# Solving Leetcode Interviews in Seconds with AI: Divide Array Into Arrays With Max Difference


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2966" 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
	> You are given an integer array nums of size n where n is a multiple of 3 and a positive integer k. Divide the array nums into n / 3 arrays of size 3 satisfying the following condition:  The difference between any two elements in one array is less than or equal to k.  Return a 2D array containing the arrays. If it is impossible to satisfy the conditions, return an empty array. And if there are multiple answers, return any of them.   Example 1:  Input: nums = [1,3,4,8,7,9,3,5,1], k = 2 Output: [[1,1,3],[3,4,5],[7,8,9]] Explanation: The difference between any two elements in each array is less than or equal to 2.  Example 2:  Input: nums = [2,4,2,2,5,2], k = 2 Output: [] Explanation: Different ways to divide nums into 2 arrays of size 3 are:  [[2,2,2],[2,4,5]] (and its permutations) [[2,2,4],[2,2,5]] (and its permutations)  Because there are four 2s there will be an array with the elements 2 and 5 no matter how we divide it. since 5 - 2 = 3 > k, the condition is not satisfied and so there is no valid division.  Example 3:  Input: nums = [4,2,9,8,2,12,7,12,10,5,8,5,5,7,9,2,5,11], k = 14 Output: [[2,2,12],[4,8,5],[5,9,7],[7,8,5],[5,9,10],[11,12,2]] Explanation: The difference between any two elements in each array is less than or equal to 14.    Constraints:  n == nums.length 1 <= n <= 105 n is a multiple of 3 1 <= nums[i] <= 105 1 <= k <= 105  

	# Explanation
	Here's the breakdown of the solution:

*   **Sort the array:** Sorting allows us to greedily group elements into triplets while maintaining the difference constraint.
*   **Greedy Grouping:** Iterate through the sorted array. Form triplets sequentially. If the difference between the largest and smallest element in a potential triplet exceeds `k`, it's impossible to form valid triplets.
*   **Result Handling:** If grouping is successful, return the 2D array of triplets. Otherwise, return an empty array.

*   **Runtime Complexity:** O(n log n) due to sorting.
*   **Storage Complexity:** O(n) to store the result array in the worst-case scenario.

	
	# Code
	```python
	def divide_array(nums: list[int], k: int) -> list[list[int]]:
    """
    Divides the array nums into n / 3 arrays of size 3 satisfying the condition:
    The difference between any two elements in one array is less than or equal to k.

    Args:
        nums: An integer array of size n where n is a multiple of 3.
        k: A positive integer.

    Returns:
        A 2D array containing the arrays. If it is impossible to satisfy the conditions, return an empty array.
    """

    n = len(nums)
    nums.sort()
    result = []
    for i in range(0, n, 3):
        if i + 2 >= n:
            return []

        if nums[i + 2] - nums[i] > k:
            return []

        result.append([nums[i], nums[i + 1], nums[i + 2]])

    return result
	```