Add solution #1330

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,234 LeetCode solutions in JavaScript
+# 1,235 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1008,6 +1008,7 @@
 1326|[Minimum Number of Taps to Open to Water a Garden](./solutions/1326-minimum-number-of-taps-to-open-to-water-a-garden.js)|Hard|
 1328|[Break a Palindrome](./solutions/1328-break-a-palindrome.js)|Medium|
 1329|[Sort the Matrix Diagonally](./solutions/1329-sort-the-matrix-diagonally.js)|Medium|
+1330|[Reverse Subarray To Maximize Array Value](./solutions/1330-reverse-subarray-to-maximize-array-value.js)|Hard|
 1331|[Rank Transform of an Array](./solutions/1331-rank-transform-of-an-array.js)|Easy|
 1332|[Remove Palindromic Subsequences](./solutions/1332-remove-palindromic-subsequences.js)|Easy|
 1333|[Filter Restaurants by Vegan-Friendly, Price and Distance](./solutions/1333-filter-restaurants-by-vegan-friendly-price-and-distance.js)|Medium|

solutions/1330-reverse-subarray-to-maximize-array-value.js

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+/**
+ * 1330. Reverse Subarray To Maximize Array Value
+ * https://leetcode.com/problems/reverse-subarray-to-maximize-array-value/
+ * Difficulty: Hard
+ *
+ * You are given an integer array nums. The value of this array is defined as the sum
+ * of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.
+ *
+ * You are allowed to select any subarray of the given array and reverse it. You can
+ * perform this operation only once.
+ *
+ * Find maximum possible value of the final array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxValueAfterReverse = function(nums) {
+  let baseSum = 0;
+  let maxGain = 0;
+  let minPair = Infinity;
+  let maxPair = -Infinity;
+  const n = nums.length;
+
+  for (let i = 0; i < n - 1; i++) {
+    baseSum += Math.abs(nums[i] - nums[i + 1]);
+    minPair = Math.min(minPair, Math.max(nums[i], nums[i + 1]));
+    maxPair = Math.max(maxPair, Math.min(nums[i], nums[i + 1]));
+  }
+
+  maxGain = Math.max(0, 2 * (maxPair - minPair));
+
+  for (let i = 0; i < n - 1; i++) {
+    const left = i > 0 ? Math.abs(nums[0] - nums[i + 1]) - Math.abs(nums[i] - nums[i + 1]) : 0;
+    const right = i < n - 2 ? Math.abs(nums[n - 1] - nums[i]) - Math.abs(nums[i] - nums[i + 1]) : 0;
+    maxGain = Math.max(maxGain, left, right);
+  }
+
+  return baseSum + maxGain;
+};