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Java-based LeetCode algorithm problem solutions, regularly updated
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496\. Next Greater Element I
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._
**Example 1:**
**Input:** nums1 = [4,1,2], nums2 = [1,3,4,2]
**Output:** [-1,3,-1]
**Explanation:**
The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
**Example 2:**
**Input:** nums1 = [2,4], nums2 = [1,2,3,4]
**Output:** [3,-1]
**Explanation:**
The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
**Constraints:**
* `1 <= nums1.length <= nums2.length <= 1000`
* 0 <= nums1[i], nums2[i] <= 104
* All integers in `nums1` and `nums2` are **unique**.
* All the integers of `nums1` also appear in `nums2`.
**Follow up:** Could you find an `O(nums1.length + nums2.length)` solution?
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