LeetCode 2688: Find Active Users Solution

Master LeetCode problem 2688 (Find Active Users), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2688. Find Active Users

Medium

Database

Problem Explanation

Explanation:

To find active users, we can use a hashtable to store the count of each user's activity. Then, we can iterate through the hashtable to find the active users based on some criteria.

Create a hashtable activityMap to store the count of each user's activity.
Iterate through the activity log and populate the activityMap.
Iterate through the activityMap to find active users based on the given criteria.
Return the list of active users.

Time Complexity:

Populating the activityMap takes O(N) where N is the number of activities.
Finding active users takes O(M) where M is the number of unique users.
Overall time complexity is O(N + M).

Space Complexity:

The hashtable activityMap will take O(M) space where M is the number of unique users.

Solution Code

import java.util.*;

class Solution {
    public List<String> findActiveUsers(String[] logs, int threshold) {
        Map<String, Integer> activityMap = new HashMap<>();
        for (String log : logs) {
            String[] logParts = log.split(" ");
            String user1 = logParts[0];
            String user2 = logParts[1];
            activityMap.put(user1, activityMap.getOrDefault(user1, 0) + 1);
            if (!user1.equals(user2)) {
                activityMap.put(user2, activityMap.getOrDefault(user2, 0) + 1);
            }
        }

        List<String> activeUsers = new ArrayList<>();
        for (Map.Entry<String, Integer> entry : activityMap.entrySet()) {
            if (entry.getValue() >= threshold) {
                activeUsers.add(entry.getKey());
            }
        }
        return activeUsers;
    }
}

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