Problem Description

You are given an integer array nums where the length n is a multiple of 3, and a positive integer k.

Your task is to divide the array nums into n/3 groups, where each group contains exactly 3 elements. The key requirement is that within each group, the difference between any two elements must be less than or equal to k.

More specifically:

• If a group contains elements [a, b, c], then |a - b| ≤ k, |b - c| ≤ k, and |a - c| ≤ k must all be true.

You need to return a 2D array containing all the groups. If it's impossible to form such groups that satisfy the condition, return an empty array. If multiple valid answers exist, you can return any one of them.

For example, if nums = [1, 3, 4, 8, 7, 9] and k = 2:

• One valid grouping could be [[1, 3, 4], [7, 8, 9]] because in the first group the maximum difference is 4 - 1 = 3 which exceeds k = 2, so this wouldn't actually work.
• The problem requires finding groups where all elements are close enough to each other (within a difference of k).

The solution approach sorts the array first, then groups consecutive elements in sets of 3. For each group of 3 consecutive elements after sorting, it checks if the difference between the largest and smallest element exceeds k. If any group violates this condition, it returns an empty array. Otherwise, it adds the valid group to the result.

Intuition

The key insight is that if we want to group elements where the maximum difference within each group is minimized, we should group elements that are close in value together.

Think about it this way: if we have a sorted array, adjacent elements will have the smallest possible differences. If we randomly pick three elements from different parts of the array, the difference between them could be very large. But if we pick three consecutive elements from a sorted array, we're guaranteed to get elements that are as close as possible to each other.

For example, consider nums = [1, 9, 3, 8, 4, 7]. If we try to group [1, 9, 4], the difference between 9 and 1 is 8, which might exceed our threshold k. But if we first sort to get [1, 3, 4, 7, 8, 9] and then group consecutive triplets [1, 3, 4] and [7, 8, 9], we minimize the maximum difference within each group.

Why does grouping consecutive elements after sorting work? In a sorted array, for any triplet [a, b, c] where a ≤ b ≤ c, the maximum difference is always c - a. Since all elements are in order, we don't need to check |a - b| or |b - c| separately - they will always be smaller than or equal to c - a.

This greedy approach of taking consecutive triplets from a sorted array is optimal because:

1. It minimizes the spread within each group
2. If this grouping doesn't work (some group has difference > k), then no other grouping will work either, since we've already minimized the differences as much as possible

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation follows a straightforward approach based on sorting and greedy grouping:

1. Sort the array: We first sort the input array nums in ascending order using nums.sort(). This ensures that elements with similar values are positioned next to each other.

2. Initialize the result: Create an empty list ans to store our groups of three elements.

3. Iterate and group: We iterate through the sorted array in steps of 3 using range(0, n, 3). This allows us to process exactly n/3 groups, where each group contains 3 consecutive elements from the sorted array.

4. Extract and validate each group: For each iteration at index i:

   • Extract three consecutive elements: t = nums[i : i + 3]
   • Check if the difference between the maximum and minimum elements in this group exceeds k: t[2] - t[0] > k
   • Since the array is sorted, t[2] is the maximum and t[0] is the minimum in the group

5. Handle invalid groups: If any group has a difference greater than k, we immediately return an empty array [], as it's impossible to satisfy the conditions.

6. Build the result: If the group is valid (difference ≤ k), we append it to our answer array: ans.append(t)

7. Return the result: After processing all groups successfully, we return the ans array containing all valid groups.

The time complexity is O(n log n) due to sorting, where n is the length of the input array. The space complexity is O(n) for storing the result array (not counting the space used by the sorting algorithm).

This greedy approach works because once the array is sorted, grouping consecutive elements guarantees the minimum possible difference within each group. If this optimal grouping fails, no other grouping can succeed.

Example Walkthrough

Let's walk through an example with nums = [3, 11, 8, 5, 10, 1] and k = 3.

Step 1: Sort the array

• Original: [3, 11, 8, 5, 10, 1]
• After sorting: [1, 3, 5, 8, 10, 11]

Step 2: Group consecutive elements in sets of 3

• Since we have 6 elements, we'll create 2 groups

Step 3: Process the first group (indices 0-2)

• Extract elements: [1, 3, 5]
• Check if valid: maximum difference = 5 - 1 = 4
• Since 4 > k = 3, this grouping violates our constraint
• Return empty array []

Let's try another example where grouping is possible: nums = [1, 3, 3, 9, 8, 8] and k = 3.

