496. Next Greater Element I (Python)

Related Topic

Stack.

Description

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Sample I/O

Example 1

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
    For number 1 in the first array, the next greater number for it in the second array is 3.
    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2

Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Methodology

The method is pretty straight forward. Put the nums2 into a dictionary, the key is value (due to no duplicates) and value is the index. Iterate the nums1, for each element, find the corresponded value in nums2 and get the it’s index. Then iterate nums2 that start from index + 1. If the value is smaller than the current element in nums1. Append the value to res stack else append -1. return res

def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        if not nums1 or not nums2: return []
        res = []
        nums2_dic = {v:i for i, v in enumerate(nums2)}
        for x in nums1:
            index=nums2_dic[x]+1
            while index < len(nums2) and x >= nums2[index]:
                index += 1
            if index == len(nums2):
                res.append(-1)
            else:
                res.append(nums2[index])
        return res

BigO

For each element in nums1 will compare every element (at most) in nums2 so it take O(n^2) time complexity.