0902HardArrayBinary SearchStackSortingHeap (Priority Queue)Monotonic Stack

The Second Taller Mast Ahead

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Mast heights along a line are given as nums.

For each mast, look rightwards for the masts strictly taller than it. Return the height of the second such mast, or -1 when fewer than two exist.

Return one answer per mast, in order.

Examples

Example 1

Input: nums = [13, 6, 19, 7, 21, 8]
Output: [21, 7, -1, 8, -1, -1]

For the mast of height 13, the taller masts ahead are 19 and 21, so the second is 21. For the 6, they are 19, 7, 21 and 8, so the second is 7. The last two masts have no second taller mast ahead.

Example 2

Input: nums = [1, 2, 3, 4, 5]
Output: [3, 4, 5, -1, -1]

Every mast is taller than the one before, so the second taller mast ahead is two places along, until the list runs out.

Example 3

Input: nums = [5, 5, 5, 5]
Output: [-1, -1, -1, -1]

No mast is strictly taller than any other, so none has an answer.

Constraints

1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def second_greater_element(nums: list[int]) -> list[int]:

Java

public int[] secondGreaterElement(int[] nums)

The intended approach

Time O(n) for n masts: each one is pushed and popped at most twice.Space O(n) for the two stacks.

A mast passes through two stages: waiting for anything taller, then waiting for a second such mast. Keep a stack for each stage, both holding masts in decreasing height, which is what lets a new mast settle a whole run of them at once. Reading left to right, a new mast first answers every mast on the second-stage stack it beats, since that is their second taller mast. It then moves every mast it beats off the first-stage stack onto the second, and finally joins the first stage itself. The move has to go on in reverse of the order the masts come off, because coming off a decreasing stack hands back the shortest first and the second stack must stay decreasing too. Each mast changes stack a bounded number of times, which keeps the whole sweep linear.

Pythonentry point: second_greater_element

def second_greater_element(nums):
    n = len(nums)
    out = [-1] * n
    waiting_first = []
    waiting_second = []
    for i in range(n):
        v = nums[i]
        while waiting_second and nums[waiting_second[-1]] < v:
            out[waiting_second.pop()] = v
        moved = []
        while waiting_first and nums[waiting_first[-1]] < v:
            moved.append(waiting_first.pop())
        while moved:
            waiting_second.append(moved.pop())
        waiting_first.append(i)
    return out

The obvious version

For each mast, walk rightwards counting the masts taller than it and stop at the second one. Quadratic in the mast count and it keeps nothing between masts.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_second_greater_element(nums):
    n = len(nums)
    out = []
    for i in range(n):
        found = 0
        answer = -1
        for j in range(i + 1, n):
            if nums[j] > nums[i]:
                found += 1
                if found == 2:
                    answer = nums[j]
                    break
        out.append(answer)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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