Leetcode Link

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.

Example 1:
Input: [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.
Note: The length of given array won't exceed 10000.

Solution:

def next_greater_element(nums):
    res = []
    stack = collections.deque()
    n = len(nums)
    for i in range(n * 2 - 1, -1, -1):
        while stack and stack[-1] <= nums[i % n]:
            stack.pop()  # remove smaller numbers
        # find next great number
        res[i % n] = stack[-1] if stack else -1
        stack.append(nums[i % n])  # push current
    return res