Add solution #2008

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,684 LeetCode solutions in JavaScript
+# 1,685 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1540,6 +1540,7 @@
 2003|[Smallest Missing Genetic Value in Each Subtree](./solutions/2003-smallest-missing-genetic-value-in-each-subtree.js)|Hard|
 2006|[Count Number of Pairs With Absolute Difference K](./solutions/2006-count-number-of-pairs-with-absolute-difference-k.js)|Easy|
 2007|[Find Original Array From Doubled Array](./solutions/2007-find-original-array-from-doubled-array.js)|Medium|
+2008|[Maximum Earnings From Taxi](./solutions/2008-maximum-earnings-from-taxi.js)|Medium|
 2011|[Final Value of Variable After Performing Operations](./solutions/2011-final-value-of-variable-after-performing-operations.js)|Easy|
 2016|[Maximum Difference Between Increasing Elements](./solutions/2016-maximum-difference-between-increasing-elements.js)|Easy|
 2017|[Grid Game](./solutions/2017-grid-game.js)|Medium|

solutions/2008-maximum-earnings-from-taxi.js

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+/**
+ * 2008. Maximum Earnings From Taxi
+ * https://leetcode.com/problems/maximum-earnings-from-taxi/
+ * Difficulty: Medium
+ *
+ * There are n points on a road you are driving your taxi on. The n points on the road are
+ * labeled from 1 to n in the direction you are going, and you want to drive from point 1
+ * to point n to make money by picking up passengers. You cannot change the direction of the taxi.
+ *
+ * The passengers are represented by a 0-indexed 2D integer array rides, where
+ * rides[i] = [starti, endi, tipi] denotes the ith passenger requesting a ride from point
+ * starti to point endi who is willing to give a tipi dollar tip.
+ *
+ * For each passenger i you pick up, you earn endi - starti + tipi dollars. You may only drive
+ * at most one passenger at a time.
+ *
+ * Given n and rides, return the maximum number of dollars you can earn by picking up the
+ * passengers optimally.
+ *
+ * Note: You may drop off a passenger and pick up a different passenger at the same point.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[][]} rides
+ * @return {number}
+ */
+var maxTaxiEarnings = function(n, rides) {
+  rides.sort((a, b) => a[1] - b[1]);
+  const dp = new Array(n + 1).fill(0);
+  let rideIndex = 0;
+
+  for (let point = 1; point <= n; point++) {
+    dp[point] = dp[point - 1];
+    while (rideIndex < rides.length && rides[rideIndex][1] === point) {
+      const [start, end, tip] = rides[rideIndex];
+      dp[point] = Math.max(dp[point], dp[start] + (end - start + tip));
+      rideIndex++;
+    }
+  }
+
+  return dp[n];
+};