Add solution #933

Author: JoshCrozier

README.md

@@ -263,6 +263,7 @@
 925|[Long Pressed Name](./0925-long-pressed-name.js)|Easy|
 926|[Flip String to Monotone Increasing](./0926-flip-string-to-monotone-increasing.js)|Medium|
 929|[Unique Email Addresses](./0929-unique-email-addresses.js)|Easy|
+933|[Number of Recent Calls](./0933-number-of-recent-calls.js)|Easy|
 966|[Vowel Spellchecker](./0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./0976-largest-perimeter-triangle.js)|Easy|

solutions/0933-number-of-recent-calls.js

@@ -0,0 +1,35 @@
+/**
+ * 933. Number of Recent Calls
+ * https://leetcode.com/problems/number-of-recent-calls/
+ * Difficulty: Easy
+ *
+ * You have a RecentCounter class which counts the number of recent requests within
+ * a certain time frame.
+ *
+ * Implement the RecentCounter class:
+ * - RecentCounter() Initializes the counter with zero recent requests.
+ * - int ping(int t) Adds a new request at time t, where t represents some time in
+ *   milliseconds, and returns the number of requests that has happened in the past
+ *   3000 milliseconds (including the new request). Specifically, return the number
+ *   of requests that have happened in the inclusive range [t - 3000, t].
+ *
+ * It is guaranteed that every call to ping uses a strictly larger value of t than
+ * the previous call.
+ */
+
+
+var RecentCounter = function() {
+  this.queue = [];
+};
+
+/**
+ * @param {number} t
+ * @return {number}
+ */
+RecentCounter.prototype.ping = function(t) {
+  this.queue.push(t);
+  while (this.queue[0] < t - 3000) {
+    this.queue.shift();
+  }
+  return this.queue.length;
+};