LeetCode #2966 — MEDIUM

Divide Array Into Arrays With Max Difference

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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: [[2,2,2],[4,5,5],[5,5,7],[7,8,8],[9,9,10],[11,12,12]]

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,3,4,8,7,9,3,5,1]
2

Example 2

[2,4,2,2,5,2]
2

Example 3

[4,2,9,8,2,12,7,12,10,5,8,5,5,7,9,2,5,11]
14

Step 02

Core Insight

What unlocks the optimal approach

Try to use a greedy approach.
Sort the array and try to group each <code>3</code> consecutive elements.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2966: Divide Array Into Arrays With Max Difference
class Solution {
    public int[][] divideArray(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int[][] ans = new int[n / 3][];
        for (int i = 0; i < n; i += 3) {
            int[] t = Arrays.copyOfRange(nums, i, i + 3);
            if (t[2] - t[0] > k) {
                return new int[][] {};
            }
            ans[i / 3] = t;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2966: Divide Array Into Arrays With Max Difference
func divideArray(nums []int, k int) [][]int {
	sort.Ints(nums)
	ans := [][]int{}
	for i := 0; i < len(nums); i += 3 {
		t := slices.Clone(nums[i : i+3])
		if t[2]-t[0] > k {
			return [][]int{}
		}
		ans = append(ans, t)
	}
	return ans
}

# Accepted solution for LeetCode #2966: Divide Array Into Arrays With Max Difference
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

// Accepted solution for LeetCode #2966: Divide Array Into Arrays With Max Difference
impl Solution {
    pub fn divide_array(mut nums: Vec<i32>, k: i32) -> Vec<Vec<i32>> {
        nums.sort();
        let mut ans = Vec::new();
        let n = nums.len();

        for i in (0..n).step_by(3) {
            if i + 2 >= n {
                return vec![];
            }

            let t = &nums[i..i + 3];
            if t[2] - t[0] > k {
                return vec![];
            }

            ans.push(t.to_vec());
        }

        ans
    }
}

// Accepted solution for LeetCode #2966: Divide Array Into Arrays With Max Difference
function divideArray(nums: number[], k: number): number[][] {
    nums.sort((a, b) => a - b);
    const ans: number[][] = [];
    for (let i = 0; i < nums.length; i += 3) {
        const t = nums.slice(i, i + 3);
        if (t[2] - t[0] > k) {
            return [];
        }
        ans.push(t);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.