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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "496" 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
	> The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2. For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1. Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.   Example 1:  Input: nums1 = [4,1,2], nums2 = [1,3,4,2] Output: [-1,3,-1] Explanation: The next greater element for each value of nums1 is as follows: - 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1. - 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3. - 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.  Example 2:  Input: nums1 = [2,4], nums2 = [1,2,3,4] Output: [3,-1] Explanation: The next greater element for each value of nums1 is as follows: - 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3. - 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.    Constraints:  1 <= nums1.length <= nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 104 All integers in nums1 and nums2 are unique. All the integers of nums1 also appear in nums2.    Follow up: Could you find an O(nums1.length + nums2.length) solution?

	# Explanation
	Here's an efficient solution to the problem, addressing the follow-up constraint of O(nums1.length + nums2.length) time complexity.

*   **Use a stack to find the next greater element for each number in `nums2`.** This allows us to process `nums2` in a single pass.
*   **Store the next greater element for each number in a hash map.** This provides O(1) lookup when processing `nums1`.
*   **Iterate through `nums1` and use the hash map to find the next greater elements.**

*   **Runtime Complexity:** O(nums1.length + nums2.length), **Storage Complexity:** O(nums2.length)

	
	# Code
	```python
	def nextGreaterElement(nums1: list[int], nums2: list[int]) -> list[int]:
    """
    Finds the next greater element for each element in nums1 within nums2.

    Args:
        nums1: The array to find next greater elements for.
        nums2: The array to search within.

    Returns:
        A list of next greater elements for each element in nums1.
    """

    next_greater = {}  # Hash map to store next greater elements
    stack = []  # Stack to keep track of elements while searching for next greater

    for num in nums2:
        while stack and num > stack[-1]:
            next_greater[stack.pop()] = num
        stack.append(num)

    # Remaining elements in the stack have no next greater element in nums2
    while stack:
        next_greater[stack.pop()] = -1

    # Build the result array for nums1
    result = [next_greater[num] for num in nums1]

    return result
	```