Description
https://leetcode.com/problems/next-greater-element-i/
You are given two integer arrays nums1 and nums2 both of unique elements, where nums1 is a 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, return -1 for this number.
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.
Constraints:
- 1 <= nums1.length <= nums2.length <= 1000
- 0 <= nums1[i], nums2[i] <= 104
- All integers in nums1andnums2are unique.
- All the integers of nums1also appear innums2.
Follow up: Could you find an O(nums1.length + nums2.length) solution?
Explanation
Count, sort, and find out the result.
Python Solution
class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        results = []
        
        for i, num1 in enumerate(nums1):
            found = False
            for num2 in nums2[nums2.index(num1):]:
                if num2 > num1:
                    results.append(num2)
                    found = True
                    break
            if not found:
                results.append(-1)
                
        return results
            - Time Complexity: O(N^2)
- Space Complexity: O(N)