Divide Array Into Arrays With Max Difference

class Solution {
public:
    vector<vector<int>> groupNumbers(vector<int>& arr, int threshold) {
        sort(arr.begin(), arr.end());
        vector<vector<int>> result;
        for (int i = 0; i < arr.size(); i += 3) {
            if (arr[i + 2] - arr[i] > threshold) {
                return {};
            }
            result.push_back({arr[i], arr[i + 1], arr[i + 2]});
        }
        return result;
    }
};

The solution involves creating groups of consecutive numbers from a sorted array such that the difference between the maximum and minimum values in each group does not exceed a specified threshold. The following steps outline how to achieve this using C++:

1. Sort the input array. This ensures that the difference between consecutive elements is minimized, which is crucial for forming valid groups.

2. Initialize an empty 2D vector to store the result. This vector will hold the groups that meet the threshold condition.

3. Iterate through the array in steps of three. The step size of three indicates that each group will contain exactly three numbers. This is a fixed group size as seen in the implementation.

4. Check if the difference between the first and the last number in the current group exceeds the threshold. If it does, immediately return an empty vector since it's impossible to form a valid group with the current threshold.

5. If the difference is within the threshold, add the group to the result vector.

6. Continue this process until all elements have been grouped or the function exits early due to an invalid group.

7. Return the result vector containing all valid groups.

This approach ensures efficient grouping by leveraging sorting, which is O(n log n), followed by a linear scan to form groups, ensuring overall efficient execution. Note that the function assumes the input array’s size is divisible by three, as it does not handle cases where the last few elements might not form a complete group of three.

class Solution {
    public int[][] partitionArray(int[] arr, int diffLimit) {
        Arrays.sort(arr);
        int[][] result = new int[arr.length / 3][3];
        for (int j = 0; j < arr.length; j += 3) {
            if (arr[j + 2] - arr[j] > diffLimit) {
                return new int[0][0];
            }
            result[j / 3] = new int[]{arr[j], arr[j + 1], arr[j + 2]};
        }
        return result;
    }
}

This Java solution involves dividing a sorted array into smaller arrays, where each subarray contains exactly three elements and must conform to a specific maximum difference (diffLimit) between the smallest and largest numbers in the subarray. The solution follows these steps:

1. Sort the input array using the Arrays.sort() method. This sorting helps in easily finding subarrays that satisfy the difference condition.
2. Initialize a 2D array result to store the subarrays. The size of this array is determined by dividing the length of the input array by 3, as each subarray is expected to contain three elements.
3. Iterate through the sorted array in steps of three to form subarrays. During each iteration, check if the difference between the first and third elements in the current segment exceeds the diffLimit.
4. If the difference condition is violated, immediately return an empty 2D array, indicating the impossibility of forming the desired subarrays.
5. If the condition is met, store the current segment in the result array.
6. Return the result array which contains all the valid subarrays.

The method is efficient and succinct, handling edge cases such as exceeding the difference limit by promptly returning an empty array. This solution ensures each group of three consecutive elements from the sorted array forms a valid subarray, strictly adhering to the specified maximum difference.

var splitArrayInTriples = function(array, threshold) {
    array.sort((x, y) => x - y);
    const result = [];
    for (let index = 0; index < array.length; index += 3) {
        if (array[index + 2] - array[index] > threshold) {
            return [];
        }
        result.push([array[index], array[index + 1], array[index + 2]]);
    }
    return result;
};

This solution addresses the problem of dividing an array into groups of three elements each, such that the maximum difference within any group does not exceed a specified threshold. The following steps outline how the JavaScript function splitArrayInTriples achieves this:

1. Sort the input array to ensure elements are arranged in ascending order, which simplifies the comparison of elements for forming triples.
2. Initialize an empty array named result to store the groups of triples that meet the condition.
3. Iterate through the array using a loop that increases the index by three each step, which naturally segments the array into groups of three.
4. During each iteration, check if the difference between the first and third elements in the current group exceeds the given threshold.
5. If the difference exceeds the threshold, the function immediately returns an empty array indicating that it is not possible to split the array under the given conditions.
6. If the difference is within the threshold, add the current triple (three consecutive elements from the array) to the result array.
7. Return the result array after the loop completes, which contains all valid triples if the array can be successfully split as per the conditions.

This approach ensures an efficient division of the array while strictly adhering to the maximum difference constraint.

class Solution:
    def partitionArray(self, array: List[int], limit: int) -> List[List[int]]:
        array.sort()
        result = []
        for i in range(0, len(array), 3):
            if array[i + 2] - array[i] > limit:
                return []
            result.append([array[i], array[i + 1], array[i + 2]])
        return result

This solution addresses the problem of partitioning an array into smaller arrays where the maximum difference between the largest and smallest elements in each subarray does not exceed a given limit. It's written in Python and employs a basic strategy to achieve this.

Start by sorting the original array. This sorting step ensures that elements are arranged in ascending order, which facilitates checking the differences within potential subarrays.

Create an empty list named result to store the subarrays that match the criteria.

Use a loop to process the sorted array in steps of three. Within each step, check the condition that the difference between the maximum and minimum element of the current triplet (i.e. subarray) is within the given limit. If any triplet exceeds the limit, return an empty list indicating the partitioning isn't possible under the constraint.

If a valid triplet is found, append this triplet to the result list.

Finally, after all suitable triplets have been processed, return the result, which contains the required subarrays.

Ensure to handle all edge cases, especially when the array length isn't a multiple of three, as the solution assumes sets of exactly three elements. Adjustments might be needed for arrays with different sizes or additional validations for cases not divisible by three.