# 1817. Finding the Users Active Minutes

<https://leetcode.com/problems/finding-the-users-active-minutes>

## Description

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]`.

## ac

```java
```


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