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]
.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);
}