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] = [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.

思路：读懂题目意思了之后，其实就是个hashmap count，hashset的size记录在array里面；

class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        HashMap<Integer, HashSet<Integer>> hashmap = new HashMap<>();
        for(int[] log: logs) {
            int id = log[0];
            int minus = log[1];
            hashmap.putIfAbsent(id, new HashSet<Integer>());
            hashmap.get(id).add(minus);
        }
        
        // userid: UAM
        int[] res = new int[k];
        for(Integer key: hashmap.keySet()) {
            int UAM = hashmap.get(key).size();
            res[UAM - 1]++;
        }
        return res;
    }
}