New Problem Solution -"1817. Finding the Users Active Minutes"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1817|[Finding the Users Active Minutes](https://leetcode.com/problems/finding-the-users-active-minutes/) | [C++](./algorithms/cpp/findingTheUsersActiveMinutes/FindingTheUsersActiveMinutes.cpp)|Medium|
 |1816|[Truncate Sentence](https://leetcode.com/problems/truncate-sentence/) | [C++](./algorithms/cpp/truncateSentence/TruncateSentence.cpp)|Easy|
 |1808|[Maximize Number of Nice Divisors](https://leetcode.com/problems/maximize-number-of-nice-divisors/) | [C++](./algorithms/cpp/maximizeNumberOfNiceDivisors/MaximizeNumberOfNiceDivisors.cpp)|Hard|
 |1807|[Evaluate the Bracket Pairs of a String](https://leetcode.com/problems/evaluate-the-bracket-pairs-of-a-string/) | [C++](./algorithms/cpp/evaluateTheBracketPairsOfAString/EvaluateTheBracketPairsOfAString.cpp)|Medium|

algorithms/cpp/findingTheUsersActiveMinutes/FindingTheUsersActiveMinutes.cpp

@@ -0,0 +1,66 @@
+// 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;
+    }
+};