Add solution #983

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,065 LeetCode solutions in JavaScript
+# 1,066 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -791,6 +791,7 @@
 980|[Unique Paths III](./solutions/0980-unique-paths-iii.js)|Hard|
 981|[Time Based Key-Value Store](./solutions/0981-time-based-key-value-store.js)|Medium|
 982|[Triples with Bitwise AND Equal To Zero](./solutions/0982-triples-with-bitwise-and-equal-to-zero.js)|Hard|
+983|[Minimum Cost For Tickets](./solutions/0983-minimum-cost-for-tickets.js)|Medium|
 985|[Sum of Even Numbers After Queries](./solutions/0985-sum-of-even-numbers-after-queries.js)|Easy|
 989|[Add to Array-Form of Integer](./solutions/0989-add-to-array-form-of-integer.js)|Easy|
 994|[Rotting Oranges](./solutions/0994-rotting-oranges.js)|Medium|

solutions/0983-minimum-cost-for-tickets.js

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+/**
+ * 983. Minimum Cost For Tickets
+ * https://leetcode.com/problems/minimum-cost-for-tickets/
+ * Difficulty: Medium
+ *
+ * You have planned some train traveling one year in advance. The days of the year in which you will
+ * travel are given as an integer array days. Each day is an integer from 1 to 365.
+ *
+ * Train tickets are sold in three different ways:
+ * - a 1-day pass is sold for costs[0] dollars,
+ * - a 7-day pass is sold for costs[1] dollars, and
+ * - a 30-day pass is sold for costs[2] dollars.
+ *
+ * The passes allow that many days of consecutive travel.
+ *
+ * For example, if we get a 7-day pass on day 2, then we can travel for 7 days: 2, 3, 4, 5, 6, 7,
+ * and 8.
+ *
+ * Return the minimum number of dollars you need to travel every day in the given list of days.
+ */
+
+/**
+ * @param {number[]} days
+ * @param {number[]} costs
+ * @return {number}
+ */
+var mincostTickets = function(days, costs) {
+  const lastDay = days[days.length - 1];
+  const travelDays = new Set(days);
+  const minCost = new Array(lastDay + 1).fill(0);
+
+  for (let day = 1; day <= lastDay; day++) {
+    if (!travelDays.has(day)) {
+      minCost[day] = minCost[day - 1];
+      continue;
+    }
+
+    const oneDayCost = minCost[day - 1] + costs[0];
+    const sevenDayCost = minCost[Math.max(0, day - 7)] + costs[1];
+    const thirtyDayCost = minCost[Math.max(0, day - 30)] + costs[2];
+
+    minCost[day] = Math.min(oneDayCost, sevenDayCost, thirtyDayCost);
+  }
+
+  return minCost[lastDay];
+};