Introduction

In this blog post, we will explore how to solve the LeetCode problem "2454" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to find the second greater element in an array:

• Monotonic Stack for First Greater: Use a stack to efficiently find the next greater element for each number.
• Secondary Stack for Second Greater: Maintain a separate stack to hold potential candidates for the second greater element. When a new element is greater than the top of the first-greater stack, it becomes a candidate for the second greater element of the elements in the secondary stack.

• Efficiently Update Results: Populate the result array as the second greater elements are discovered, avoiding redundant searches.

• Runtime Complexity: O(n), where n is the length of the input array.

• Storage Complexity: O(n)

Code

    def secondGreaterElement(nums):
    n = len(nums)
    result = [-1] * n
    stack1 = []  # For finding the first greater element
    stack2 = []  # For storing potential second greater elements

    for i, num in enumerate(nums):
        # Find second greater elements for elements in stack2
        while stack2 and nums[stack2[-1][0]] < num:
            index, _ = stack2.pop()
            result[index] = num

        # Find first greater elements, and move potential ones to stack2
        temp_stack = []
        while stack1 and nums[stack1[-1][0]] < num:
            index, _ = stack1.pop()
            temp_stack.append((index, nums[index]))

        # Push back to stack1 to maintain order
        while temp_stack:
            stack2.append(temp_stack.pop())

        stack1.append((i, num))

    return result