Add solution #1911

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,624 LeetCode solutions in JavaScript
+# 1,625 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
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 1906|[Minimum Absolute Difference Queries](./solutions/1906-minimum-absolute-difference-queries.js)|Medium|
 1909|[Remove One Element to Make the Array Strictly Increasing](./solutions/1909-remove-one-element-to-make-the-array-strictly-increasing.js)|Easy|
 1910|[Remove All Occurrences of a Substring](./solutions/1910-remove-all-occurrences-of-a-substring.js)|Medium|
+1911|[Maximum Alternating Subsequence Sum](./solutions/1911-maximum-alternating-subsequence-sum.js)|Medium|
 1920|[Build Array from Permutation](./solutions/1920-build-array-from-permutation.js)|Easy|
 1922|[Count Good Numbers](./solutions/1922-count-good-numbers.js)|Medium|
 1926|[Nearest Exit from Entrance in Maze](./solutions/1926-nearest-exit-from-entrance-in-maze.js)|Medium|

solutions/1911-maximum-alternating-subsequence-sum.js

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+/**
+ * 1911. Maximum Alternating Subsequence Sum
+ * https://leetcode.com/problems/maximum-alternating-subsequence-sum/
+ * Difficulty: Medium
+ *
+ * The alternating sum of a 0-indexed array is defined as the sum of the elements at even
+ * indices minus the sum of the elements at odd indices.
+ * - For example, the alternating sum of [4,2,5,3] is (4 + 5) - (2 + 3) = 4.
+ *
+ * Given an array nums, return the maximum alternating sum of any subsequence of nums (after
+ * reindexing the elements of the subsequence).
+ *
+ * A subsequence of an array is a new array generated from the original array by deleting some
+ * elements (possibly none) without changing the remaining elements' relative order. For
+ * example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements),
+ * while [2,4,2] is not.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxAlternatingSum = function(nums) {
+  let result = 0;
+  let oddSum = 0;
+
+  for (const num of nums) {
+    const prevEven = result;
+    result = Math.max(result, oddSum + num);
+    oddSum = Math.max(oddSum, prevEven - num);
+  }
+
+  return result;
+};