# 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] = \[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 <= 104 0 <= IDi <= 109 1 <= timei <= 105 k is in the range \[The maximum UAM for a user, 105].

## Solutions

1. **hashmap**

```cpp
class Solution {
public:
    vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
        unordered_map<int, unordered_set<int>> umas;
        for (auto & log : logs)
            umas[log[0]].insert(log[1]);

        vector<int> res(k);
        for (auto [u, times] : umas)
            res[times.size() - 1]++;

        return res;
    }
};
```