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
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
1 <= j <= k),
answer[j] is the number of users whose UAM equals
Return the array
answer as described above.
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 is 2, and the remaining answer[j] values are 0.
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, answer = 1, and the remaining values are 0.
1 <= logs.length <= 104
0 <= IDi <= 109
1 <= timei <= 105
kis in the range
[The maximum UAM for a user, 105].