"""
496. Next Greater Element I
Difficulty: Easy
https://leetcode.com/problems/next-greater-element-i/

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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.



Follow up: Could you find an O(nums1.length + nums2.length) solution?
"""


class Solution:
    def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
        mono_stack = []
        next_greater = {}

        for cur in nums2:
            while mono_stack and cur > mono_stack[-1]:
                val = mono_stack.pop()
                next_greater[val] = cur
            mono_stack.append(cur)

        return [next_greater.get(x, -1) for x in nums1]