Add solution #1348

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,247 LeetCode solutions in JavaScript
+# 1,248 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1024,6 +1024,7 @@
 1345|[Jump Game IV](./solutions/1345-jump-game-iv.js)|Hard|
 1346|[Check If N and Its Double Exist](./solutions/1346-check-if-n-and-its-double-exist.js)|Easy|
 1347|[Minimum Number of Steps to Make Two Strings Anagram](./solutions/1347-minimum-number-of-steps-to-make-two-strings-anagram.js)|Medium|
+1348|[Tweet Counts Per Frequency](./solutions/1348-tweet-counts-per-frequency.js)|Medium|
 1351|[Count Negative Numbers in a Sorted Matrix](./solutions/1351-count-negative-numbers-in-a-sorted-matrix.js)|Easy|
 1352|[Product of the Last K Numbers](./solutions/1352-product-of-the-last-k-numbers.js)|Medium|
 1356|[Sort Integers by The Number of 1 Bits](./solutions/1356-sort-integers-by-the-number-of-1-bits.js)|Easy|

solutions/1348-tweet-counts-per-frequency.js

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+/**
+ * 1348. Tweet Counts Per Frequency
+ * https://leetcode.com/problems/tweet-counts-per-frequency/
+ * Difficulty: Medium
+ *
+ * A social media company is trying to monitor activity on their site by analyzing the number
+ * of tweets that occur in select periods of time. These periods can be partitioned into
+ * smaller time chunks based on a certain frequency (every minute, hour, or day).
+ *
+ * For example, the period [10, 10000] (in seconds) would be partitioned into the following
+ * time chunks with these frequencies:
+ * - Every minute (60-second chunks): [10,69], [70,129], [130,189], ..., [9970,10000]
+ * - Every hour (3600-second chunks): [10,3609], [3610,7209], [7210,10000]
+ * - Every day (86400-second chunks): [10,10000]
+ *
+ * Notice that the last chunk may be shorter than the specified frequency's chunk size and will
+ * always end with the end time of the period (10000 in the above example).
+ *
+ * Design and implement an API to help the company with their analysis.
+ *
+ * Implement the TweetCounts class:
+ * - TweetCounts() Initializes the TweetCounts object.
+ * - void recordTweet(String tweetName, int time) Stores the tweetName at the recorded time
+ *   (in seconds).
+ * - List<Integer> getTweetCountsPerFrequency(String freq, String tweetName, int startTime,
+ *   int endTime) Returns a list of integers representing the number of tweets with tweetName
+ *   in each time chunk for the given period of time [startTime, endTime] (in seconds) and
+ *   frequency freq.
+ * - freq is one of "minute", "hour", or "day" representing a frequency of every minute, hour,
+ *   or day respectively.
+ */
+
+var TweetCounts = function() {
+  this.tweets = new Map();
+};
+
+/**
+ * @param {string} tweetName
+ * @param {number} time
+ * @return {void}
+ */
+TweetCounts.prototype.recordTweet = function(tweetName, time) {
+  if (!this.tweets.has(tweetName)) {
+    this.tweets.set(tweetName, []);
+  }
+  this.tweets.get(tweetName).push(time);
+};
+
+/**
+ * @param {string} freq
+ * @param {string} tweetName
+ * @param {number} startTime
+ * @param {number} endTime
+ * @return {number[]}
+ */
+TweetCounts.prototype.getTweetCountsPerFrequency = function(freq, tweetName, startTime, endTime) {
+  let interval;
+  if (freq === 'minute') interval = 60;
+  else if (freq === 'hour') interval = 3600;
+  else interval = 86400;
+
+  const count = Math.floor((endTime - startTime) / interval) + 1;
+  const result = new Array(count).fill(0);
+
+  if (this.tweets.has(tweetName)) {
+    for (const time of this.tweets.get(tweetName)) {
+      if (time >= startTime && time <= endTime) {
+        const index = Math.floor((time - startTime) / interval);
+        result[index]++;
+      }
+    }
+  }
+
+  return result;
+};