Add solution #496

Author: JoshCrozier

README.md

@@ -202,6 +202,7 @@
 485|[Max Consecutive Ones](./0485-max-consecutive-ones.js)|Easy|
 491|[Non-decreasing Subsequences](./0491-non-decreasing-subsequences.js)|Medium|
 492|[Construct the Rectangle](./0492-construct-the-rectangle.js)|Easy|
+496|[Next Greater Element I](./0496-next-greater-element-i.js)|Easy|
 500|[Keyboard Row](./0500-keyboard-row.js)|Easy|
 501|[Find Mode in Binary Search Tree](./0501-find-mode-in-binary-search-tree.js)|Easy|
 502|[IPO](./0502-ipo.js)|Hard|

solutions/0496-next-greater-element-i.js

@@ -0,0 +1,33 @@
+/**
+ * 496. Next Greater Element I
+ * https://leetcode.com/problems/next-greater-element-i/
+ * Difficulty: Easy
+ *
+ * The next greater element of some element x in an array is the first greater element
+ * that is to the right of x in the same array.
+ *
+ * You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is
+ * a subset of nums2.
+ *
+ * For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and
+ * determine the next greater element of nums2[j] in nums2. If there is no next greater
+ * element, then the answer for this query is -1.
+ *
+ * Return an array ans of length nums1.length such that ans[i] is the next greater
+ * element as described above.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number[]}
+ */
+var nextGreaterElement = function(nums1, nums2) {
+  return nums1.map(n => {
+    let index = nums2.indexOf(n);
+    while (nums2[index] <= n) {
+      index++;
+    }
+    return index >= nums2.length ? -1 : nums2[index];
+  });
+};