0503-Next-Greater-Element-II.py

'''
 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. 
'''
# Time Complexity: O(n)
# Space Complexity: O(n)
class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        res = [-1] * len(nums)
        stack = [nums[n] for n in range(len(nums) - 1, -1, -1)]
        for i in range(len(nums) - 1, -1, -1):
            while stack and stack[-1] <= nums[i]:
                stack.pop()
            if stack:
                res[i] = stack[-1]
            stack.append(nums[i])
        return res