67 lines
2.5 KiB
C++
67 lines
2.5 KiB
C++
// Source : https://leetcode.com/problems/finding-the-users-active-minutes/
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// Author : Hao Chen
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// Date : 2021-04-05
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/*****************************************************************************************************
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*
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* You are given the logs for users' actions on LeetCode, and an integer k. The logs are represented
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* by a 2D integer array logs where each logs[i] = [IDi, timei] indicates that the user with IDi
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* performed an action at the minute timei.
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*
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* Multiple users can perform actions simultaneously, and a single user can perform multiple actions
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* in the same minute.
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*
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* The user active minutes (UAM) for a given user is defined as the number of unique minutes in which
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* the user performed an action on LeetCode. A minute can only be counted once, even if multiple
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* actions occur during it.
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*
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* You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k),
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* answer[j] is the number of users whose UAM equals j.
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*
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* Return the array answer as described above.
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*
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* Example 1:
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*
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* Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
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* Output: [0,2,0,0,0]
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* Explanation:
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* The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2
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* (minute 5 is only counted once).
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* The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
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* Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.
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*
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* Example 2:
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*
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* Input: logs = [[1,1],[2,2],[2,3]], k = 4
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* Output: [1,1,0,0]
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* Explanation:
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* The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.
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* The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
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* There is one user with a UAM of 1 and one with a UAM of 2.
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* Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.
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*
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* Constraints:
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*
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* 1 <= logs.length <= 10^4
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* 0 <= IDi <= 10^9
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* 1 <= timei <= 10^5
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* k is in the range [The maximum UAM for a user, 10^5].
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******************************************************************************************************/
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class Solution {
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public:
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vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
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vector<int> result(k, 0);
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unordered_map<int, set<int>> uam;
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for (auto& log : logs) {
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uam[log[0]].insert(log[1]);
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}
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for (auto& [id, t] : uam) {
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if (t.size() <= k) {
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result[t.size()-1]++;
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}
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}
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return result;
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}
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};
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