1817. 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].

 

Solutions

Solution: Hash Map

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} logs
 * @param {number} k
 * @return {number[]}
 */
const findingUsersActiveMinutes = function (logs, k) {
  const activeMap = logs.reduce((map, [id, time]) => {
    return (map[id] = (map[id] ?? new Set()).add(time)), map;
  }, {});
  const result = new Array(k).fill(0);
  const actives = Object.values(activeMap);

  for (const active of actives) {
    result[active.size - 1] += 1;
  }
  return result;
};