# LeetCode 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 <= 10^4`
* `0 <= IDi <= 10^9`
* `1 <= timei <= 10^5`
* `k` is in the range `[The maximum`` `**`UAM`**` ``for a user, 10^5]`.

## Solution

```
class Solution {
public:
    vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
        vector<int> res(k, 0);
        
        map<int, set<int>> activities;
        for (const vector<int>& log : logs) {
            activities[log[0]].insert(log[1]);
        }
        
        for (auto [id, activity] : activities) {
            res[activity.size() - 1]++;
        }
        
        return res;
    }
};
```
