1817. Finding the Users Active Minutes LeetCode Solution

Last updated: February 25, 2025

In this guide, you will get 1817. Finding the Users Active Minutes LeetCode Solution with the best time and space complexity. The solution to Finding the Users Active Minutes problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Finding the Users Active Minutes solution in C++
Finding the Users Active Minutes solution in Java
Finding the Users Active Minutes solution in Python
Additional Resources

1817. Finding the Users Active Minutes LeetCode Solution image

Problem Statement of 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].

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

1817. Finding the Users Active Minutes LeetCode Solution in C++

class Solution {
 public:
  vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
    vector<int> ans(k);
    unordered_map<int, unordered_set<int>> idToTimes;

    for (const vector<int>& log : logs)
      idToTimes[log[0]].insert(log[1]);

    for (const auto& [_, mins] : idToTimes)
      ++ans[mins.size() - 1];

    return ans;
  }
};
/* code provided by PROGIEZ */

1817. Finding the Users Active Minutes LeetCode Solution in Java

class Solution {
  public int[] findingUsersActiveMinutes(int[][] logs, int k) {
    int[] ans = new int[k];
    Map<Integer, Set<Integer>> idToTimes = new HashMap<>();

    for (int[] log : logs) {
      idToTimes.putIfAbsent(log[0], new HashSet<>());
      idToTimes.get(log[0]).add(log[1]);
    }

    for (final int id : idToTimes.keySet())
      ++ans[idToTimes.get(id).size() - 1];

    return ans;
  }
}
// code provided by PROGIEZ

1817. Finding the Users Active Minutes LeetCode Solution in Python

class Solution:
  def findingUsersActiveMinutes(
      self,
      logs: list[list[int]],
      k: int,
  ) -> list[int]:
    idToTimes = collections.defaultdict(set)

    for id, time in logs:
      idToTimes[id].add(time)

    c = collections.Counter(len(times) for times in idToTimes.values())
    return [c[i] for i in range(1, k + 1)]
 # code by PROGIEZ

Additional Resources

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