LeetCode 496. Next Greater Element I

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 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?

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)

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