts Per Frequency

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode
The Problem

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<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.
Patterns Used
🔍Modified Binary Search→🏗Design Patterns→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Binary Search · Design · Segment Tree

Example 1

["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]]

Related Problems

  • Design Video Sharing Platform (design-video-sharing-platform)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1348: Tweet Counts Per Frequency
class TweetCounts {
    private Map<String, TreeMap<Integer, Integer>> data = new HashMap<>();

    public TweetCounts() {
    }

    public void recordTweet(String tweetName, int time) {
        data.putIfAbsent(tweetName, new TreeMap<>());
        var tm = data.get(tweetName);
        tm.put(time, tm.getOrDefault(time, 0) + 1);
    }

    public List<Integer> getTweetCountsPerFrequency(
        String freq, String tweetName, int startTime, int endTime) {
        int f = 60;
        if ("hour".equals(freq)) {
            f = 3600;
        } else if ("day".equals(freq)) {
            f = 86400;
        }
        var tm = data.get(tweetName);
        List<Integer> ans = new ArrayList<>();
        for (int i = startTime; i <= endTime; i += f) {
            int s = 0;
            int end = Math.min(i + f, endTime + 1);
            for (int v : tm.subMap(i, end).values()) {
                s += v;
            }
            ans.add(s);
        }
        return ans;
    }
}

/**
 * Your TweetCounts object will be instantiated and called as such:
 * TweetCounts obj = new TweetCounts();
 * obj.recordTweet(tweetName,time);
 * List<Integer> param_2 = obj.getTweetCountsPerFrequency(freq,tweetName,startTime,endTime);
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#352Data Stream as Disjoint Intervalshard→#2080Range Frequency Queriesmedium→#2424Longest Uploaded Prefixmedium→#3508Implement Routermedium→#729My Calendar Imedium→