Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #2966: Divide Array Into Arrays With Max Difference

In this guide, we solve Leetcode #2966 Divide Array Into Arrays With Max Difference in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Greedy, Array, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Python Solution

class Solution: def divideArray(self, nums: List[int], k: int) -> List[List[int]]: nums.sort() ans = [] n = len(nums) for i in range(0, n, 3): t = nums[i : i + 3] if t[2] - t[0] > k: return [] ans.append(t) return ans

Complexity

The time complexity is O(n×log⁡n)O(n \times \log n)O(n×logn), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy