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

查找用户活跃分钟数。

给你用户在 LeetCode 的操作日志，和一个整数 k 。日志用一个二维整数数组 logs 表示，其中每个 logs[i] = [IDi, timei] 表示 ID 为 IDi 的用户在 timei 分钟时执行了某个操作。
多个用户 可以同时执行操作，单个用户可以在同一分钟内执行 多个操作 。
指定用户的 用户活跃分钟数（user active minutes，UAM） 定义为用户对 LeetCode 执行操作的 唯一分钟数 。 即使一分钟内执行多个操作，也只能按一分钟计数。
请你统计用户活跃分钟数的分布情况，统计结果是一个长度为 k 且 下标从 1 开始计数 的数组 answer ，对于每个 j（1 <= j <= k），answer[j] 表示 用户活跃分钟数 等于 j 的用户数。
返回上面描述的答案数组 answer 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/finding-the-users-active-minutes
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是用一个 HashMap<Integer, Set<Integer>> 记录每个用户都在哪些分钟执行了操作。注意活跃分钟数的定义是只要用户在某个分钟有了操作，无论操作多少次，都只按一分钟计数。所以这样记录下来之后，理论上每个用户对应的 hashset 里存的是一堆 unique 的分钟数。

最后要求返回的是一个长度为 K 的数组，表示拥有X个活跃分钟数的用户有几个。注意最后数组 index 的表达，题目说下标从 1 开始计数，是需要我们把活跃分钟数 - 1当做数组的下标。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public int[] findingUsersActiveMinutes(int[][] logs, int k) {
 3         HashMap<Integer, Set<Integer>> map = new HashMap<>();
 4         for (int[] log : logs) {
 5             if (!map.containsKey(log[0])) {
 6                 map.put(log[0], new HashSet<>());
 7             }
 8             map.get(log[0]).add(log[1]);
 9         }
10         
11         int[] res = new int[k];
12         for (int key : map.keySet()) {
13             int size = map.get(key).size();
14             res[size - 1]++;
15         }
16         return res;
17     }
18 }

LeetCode 题目总结