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Abhishek Chaudhary

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# 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?SOLUTION:

``````class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
n = len(nums2)
greater = {}
for i in range(n):
for j in range(i + 1, n):
if nums2[j] > nums2[i]:
greater[nums2[i]] = nums2[j]
break
return [greater.get(num, -1) for num in nums1]
``````