# 496. Next Greater Element I

## Problem

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] <= 10^4
- All integers in nums1 and nums2 are unique.
- All the integers of nums1 also appear in nums2.

## Code

```
class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<>();
Stack<Integer> stack = new Stack<>();
for(int num : nums2){
while(!stack.isEmpty() && stack.peek() < num){
map.put(stack.pop(), num);
}
stack.push(num);
}
for(int i = 0; i < nums1.length; i++){
nums1[i] = map.getOrDefault(nums1[i], -1);
}
return nums1;
}
}
```

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