496. Next Greater Element I

Description

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

next-greater-element-i.py

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        n1, n2 = len(nums1), len(nums2)
        index = {x: i for i, x in enumerate(nums1)}
        res = [-1] * n1

        stack = []

        for i, x in enumerate(nums2):
            while stack and x > nums2[stack[-1]]:
                num = nums2[stack.pop()]
                if num in index:
                    res[index[num]] = x
            stack.append(i)

        return res