503.next-greater-element-ii.py

#  Tag: Array, Stack, Monotonic Stack
#  Time: O(N)
#  Space: O(N)
#  Ref: -
#  Note: -

#  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
#  
#  

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        
        n = len(nums)
        res = [-1 for i in range(n)]
        stack = []
        nums = nums

        for i in range(n * 2):
            cur = i % n
            while len(stack) > 0 and nums[cur] > nums[stack[-1]]:
                j = stack[-1]
                stack.pop()
                res[j] = nums[cur]

            stack.append(cur)

        return res

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        
        n = len(nums)
        res = [-1 for i in range(n)]
        stack = []
        nums = nums

        for i in range(n):
            while len(stack) > 0 and nums[i] > nums[stack[-1]]:
                j = stack[-1]
                stack.pop()
                res[j] = nums[i]

            stack.append(i)

        for i in range(n):
            while len(stack) > 0 and nums[i] > nums[stack[-1]]:
                j = stack[-1]
                stack.pop()
                res[j] = nums[i]

        return res