Next Greater Element II

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number.

Example 1:

Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number.
The second 1's next greater number needs to search circularly, which is also 2.

Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]

Constraints:

• 1 <= nums.length <= 104
• -109 <= nums[i] <= 109

SOLUTION:

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        n = len(nums)
        op = []
        for i in range(n):
            found = False
            for j in range(i + 1, i + n + 1):
                if nums[j % n] > nums[i]:
                    op.append(nums[j % n])
                    found = True
                    break
            if not found:
                op.append(-1)
        return op