Divide Array Into Arrays With Max Difference

• 1. Define a recursive backtracking function, e.g., canPartition(nums, k, used_mask, current_partition).
• 2. The base case for the recursion is when all elements have been placed into valid groups. In this case, a solution has been found.
• 3. In the recursive step, find the first available element in nums.
• 4. Iterate through all possible pairs of other available elements to form a group of three.
• 5. For each potential group, check if it satisfies the condition max(group) - min(group) <= k.
• 6. If the group is valid, add it to the current partition, mark its elements as used, and make a recursive call for the remaining elements.
• 7. If the recursive call returns false, backtrack by removing the group and unmarking its elements, then try the next combination.
• 8. If all combinations for the current element are exhausted without success, return false.

The fundamental idea is to systematically try every possible way to group the n numbers into n/3 sets of three. A recursive function can be used to implement this. To manage the combinations, we can use a boolean array used to keep track of elements that have already been assigned to a group.

The backtracking function would work as follows: find the first unused element, assign it to a new group, then search for two other unused elements to complete the group. For each potential group, we check if it's valid (i.e., max_element - min_element <= k). If it is, we add it to our current partition and recurse on the remaining elements. If the recursive call fails to find a complete solution, we backtrack by undoing our choice and trying a different combination of elements for the current group.

While this approach is guaranteed to find a solution if one exists, its performance is extremely poor due to the combinatorial explosion of possibilities. For an array of size n, the number of partitions is enormous, making this approach impractical for the given constraints.

Here is a conceptual code skeleton for the backtracking approach:

1// This is a conceptual implementation and will be too slow for the given constraints.2class Solution {3    public int[][] divideArray(int[] nums, int k) {4        int n = nums.length;5        Arrays.sort(nums); // Sorting helps to prune some branches but doesn't change the worst-case complexity.6        List<int[]> resultGroups = new ArrayList<>();7        boolean[] used = new boolean[n];8        if (canPartition(nums, k, used, resultGroups, 0)) {9            return resultGroups.toArray(new int[0][]);10        } else {11            return new int[0][0];12        }13    }14 15    private boolean canPartition(int[] nums, int k, boolean[] used, List<int[]> resultGroups, int start) {16        if (resultGroups.size() == nums.length / 3) {17            return true; // Successfully partitioned the whole array.18        }19 20        // Find the first element not yet used.21        int first_idx = -1;22        for (int i = start; i < nums.length; i++) {23            if (!used[i]) {24                first_idx = i;25                break;26            }27        }28 29        if (first_idx == -1) return true; // All elements used.30 31        used[first_idx] = true;32        // Find two other elements to form a group.33        for (int j = first_idx + 1; j < nums.length; j++) {34            if (used[j]) continue;35            for (int l = j + 1; l < nums.length; l++) {36                if (used[l]) continue;37 38                // Check if the group is valid.39                if (nums[l] - nums[first_idx] <= k) {40                    used[j] = true;41                    used[l] = true;42                    resultGroups.add(new int[]{nums[first_idx], nums[j], nums[l]});43 44                    if (canPartition(nums, k, used, resultGroups, first_idx + 1)) {45                        return true;46                    }47 48                    // Backtrack49                    resultGroups.remove(resultGroups.size() - 1);50                    used[l] = false;51                    used[j] = false;52                }53            }54        }55 56        used[first_idx] = false; // Backtrack57        return false;58    }59}

• 1. Sort the input array nums in non-decreasing order.
• 2. Create a 2D array result of size (n/3) x 3 to store the answer.
• 3. Iterate through the sorted array with a step of 3, from i = 0 to n-1.
• 4. In each iteration, consider the triplet (nums[i], nums[i+1], nums[i+2]).
• 5. Check if the difference between the largest and smallest element, nums[i+2] - nums[i], is greater than k.
• 6. If it is, a valid partition is impossible. Return an empty 2D array.
• 7. If the difference is less than or equal to k, the triplet is a valid group. Add it to the result array.
• 8. After the loop finishes, return the result array.

The core insight is that to satisfy the condition max - min <= k for every group, we should try to make the difference within each group as small as possible. Sorting the array nums helps achieve this. Once the array is sorted, the smallest elements are at the beginning and the largest at the end.

