1817. Finding the Users Active Minutes

Problem

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] = [IDi, timei] indicates that the user with IDi performed an action at the minute timei.

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]
Explanation:
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]
Explanation:
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.

  Constraints:

• 1 <= logs.length <= 10^4

• 0 <= IDi <= 10^9

• 1 <= timei <= 10^5

• k is in the range [The maximum **UAM** for a user, 10^5].

Solution (Java)

class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        if (logs.length == 1) {
            int[] res = new int[k];
            res[0] = 1;
            return res;
        }
        Arrays.sort(logs, Comparator.comparingInt((int[] a) -> a[0]).thenComparingInt(a -> a[1]));
        int[] result = new int[k];
        int start = 1;
        int prevUser = logs[0][0];
        int prevMin = logs[0][1];
        int count = 1;
        while (true) {
            while (start < logs.length && prevUser == logs[start][0]) {
                if (prevMin != logs[start][1]) {
                    count++;
                }
                prevMin = logs[start][1];
                start++;
            }
            result[count - 1]++;
            if (start >= logs.length) {
                break;
            }
            count = 1;
            prevUser = logs[start][0];
            prevMin = logs[start][1];
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).