Formatted question description:

1817. Finding the Users Active Minutes




You are given the logs for users’ actions on LeetCode, and an integer k. The logs are represented by a 2D integer array logs where each logs[i] = [ID_i, time_i] indicates that the user with ID_i performed an action at the minute time_i.

Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute.

The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.

You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k), answer[j] is the number of users whose UAM equals j.

Return the array answer as described above.

Example 1:

Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5

Output: [0,2,0,0,0]


The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).

The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.

Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.

Example 2:

Input: logs = [[1,1],[2,2],[2,3]], k = 4

Output: [1,1,0,0]


The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.

The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.

There is one user with a UAM of 1 and one with a UAM of 2.

Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.


  • 1 <= logs.length <= 10^4
  • 0 <= ID_i <= 10^9
  • 1 <= time_i <= 10^5
  • k is in the range [The maximum UAM for a user, 10^5].


Use a map to store each user’s active minutes, where the active minutes are stored in a set. Then for each user, obtain the number of active minutes, and update the array answer accordingly. Finally, return answer.

class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        int[] answer = new int[k];
        Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
        for (int[] log : logs) {
            int id = log[0], time = log[1];
            Set<Integer> set = map.getOrDefault(id, new HashSet<Integer>());
            map.put(id, set);
        for (Map.Entry<Integer, Set<Integer>> entry : map.entrySet()) {
            int size = entry.getValue().size();
            answer[size - 1]++;
        return answer;

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