503-next-greater-element-ii.py

"""
Problem Link: https://leetcode.com/problems/next-greater-element-ii/

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.
"""
class Solution:
    # Space Complexity: O(N)
    # Time Complexity: O(N)
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
      stack, res = [], [-1] * len(nums)
      for i in range(2*len(nums)-1, -1, -1):
        index = i % len(nums)
        while stack and nums[stack[-1]] <= nums[index]:
          stack.pop()
          
        res[index] = nums[stack[-1]] if stack else -1
        stack.append(index)
      return res