Divide Array Into Arrays With Max Difference

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Salesforce

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ArraysGreedy Algorithms

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

Divide Array Into Arrays With Max Difference

Medium

Salesforce

0 views

Topics:

ArraysGreedy Algorithms

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array `nums`?
Is the input array guaranteed to be non-empty?
What should I return if it is not possible to divide the array into subarrays that satisfy the maximum difference condition?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when forming the subarrays?
If multiple valid groupings of subarrays are possible, is any valid grouping acceptable, or is there a specific criteria for the groupings I should prioritize?

Brute Force Solution

Approach

The brute force strategy for this array problem involves checking every single possible way to split the numbers into smaller groups. We explore all combinations to find valid groups that meet the difference requirement.

Here's how the algorithm would work step-by-step:

First, consider every way to start the first group of numbers.
For each of these starting groups, check if the largest and smallest number in the group have a difference that is too big.
If the difference is too big, forget about that group and try a different one.
If the difference is small enough, then try all possible ways to form the next group from the remaining numbers.
Again, check if the largest and smallest number in this new group have a difference that is too big.
Keep doing this until all the numbers have been assigned to a group.
Remember that we're trying every single possible way to make these groups.
Finally, if a complete set of groups is found where every group's difference is small enough, we have found a solution.

Code Implementation

def divide_array_brute_force(numbers, max_difference):
    result_sets = []

    def find_all_possible_groups(remaining_numbers, current_groups):
        # If no numbers are left, a complete set of groups has been formed
        if not remaining_numbers:
            result_sets.append(current_groups)
            return

        for i in range(1, len(remaining_numbers) + 1):
            first_group = remaining_numbers[:i]

            maximum_value = max(first_group)
            minimum_value = min(first_group)

            # Filter out invalid groups that don't meet the difference requirement.
            if maximum_value - minimum_value > max_difference:
                continue

            remaining_numbers_after_group = remaining_numbers[i:]
            find_all_possible_groups(remaining_numbers_after_group, current_groups + [first_group])

    find_all_possible_groups(numbers, [])

    # Return the first valid result found
    for possible_grouping in result_sets:
        return possible_grouping
    return []

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Big(O) Analysis

Time Complexity

O(n!) – The described brute force solution explores all possible ways to divide the array into subarrays. The number of ways to partition an array of size n grows factorially. Specifically, the algorithm considers all possible starting groups, then for each valid group it considers all possible ways to form the next group from the remaining numbers, and so on. Therefore, the time complexity is proportional to the number of possible partitions, which is O(n!).

Space Complexity

O(N) – The brute force approach, as described, explores all possible ways to split the array. In the worst-case scenario, the recursion depth could reach N, where N is the number of elements in the input array, because each recursive call considers forming a new group. Each level of recursion stores function call context, leading to a recursion stack of maximum depth N. Therefore, the auxiliary space used by the recursion stack grows linearly with the input size N, resulting in O(N) space complexity.

Optimal Solution

Approach

The best way to solve this is to first organize the input data in ascending order. Then, you can efficiently create new groups by iterating through the organized data and checking if the difference between the largest and smallest element in each group is within the limit.

Here's how the algorithm would work step-by-step:

Begin by arranging the numbers from smallest to largest.
Take the three smallest numbers to create the first potential group.
Examine the largest and smallest numbers in that group. If their difference isn't bigger than the maximum allowed difference, then you have a valid group, and you can save that group.
If the difference is too big, it's impossible to create valid groups, so return nothing.
Repeat this process by taking the next three smallest numbers that haven't been grouped yet to form another potential group.
Continue until all the numbers have been assigned to groups, or you discover that forming valid groups is impossible.

Code Implementation

def divide_array_into_arrays_with_max_difference(numbers, maximum_difference):
    numbers.sort()
    result = []
    group = []

    for number in numbers:
        group.append(number)
        if len(group) == 3:
            # Check if the difference within the group exceeds limit
            if group[-1] - group[0] > maximum_difference:
                return []

            result.append(group)
            group = []

    # If any numbers remain, it's impossible to form groups of 3
    if len(group) > 0:
        return []

    # All numbers have been successfully grouped
    return result

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is sorting the input array of size n. Common efficient sorting algorithms, like merge sort or quicksort, have a time complexity of O(n log n). The subsequent grouping and difference checking involves iterating through the sorted array once, which takes O(n) time. Since O(n log n) grows faster than O(n) as n increases, the overall time complexity is determined by the sorting step. Therefore, the time complexity is O(n log n).

Space Complexity

O(1) – The provided solution primarily sorts the input array in place. Beyond the input array itself, the algorithm only creates a few variables for tracking the group index and potentially a temporary list or array to store the resulting groups. However, since the question states to return nothing if valid groups can't be formed and the example solution only takes the three smallest numbers at a time to check if they are a group, the temporary group storage can be done with a fixed number of variables. The number of variables for group index and checking elements do not scale with the input size N. Therefore, the auxiliary space complexity is constant, O(1).

Edge Cases

Case	How to Handle
Empty input array	Return an empty list as there are no elements to form sub-arrays.
Input array with size less than 3	If the array size is less than 3, it's impossible to form sub-arrays of size 3; return an empty list.
Input array already sorted in ascending order	The sorting step might become redundant, but it still produces the correct result as per the problem statement.
Input array with all elements being equal	If max difference is at least 0, a single sub-array with those same numbers can be returned.
The array contains duplicate numbers	The algorithm should still function correctly as long as the sub-arrays are formed respecting the max difference requirement.
No valid subarrays exist because the difference between elements exceeds the limit.	Return an empty list, indicating no valid subarrays could be formed.
Integer overflow if array elements are very large	Use a data type capable of holding the difference between the largest numbers without overflowing (e.g., long in Java).
Large input array (scalability concerns)	Ensure the solution has an optimal time complexity (ideally O(n log n) due to sorting) to handle large arrays efficiently.