FindingTheUsersActiveMinutes.cpp

// Source : https://leetcode.com/problems/finding-the-users-active-minutes/
// Author : Hao Chen
// Date   : 2021-04-05

/***************************************************************************************************** 
 *
 * 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 <= 10^4
 * 	0 <= IDi <= 10^9
 * 	1 <= timei <= 10^5
 * 	k is in the range [The maximum UAM for a user, 10^5].
 ******************************************************************************************************/

class Solution {
public:
    vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
        vector<int> result(k, 0);
        unordered_map<int, set<int>> uam;
        for (auto& log : logs) {
            uam[log[0]].insert(log[1]);
        }
        for (auto& [id, t] : uam) {
            if (t.size() <= k) {
                result[t.size()-1]++;
            }
        }
        return result;
    }
};