Add solution #1561

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,369 LeetCode solutions in JavaScript
+# 1,370 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1192,6 +1192,7 @@
 1558|[Minimum Numbers of Function Calls to Make Target Array](./solutions/1558-minimum-numbers-of-function-calls-to-make-target-array.js)|Medium|
 1559|[Detect Cycles in 2D Grid](./solutions/1559-detect-cycles-in-2d-grid.js)|Medium|
 1560|[Most Visited Sector in a Circular Track](./solutions/1560-most-visited-sector-in-a-circular-track.js)|Easy|
+1561|[Maximum Number of Coins You Can Get](./solutions/1561-maximum-number-of-coins-you-can-get.js)|Medium|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|
 1576|[Replace All ?'s to Avoid Consecutive Repeating Characters](./solutions/1576-replace-all-s-to-avoid-consecutive-repeating-characters.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|

solutions/1561-maximum-number-of-coins-you-can-get.js

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+/**
+ * 1561. Maximum Number of Coins You Can Get
+ * https://leetcode.com/problems/maximum-number-of-coins-you-can-get/
+ * Difficulty: Medium
+ *
+ * There are 3n piles of coins of varying size, you and your friends will take piles of coins as
+ * follows:
+ * - In each step, you will choose any 3 piles of coins (not necessarily consecutive).
+ * - Of your choice, Alice will pick the pile with the maximum number of coins.
+ * - You will pick the next pile with the maximum number of coins.
+ * - Your friend Bob will pick the last pile.
+ * - Repeat until there are no more piles of coins.
+ *
+ * Given an array of integers piles where piles[i] is the number of coins in the ith pile.
+ *
+ * Return the maximum number of coins that you can have.
+ */
+
+/**
+ * @param {number[]} piles
+ * @return {number}
+ */
+var maxCoins = function(piles) {
+  piles.sort((a, b) => b - a);
+  let result = 0;
+  const rounds = piles.length / 3;
+
+  for (let i = 1; i < piles.length - rounds; i += 2) {
+    result += piles[i];
+  }
+
+  return result;
+};