LeetCode Q.2454

[StarsExpress]

https://github.com/doocs/leetcode/blob/main/solution/2400-2499/2454.Next%20Greater%20Element%20IV/README.md

Time complexity can achieve O(n) by double stacks.

class Solution:
    def secondGreaterElement(self, nums: List[int]) -> List[int]:
        total_nums = len(nums)
        second_next_greater = [-1] * total_nums

        # Decreasing monotonic stacks: (num, idx).
        stack_1, stack_2 = [(nums[0], 0)], []
        # Increasing monotonic list to transport tuples from stack 1 to stack 2.
        transporter = []

        for current_idx in range(1, total_nums):
            while stack_2 and stack_2[-1][0] < nums[current_idx]:
                _, past_idx = stack_2.pop(-1)
                second_next_greater[past_idx] = nums[current_idx]

            while stack_1 and stack_1[-1][0] < nums[current_idx]:
                transporter.append(stack_1.pop(-1))
            
            stack_2.extend(transporter[::-1])  # Ensure decreasing monotonicity.
            transporter.clear()
            stack_1.append((nums[current_idx], current_idx))

        return second_next_greater

[HARIHARANS24]

here is the solution to solve the above problem

class Solution:
def secondGreaterElement(self, nums: List[int]) -> List[int]:
total_nums = len(nums)
second_next_greater = [-1] * total_nums

    # Stack 1: elements waiting for their first greater
    # Stack 2: elements waiting for their second greater
    stack_1, stack_2 = [(nums[0], 0)], []
    transporter = []

    for current_idx in range(1, total_nums):
        # Process second greater
        while stack_2 and stack_2[-1][0] < nums[current_idx]:
            _, past_idx = stack_2.pop()
            second_next_greater[past_idx] = nums[current_idx]

        # Process first greater and move to transporter
        while stack_1 and stack_1[-1][0] < nums[current_idx]:
            transporter.append(stack_1.pop())

        # Move from transporter to stack_2
        stack_2.extend(reversed(transporter))
        transporter.clear()

        # Push current element to stack_1
        stack_1.append((nums[current_idx], current_idx))

    return second_next_greater