Add solution #911

Author: JoshCrozier

README.md

@@ -721,6 +721,7 @@
 908|[Smallest Range I](./solutions/0908-smallest-range-i.js)|Easy|
 909|[Snakes and Ladders](./solutions/0909-snakes-and-ladders.js)|Medium|
 910|[Smallest Range II](./solutions/0910-smallest-range-ii.js)|Medium|
+911|[Online Election](./solutions/0911-online-election.js)|Medium|
 912|[Sort an Array](./solutions/0912-sort-an-array.js)|Medium|
 914|[X of a Kind in a Deck of Cards](./solutions/0914-x-of-a-kind-in-a-deck-of-cards.js)|Medium|
 916|[Word Subsets](./solutions/0916-word-subsets.js)|Medium|

solutions/0911-online-election.js

@@ -0,0 +1,61 @@
+/**
+ * 911. Online Election
+ * https://leetcode.com/problems/online-election/
+ * Difficulty: Medium
+ *
+ * You are given two integer arrays persons and times. In an election, the ith vote was cast for
+ * persons[i] at time times[i].
+ *
+ * For each query at a time t, find the person that was leading the election at time t. Votes
+ * cast at time t will count towards our query. In the case of a tie, the most recent vote (among
+ * tied candidates) wins.
+ *
+ * Implement the TopVotedCandidate class:
+ * - TopVotedCandidate(int[] persons, int[] times) Initializes the object with the persons and
+ *   times arrays.
+ * - int q(int t) Returns the number of the person that was leading the election at time t according
+ *   to the mentioned rules.
+ */
+
+/**
+ * @param {number[]} persons
+ * @param {number[]} times
+ */
+var TopVotedCandidate = function(persons, times) {
+  this.times = times;
+  this.leaders = [];
+  const voteCount = new Map();
+  let maxVotes = 0;
+  let currentLeader = 0;
+
+  for (let i = 0; i < persons.length; i++) {
+    const votes = (voteCount.get(persons[i]) || 0) + 1;
+    voteCount.set(persons[i], votes);
+
+    if (votes >= maxVotes) {
+      maxVotes = votes;
+      currentLeader = persons[i];
+    }
+    this.leaders[i] = currentLeader;
+  }
+};
+
+/**
+ * @param {number} t
+ * @return {number}
+ */
+TopVotedCandidate.prototype.q = function(t) {
+  let left = 0;
+  let right = this.times.length - 1;
+
+  while (left <= right) {
+    const middle = Math.floor((left + right) / 2);
+    if (this.times[middle] <= t) {
+      left = middle + 1;
+    } else {
+      right = middle - 1;
+    }
+  }
+
+  return this.leaders[right];
+};