496. Next Greater Element I

#Problem

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.

https://leetcode.com/problems/next-greater-element-i/

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?

#Test Cases

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:

import pytest

from solution import Solution
from solution2 import Solution as Solution2


@pytest.mark.parametrize('nums1, nums2, expected', [
    ([4,1,2], [1,3,4,2], [-1,3,-1]),
    ([2,4], [1,2,3,4], [3,-1]),
])
@pytest.mark.parametrize('sol', [Solution(), Solution2()])
class Test:
    def test_solution(self, sol, nums1, nums2, expected):
        assert sol.nextGreaterElement(nums1.copy(), nums2) == expected

#Thoughts

在 1475. Final Prices With a Special Discount in a Shop 中提到这类找左侧/右侧第一个比当前元素小/大的问题，都可以使用单调栈，线性时间可解。

本题可以先对 nums2，利用单调栈计算每个元素的 next greater 元素，用哈希表保存结果。然后遍历 nums1，从哈希表中查到对应的结果。

时间复杂度 O(n)，空间复杂度 O(n)。

#Code

=Backward Iteration

class Solution:
    def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
        answers: dict[int, int] = {} # {val: val's next greater element}
        stack = []
        for i in range(len(nums2) - 1, -1, -1):
            val = nums2[i]
            while stack and stack[-1] < val:
                stack.pop()

            answers[val] = stack[-1] if stack else -1
            stack.append(val) # val is greater than stack[-1].

        for i, val in enumerate(nums1):
            nums1[i] = answers[val]

        return nums1

=Forward Iteration

class Solution:
    def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
        answers = {val: -1 for val in nums1} # {val: val's next greater element}
        stack = []
        for i, val in enumerate(nums2):
            while stack and nums2[stack[-1]] < val:
                answers[nums2[stack.pop()]] = val
            stack.append(i)

        for i, val in enumerate(nums1):
            nums1[i] = answers[val]

        return nums1