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
andnums2
are unique. - All the integers of
nums1
also 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)