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feat(work): add solution for Next Greater Element I
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496. Next Greater Element I
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Difficulty: Easy
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https://leetcode.com/problems/next-greater-element-i/
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──────────────────────────────────────────────────
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The next greater element of some element x in an array is the first
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greater element that is to the right of x in the same array.
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You are given two distinct 0-indexed integer arrays nums1 and nums2,
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where nums1 is a subset of nums2.
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For each 0 <= i < nums1.length, find the index j such that nums1[i]
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== nums2[j] and determine the next greater element of nums2[j] in
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nums2. If there is no next greater element, then the answer for this
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query is -1.
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Return an array ans of length nums1.length such that ans[i] is the
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next greater element as described above.
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Example 1:
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Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
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Output: [-1,3,-1]
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Explanation: The next greater element for each value of nums1 is as
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follows:
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- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater
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element, so the answer is -1.
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- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
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- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater
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element, so the answer is -1.
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Example 2:
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Input: nums1 = [2,4], nums2 = [1,2,3,4]
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Output: [3,-1]
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Explanation: The next greater element for each value of nums1 is as
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follows:
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- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
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- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater
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element, so the answer is -1.
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Constraints:
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• 1 <= nums1.length <= nums2.length <= 1000
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• 0 <= nums1[i], nums2[i] <= 10^4
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• All integers in nums1 and nums2 are unique.
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• All the integers of nums1 also appear in nums2.
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Follow up: Could you find an O(nums1.length + nums2.length) solution?
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"""
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class Solution:
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def nextGreaterElement(self, nums1: list[int], nums2: list[int]) -> list[int]:
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mono_stack = []
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next_greater = {}
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for cur in nums2:
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while mono_stack and cur > mono_stack[-1]:
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val = mono_stack.pop()
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next_greater[val] = cur
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mono_stack.append(cur)
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return [next_greater.get(x, -1) for x in nums1]
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