How do you solve Divide Array Into Arrays With Max Difference on LeetCode?

Divide Array Into Arrays With Max Difference (LeetCode #2966) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the array allows us to easily group elements with minimal differences.

What is the time complexity of Divide Array Into Arrays With Max Difference?

The optimal solution for Divide Array Into Arrays With Max Difference runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and we iterate through the array in O(n), making this approach efficient.

What is the brute force solution for Divide Array Into Arrays With Max Difference?

The brute force approach for Divide Array Into Arrays With Max Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of triplets in the array to see if they satisfy the condition of having a maximum difference of k. This is simple but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Divide Array Into Arrays With Max Difference?

Divide Array Into Arrays With Max Difference uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting, Array.

What are the interview tips for Divide Array Into Arrays With Max Difference?

Always consider sorting as a first step when dealing with range constraints. Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice dry runs on paper to solidify your understanding of the algorithm.

What are common mistakes when solving Divide Array Into Arrays With Max Difference?

Not sorting the array first, which leads to incorrect triplet checks. Failing to handle edge cases where the array cannot be divided correctly.

#2966

Divide Array Into Arrays With Max Difference

Medium

ArrayGreedySortingGreedySortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting and a greedy approach. By sorting the array first, we can ensure that any three consecutive elements will have the smallest possible difference, making it easier to check the condition.

⚙️

Algorithm

3 steps

1Step 1: Sort the array.
2Step 2: Iterate through the sorted array in steps of 3, checking if the difference between the first and last element of each triplet is less than or equal to k.
3Step 3: If valid, add the triplet to the result; if not, return an empty array.

solution.py11 lines

1# Full working Python code
2
3def divide_array(nums, k):
4    nums.sort()
5    result = []
6    for i in range(0, len(nums), 3):
7        if nums[i + 2] - nums[i] > k:
8            return []
9        result.append([nums[i], nums[i + 1], nums[i + 2]])
10    return result
11

ℹ

Complexity note: The sorting step takes O(n log n), and we iterate through the array in O(n), making this approach efficient.

1Sorting the array allows us to easily group elements with minimal differences.
2Using a greedy approach ensures that we can quickly check the conditions for triplets.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.