Step 1: Sort the array

• Original: [1, 3, 3, 9, 8, 8]
• After sorting: [1, 3, 3, 8, 8, 9]

Step 2: Process the first group (indices 0-2)

• Extract elements: [1, 3, 3]
• Check if valid: maximum difference = 3 - 1 = 2
• Since 2 ≤ k = 3, this group is valid
• Add [1, 3, 3] to result

Step 3: Process the second group (indices 3-5)

• Extract elements: [8, 8, 9]
• Check if valid: maximum difference = 9 - 8 = 1
• Since 1 ≤ k = 3, this group is valid
• Add [8, 8, 9] to result

Step 4: Return the result

• Final answer: [[1, 3, 3], [8, 8, 9]]

The key insight is that by sorting first and taking consecutive triplets, we ensure that elements within each group have the smallest possible differences. If this greedy approach fails (as in the first example), then no valid grouping exists.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The dominant operation is the sorting of the array using nums.sort(), which has a time complexity of O(n × log n) where n is the length of the input array. After sorting, the code iterates through the array once in steps of 3, performing constant-time operations (slicing 3 elements, comparison, and appending to the result list). This iteration takes O(n) time. Therefore, the overall time complexity is O(n × log n) + O(n) = O(n × log n).

Space Complexity: O(n)

The space complexity consists of:

• The sorting operation may require O(log n) auxiliary space for the sorting algorithm (typically Timsort in Python)
• The answer list ans stores all elements from the input array, organized into subarrays of size 3, requiring O(n) space
• The temporary list t holds at most 3 elements at a time, which is O(1)

The dominant space requirement is the output array ans which contains all n elements, giving us an overall space complexity of O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Checking Array Length Before Processing

A common pitfall is assuming the array length is always a valid multiple of 3 without verification. While the problem states this constraint, defensive programming suggests validating input.

Issue: If len(nums) is not divisible by 3, the slicing operation nums[i:i+3] might create incomplete groups with fewer than 3 elements, leading to index errors when accessing triplet[2].

Solution:

class Solution:
    def divideArray(self, nums: List[int], k: int) -> List[List[int]]:
        n = len(nums)
      
        # Validate that array length is divisible by 3
        if n % 3 != 0:
            return []
      
        nums.sort()
        result = []
      
        for i in range(0, n, 3):
            triplet = nums[i:i + 3]
            if triplet[2] - triplet[0] > k:
                return []
            result.append(triplet)
      
        return result

2. Modifying the Original Input Array

The current solution sorts nums in-place using nums.sort(), which modifies the original array. This could be problematic if the caller expects the input to remain unchanged.

Issue: Side effects on input data can cause bugs in calling code that relies on the original order.

Solution:

class Solution:
    def divideArray(self, nums: List[int], k: int) -> List[List[int]]:
        # Create a copy to avoid modifying the original array
        sorted_nums = sorted(nums)
      
        result = []
        n = len(sorted_nums)
      
        for i in range(0, n, 3):
            triplet = sorted_nums[i:i + 3]
            if triplet[2] - triplet[0] > k:
                return []
            result.append(triplet)
      
        return result

3. Edge Case: Empty Array or Array with Less Than 3 Elements

Not handling edge cases where the input array is empty or has fewer than 3 elements can cause runtime errors.

Issue: Accessing triplet[2] when the triplet has fewer than 3 elements will raise an IndexError.

Solution:

class Solution:
    def divideArray(self, nums: List[int], k: int) -> List[List[int]]:
        # Handle edge cases
        if not nums or len(nums) < 3:
            return []
      
        n = len(nums)
        if n % 3 != 0:
            return []
      
        nums.sort()
        result = []
      
        for i in range(0, n, 3):
            triplet = nums[i:i + 3]
            # Additional safety check
            if len(triplet) != 3:
                return []
            if triplet[2] - triplet[0] > k:
                return []
            result.append(triplet)
      
        return result