Add solution #2012

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,686 LeetCode solutions in JavaScript
+# 1,687 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1543,6 +1543,7 @@
 2008|[Maximum Earnings From Taxi](./solutions/2008-maximum-earnings-from-taxi.js)|Medium|
 2009|[Minimum Number of Operations to Make Array Continuous](./solutions/2009-minimum-number-of-operations-to-make-array-continuous.js)|Hard|
 2011|[Final Value of Variable After Performing Operations](./solutions/2011-final-value-of-variable-after-performing-operations.js)|Easy|
+2012|[Sum of Beauty in the Array](./solutions/2012-sum-of-beauty-in-the-array.js)|Medium|
 2016|[Maximum Difference Between Increasing Elements](./solutions/2016-maximum-difference-between-increasing-elements.js)|Easy|
 2017|[Grid Game](./solutions/2017-grid-game.js)|Medium|
 2027|[Minimum Moves to Convert String](./solutions/2027-minimum-moves-to-convert-string.js)|Easy|

solutions/2012-sum-of-beauty-in-the-array.js

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+/**
+ * 2012. Sum of Beauty in the Array
+ * https://leetcode.com/problems/sum-of-beauty-in-the-array/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2)
+ * the beauty of nums[i] equals:
+ * - 2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1.
+ * - 1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied.
+ * - 0, if none of the previous conditions holds.
+ *
+ * Return the sum of beauty of all nums[i] where 1 <= i <= nums.length - 2.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var sumOfBeauties = function(nums) {
+  const n = nums.length;
+  let result = 0;
+  const leftMax = new Array(n).fill(nums[0]);
+  const rightMin = new Array(n).fill(nums[n - 1]);
+
+  for (let i = 1; i < n; i++) {
+    leftMax[i] = Math.max(leftMax[i - 1], nums[i - 1]);
+  }
+
+  for (let i = n - 2; i >= 0; i--) {
+    rightMin[i] = Math.min(rightMin[i + 1], nums[i + 1]);
+  }
+
+  for (let i = 1; i < n - 1; i++) {
+    if (leftMax[i] < nums[i] && nums[i] < rightMin[i]) {
+      result += 2;
+    } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
+      result += 1;
+    }
+  }
+
+  return result;
+};