# 496 Next Greater Element I

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?

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 `````` ``````class Solution: def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: stack = [] m = {} for num in nums2: while stack and stack[-1] < num: m[stack[-1]] = num stack.pop() stack.append(num) for num in stack: m[num] = -1 res = [] for num in nums1: res.append(m[num]) return res``````