Add solution #1817

Author: JoshCrozier

README.md

@@ -282,6 +282,7 @@
 1780|[Check if Number is a Sum of Powers of Three](./1780-check-if-number-is-a-sum-of-powers-of-three.js)|Medium|
 1791|[Find Center of Star Graph](./1791-find-center-of-star-graph.js)|Easy|
 1812|[Determine Color of a Chessboard Square](./1812-determine-color-of-a-chessboard-square.js)|Easy|
+1817|[Finding the Users Active Minutes](./1817-finding-the-users-active-minutes.js)|Medium|
 1832|[Check if the Sentence Is Pangram](./1832-check-if-the-sentence-is-pangram.js)|Easy|
 1880|[Check if Word Equals Summation of Two Words](./1880-check-if-word-equals-summation-of-two-words.js)|Easy|
 1886|[Determine Whether Matrix Can Be Obtained By Rotation](./1886-determine-whether-matrix-can-be-obtained-by-rotation.js)|Easy|

solutions/1817-finding-the-users-active-minutes.js

@@ -0,0 +1,39 @@
+/**
+ * 1817. Finding the Users Active Minutes
+ * https://leetcode.com/problems/finding-the-users-active-minutes/
+ * Difficulty: Medium
+ *
+ * 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.
+ */
+
+/**
+ * @param {number[][]} logs
+ * @param {number} k
+ * @return {number[]}
+ */
+var findingUsersActiveMinutes = function(logs, k) {
+  const map = new Map();
+  logs.forEach(([id, time]) => {
+    map.set(id, map.get(id) || new Set());
+    map.get(id).add(time);
+  });
+
+  const result = new Array(k).fill(0);
+  [...map.values()].forEach(({ size }) => result[size - 1]++);
+
+  return result;
+};