A greedy strategy can then be applied: iterate through the sorted array and form groups from consecutive elements. We take the first three elements nums[0], nums[1], nums[2] to form the first group. Since the array is sorted, nums[0] is the minimum and nums[2] is the maximum in this group. We only need to check if nums[2] - nums[0] <= k.

If this condition fails, no solution is possible. Why? Because nums[0] is the smallest element overall. To form a valid group with nums[0], we must pick two other elements. To minimize the group's difference, we should pick the two smallest available elements, which are nums[1] and nums[2]. If even this best-case group for nums[0] is invalid, no other group containing nums[0] can be valid, as any other element choices would lead to an even larger difference. Thus, we can safely conclude no solution exists.

If the condition holds, we form the group and move on to the next three elements nums[3], nums[4], nums[5] and repeat the process. If we successfully process the entire array this way, we have found a valid partition.

1import java.util.Arrays;2 3class Solution {4    public int[][] divideArray(int[] nums, int k) {5        int n = nums.length;6        // The problem statement guarantees n is a multiple of 3.7 8        // Sort the array to enable the greedy approach.9        Arrays.sort(nums);10 11        int[][] result = new int[n / 3][3];12 13        // Iterate through the sorted array, taking elements three at a time.14        for (int i = 0; i < n; i += 3) {15            // For a group {nums[i], nums[i+1], nums[i+2]}, the min is nums[i]16            // and the max is nums[i+2] because the array is sorted.17            if (nums[i + 2] - nums[i] > k) {18                // If the condition is not met for this group, it's impossible to partition.19                return new int[0][0]; // Return an empty array.20            }21            // If the condition is met, form the group.22            result[i / 3] = new int[]{nums[i], nums[i + 1], nums[i + 2]};23        }24 25        return result;26    }27}

• 1. Determine the range of values in nums. Let the maximum value be maxVal.
• 2. Create a frequency array, counts, of size maxVal + 1.
• 3. Populate the counts array by iterating through nums.
• 4. Overwrite the original nums array (or create a new one) with the sorted elements by iterating through the counts array. This is the Counting Sort algorithm.
• 5. Proceed with the same greedy grouping strategy as the previous approach: iterate through the sorted nums array with a step of 3.
• 6. For each triplet (nums[i], nums[i+1], nums[i+2]), check if nums[i+2] - nums[i] > k.
• 7. If the condition is violated, return an empty array.
• 8. Otherwise, add the valid group to the result.
• 9. Return the final result array.

This approach follows the same greedy logic as the previous one: sort the array and then form groups of three from consecutive elements. The only difference is the sorting algorithm used.

Given that the numbers in nums are within a known range (e.g., 1 to 10^5 as per constraints), we can use Counting Sort for a more efficient sorting step.

Counting Sort Steps:

1. Find the maximum value (maxVal) in nums.
2. Create a frequency map (or an array counts) of size maxVal + 1 to store the count of each number in nums.
3. Iterate through nums and populate the counts array.
4. Reconstruct the sorted array by iterating through the counts array. For each number i from 0 to maxVal, add i to the sorted array counts[i] times.

After sorting nums in O(N + M) time (where M is the range of values), the rest of the algorithm is identical to the previous greedy approach. We iterate through the now-sorted array, taking three elements at a time, and check if nums[i+2] - nums[i] <= k.

This optimization reduces the overall time complexity from O(N log N) to O(N + M), which is linear.

1import java.util.Arrays;2 3class Solution {4    public int[][] divideArray(int[] nums, int k) {5        int n = nums.length;6        int maxVal = 0;7        for (int num : nums) {8            if (num > maxVal) {9                maxVal = num;10            }11        }12 13        // Use Counting Sort14        int[] counts = new int[maxVal + 1];15        for (int num : nums) {16            counts[num]++;17        }18 19        int[] sortedNums = new int[n];20        int index = 0;21        for (int i = 0; i <= maxVal; i++) {22            for (int j = 0; j < counts[i]; j++) {23                sortedNums[index++] = i;24            }25        }26 27        // The rest of the logic is the same as the standard greedy approach28        int[][] result = new int[n / 3][3];29        for (int i = 0; i < n; i += 3) {30            if (sortedNums[i + 2] - sortedNums[i] > k) {31                return new int[0][0];32            }33            result[i / 3] = new int[]{sortedNums[i], sortedNums[i + 1], sortedNums[i + 2]};34        }35 36        return result;37    }38}