# Solving Leetcode Interviews in Seconds with AI: Next Greater Element II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "503" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> 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  

	# Explanation
	Here's the solution:

*   **Monotonic Stack:** Utilize a stack to maintain a decreasing order of indices. This allows us to efficiently find the next greater element for each number.
*   **Circular Traversal:** Simulate a circular traversal by iterating through the array twice. This ensures that we can find the next greater element even if it wraps around to the beginning of the array.

*   **Runtime Complexity:** O(N), **Storage Complexity:** O(N)

	
	# Code
	```python
	def nextGreaterElements(nums):
    n = len(nums)
    result = [-1] * n
    stack = []  # Stack to store indices

    for i in range(2 * n):
        num = nums[i % n]  # Circular traversal using modulo operator

        while stack and nums[stack[-1]] < num:
            result[stack.pop()] = num

        if i < n:
            stack.append(i)

    return result
	```