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Problem solution · C++

Count Mentions Per User

Count Mentions Per User: a C++ solution using heap or priority queue. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Heap or priority queue
Source
walkccc LeetCode Solutions
Length
66 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

Heap or priority queue

For Count Mentions Per User, the implementation repeatedly takes the currently best candidate from a heap while inserting newly available choices.

  1. Define the priority key and whether the smallest or largest item should lead.
  2. Push each candidate when it becomes eligible.
  3. Discard stale entries when necessary and process the best live candidate.

Code notes

  • 66 lines of C++ from the credited upstream file 3433.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 6 loop blocks detected.

Complexity

Count heap pushes and pops; each normally contributes a logarithmic factor in the heap size.

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 codeCount Mentions Per User · C++C++
Use this to learn the idea, then write your own version.
struct OfflineUser {  int returnTimestamp;  int userId;  bool operator>(const OfflineUser& other) const {    return returnTimestamp > other.returnTimestamp;  }}; class Solution { public:  vector<int> countMentions(int numberOfUsers, vector<vector<string>>& events) {    vector<int> ans(numberOfUsers);    vector<int> online(numberOfUsers, true);    // min-heap to track users that are offline    priority_queue<OfflineUser, vector<OfflineUser>, greater<>> offlineQueue;    int allMentionsCount = 0;     ranges::sort(events, ranges::less{}, [](const vector<string>& event) {      const int timestamp = stoi(event[1]);      const char eventType = event[0][0];      return pair<int, char>{timestamp, -eventType};    });     for (const vector<string>& event : events) {      const string eventType = event[0];      const int timestamp = stoi(event[1]);      // Bring users back online if their offline period has ended.      while (!offlineQueue.empty() &&             offlineQueue.top().returnTimestamp <= timestamp)        online[offlineQueue.top().userId] = true, offlineQueue.pop();      if (eventType == "MESSAGE") {        const string mentionsString = event[2];        if (mentionsString == "ALL") {          ++allMentionsCount;        } else if (mentionsString == "HERE") {          for (int userId = 0; userId < numberOfUsers; ++userId)            if (online[userId])              ++ans[userId];        } else {          for (const int userId : getUserIds(mentionsString))            ++ans[userId];        }      } else if (eventType == "OFFLINE") {        const int userId = stoi(event[2]);        online[userId] = false;        // Add to queue to bring back online after 60 units.        offlineQueue.emplace(timestamp + 60, userId);      }    }     // Add the "ALL" mentions to all users.    for (int userId = 0; userId < numberOfUsers; ++userId)      ans[userId] += allMentionsCount;    return ans;  }  private:  vector<int> getUserIds(const string& s) {    vector<int> integers;    istringstream iss(s);    for (string id; iss >> id;)      integers.push_back(stoi(id.substr(2)));    return integers;  }};

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