### Problem Statement:

Given an array of integers and a threshold value *k*, the task is to divide the array into triplets such that the absolute difference between adjacent elements in each triplet does not exceed *k*. If such division is possible, return the triplets; otherwise, return an empty array.

### Approach:

To efficiently divide the array into triplets while satisfying the maximum difference constraint, we can follow these steps:

- Sort the array in non-decreasing order to facilitate easier traversal.
- Initialize variables to track the total number of possible triplets and an array to store the triplets.
- Iterate over the sorted array, forming triplets at each step.
- For each triplet, ensure that the absolute difference between adjacent elements does not exceed the given threshold
*k*. - If a valid triplet is formed, store it in the result array; otherwise, return an empty array indicating that such division is not possible.

https://leetcode.com/problems/divide-array-into-arrays-with-max-difference

```
class Solution {
public int[][] divideArray(int[] nums, int k) {
// Sort Array
Arrays.sort(nums);
int totalPossibleArrays = nums.length/3;
int answer[][] = new int[totalPossibleArrays][3];
int counter = 0;
for(int i=0;i<totalPossibleArrays;i++) {
answer[i][0] = nums[counter++];
if ((nums[counter]-answer[i][0]) <= k) {
answer[i][1] = nums[counter++];
} else {
return new int[0][0];
}
if ((nums[counter]-answer[i][0]) <= k) {
answer[i][2] = nums[counter++];
} else {
return new int[0][0];
}
}
return answer;
}
}
```

### Conclusion:

Dividing an array into triplets while ensuring a maximum difference constraint can be efficiently achieved using the algorithm described above. By sorting the array and carefully forming triplets, we can divide the array in *O*(*n*log*n*) time complexity, where *n* is the size of the input array. This approach provides an effective solution to this common problem in algorithmic programming.