# 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.

• code
``````class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
d = defaultdict(lambda: -1)
stack = []
for v in nums2:
while stack and v > stack[-1]:
d[stack.pop()] = v
stack.append(v)
return [d[v] for v in nums1]
``````
• code
``````class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
d = defaultdict(lambda: -1)
for i in range(len(nums2)):
d[nums2[i]] = -1
for k in d.keys():
if d[k] == -1 and nums2[i] > k:
d[k] = nums2[i]
return [d[v] for v in nums1]
``````