# 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