2454. Next Greater Element IV

Description

You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.

The second greater integer of nums[i] is nums[j] such that:

• j > i
• nums[j] > nums[i]
• There exists exactly one index k such that nums[k] > nums[i] and i < k < j.

If there is no such nums[j], the second greater integer is considered to be -1.

• For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1.

Return an integer array answer, where answer[i] is the second greater integer of nums[i].

 

Example 1:

Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].

Example 2:

Input: nums = [3,3]
Output: [-1,-1]
Explanation:
We return [-1,-1] since neither integer has any integer greater than it.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 109

Solution

next-greater-element-iv.py

class Solution:
    def secondGreaterElement(self, nums: List[int]) -> List[int]:
        N = len(nums)
        res = [-1] * N
        stack = []
        pq = []

        for i, x in enumerate(nums):
            while pq and x > pq[0][0]:
                rx, ri = heappop(pq)
                res[ri] = x

            while stack and x > nums[stack[-1]]:
                popIndex = stack.pop()
                heappush(pq, (nums[popIndex], popIndex))

            stack.append(i)

        return res