# 0496. Next Greater Element I

<https://leetcode.com/problems/next-greater-element-i>

## Description

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?

## ac

```java
class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        // edge cases
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0 || nums1.length > nums2.length)
            return new int[0];

        Stack<Integer> stack = new Stack<Integer>();
        Map<Integer, Integer> map = new HashMap<>();
        stack.push(nums2[0]);
        for (int i = 1; i < nums2.length; i++) {
            while (!stack.isEmpty() && nums2[i] > stack.peek()) {
                map.put(stack.pop(), nums2[i]);
            }
            stack.push(nums2[i]);
        }

        // get result
        int[] res = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            res[i] = map.getOrDefault(nums1[i], -1);
        }

        return res;
    }
}
```