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, 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

    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