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].

1817. Finding the Users Active Minutes
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn finding_users_active_minutes(logs: Vec<Vec<i32>>, k: i32) -> Vec<i32> {
        let mut users: HashMap<i32, HashSet<i32>> = HashMap::new();
        let k = k as usize;
        let mut res = vec![0; k];
        for log in logs {
            let id = log[0];
            let time = log[1];
            users.entry(id).or_default().insert(time);
        }
        for times in users.values() {
            let j = times.len() - 1;
            if j < k {
                res[j] += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let logs = vec_vec_i32![[0, 5], [1, 2], [0, 2], [0, 5], [1, 3]];
    let k = 5;
    let res = vec![0, 2, 0, 0, 0];
    assert_eq!(Solution::finding_users_active_minutes(logs, k), res);
    let logs = vec_vec_i32![[1, 1], [2, 2], [2, 3]];
    let k = 4;
    let res = vec![1, 1, 0, 0];
    assert_eq!(Solution::finding_users_active_minutes(logs, k), res);
}