Add solution #1599

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,409 LeetCode solutions in JavaScript
+# 1,410 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1234,6 +1234,7 @@
 1594|[Maximum Non Negative Product in a Matrix](./solutions/1594-maximum-non-negative-product-in-a-matrix.js)|Medium|
 1595|[Minimum Cost to Connect Two Groups of Points](./solutions/1595-minimum-cost-to-connect-two-groups-of-points.js)|Hard|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
+1599|[Maximum Profit of Operating a Centennial Wheel](./solutions/1599-maximum-profit-of-operating-a-centennial-wheel.js)|Medium|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1599-maximum-profit-of-operating-a-centennial-wheel.js

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+/**
+ * 1599. Maximum Profit of Operating a Centennial Wheel
+ * https://leetcode.com/problems/maximum-profit-of-operating-a-centennial-wheel/
+ * Difficulty: Medium
+ *
+ * You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for
+ * up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you
+ * runningCost dollars.
+ *
+ * You are given an array customers of length n where customers[i] is the number of new customers
+ * arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times
+ * before the customers[i] customers arrive. You cannot make customers wait if there is room in the
+ * gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the
+ * ground and will exit once that gondola reaches the ground again.
+ *
+ * You can stop the wheel at any time, including before serving all customers. If you decide to stop
+ * serving customers, all subsequent rotations are free in order to get all the customers down
+ * safely. Note that if there are currently more than four customers waiting at the wheel, only
+ * four will board the gondola, and the rest will wait for the next rotation.
+ *
+ * Return the minimum number of rotations you need to perform to maximize your profit. If there is
+ * no scenario where the profit is positive, return -1.
+ */
+
+/**
+ * @param {number[]} customers
+ * @param {number} boardingCost
+ * @param {number} runningCost
+ * @return {number}
+ */
+var minOperationsMaxProfit = function(customers, boardingCost, runningCost) {
+  let waitingCustomers = 0;
+  let totalBoarded = 0;
+  let maxProfit = -Infinity;
+  let rotationAtMaxProfit = -1;
+  let currentProfit = 0;
+  let rotations = 0;
+
+  const gondolaCapacity = 4;
+
+  for (let i = 0; i < customers.length || waitingCustomers > 0; i++) {
+    rotations++;
+
+    if (i < customers.length) {
+      waitingCustomers += customers[i];
+    }
+
+    const boarding = Math.min(waitingCustomers, gondolaCapacity);
+    waitingCustomers -= boarding;
+    totalBoarded += boarding;
+
+    currentProfit = totalBoarded * boardingCost - rotations * runningCost;
+
+    if (currentProfit > maxProfit) {
+      maxProfit = currentProfit;
+      rotationAtMaxProfit = rotations;
+    }
+  }
+
+  return maxProfit > 0 ? rotationAtMaxProfit : -1;
+};