Partition Array Such That Maximum Difference Is K

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ArraysGreedy Algorithms

You are given an integer array nums and an integer k. You may partition nums into one or more subsequences such that each element in nums appears in exactly one of the subsequences.

Return the minimum number of subsequences needed such that the difference between the maximum and minimum values in each subsequence is at most k.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [3,6,1,2,5], k = 2
Output: 2
Explanation:
We can partition nums into the two subsequences [3,1,2] and [6,5].
The difference between the maximum and minimum value in the first subsequence is 3 - 1 = 2.
The difference between the maximum and minimum value in the second subsequence is 6 - 5 = 1.
Since two subsequences were created, we return 2. It can be shown that 2 is the minimum number of subsequences needed.

Example 2:

Input: nums = [1,2,3], k = 1
Output: 2
Explanation:
We can partition nums into the two subsequences [1,2] and [3].
The difference between the maximum and minimum value in the first subsequence is 2 - 1 = 1.
The difference between the maximum and minimum value in the second subsequence is 3 - 3 = 0.
Since two subsequences were created, we return 2. Note that another optimal solution is to partition nums into the two subsequences [1] and [2,3].

Example 3:

Input: nums = [2,2,4,5], k = 0
Output: 3
Explanation:
We can partition nums into the three subsequences [2,2], [4], and [5].
The difference between the maximum and minimum value in the first subsequences is 2 - 2 = 0.
The difference between the maximum and minimum value in the second subsequences is 4 - 4 = 0.
The difference between the maximum and minimum value in the third subsequences is 5 - 5 = 0.
Since three subsequences were created, we return 3. It can be shown that 3 is the minimum number of subsequences needed.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵
0 <= k <= 10⁵

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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 are the possible ranges and data types of the numbers in the input array? Can I expect negative numbers, zeros, or floating-point numbers?
What should the function return if it is not possible to partition the array such that the maximum difference between any two numbers in the same partition is no more than K?
Is K guaranteed to be non-negative? Can K be zero?
Can the input array be empty or null? If so, what should I return?
If there are multiple valid ways to partition the array, is any valid partition acceptable, or is there a specific criteria for selecting the optimal one?

Brute Force Solution

Approach

The brute force method for this problem involves checking absolutely every possible way to split the numbers into groups. We want to find at least one split where, within each group, the biggest and smallest numbers are no more than a set amount apart. If no group exceeds that threshold, that partitioning is a candidate.

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

Consider the first possible split: putting all the numbers into a single group.
Check if the biggest and smallest numbers in this single group are within the specified distance from each other.
If they are, we've found a valid solution. If not, move on to the next possibility.
Now, try splitting the numbers into two groups. Consider all possible ways to do this: the first group has one number, and the second group has the rest; the first group has two numbers, and the second group has the rest; and so on.
For each of these two-group splits, check if the biggest and smallest numbers in EACH group are within the specified distance.
If BOTH groups meet this requirement, we have found a valid partitioning.
Repeat the same process by considering all possible ways to split the numbers into three groups, four groups, and so on, up to having each number in its own group.
For each way of splitting, check if the difference between the biggest and smallest value in each of the groups is less than or equal to the required distance.
If you find a splitting that satisfies the requirements, remember that splitting as one potential answer.
Continue until you've explored all possible ways of splitting the initial numbers into groups.
If you have found one or more ways to split the numbers while following the rules, we've solved the problem. Otherwise, there is no valid partitioning.

Code Implementation

def can_partition_brute_force(numbers, max_difference): 
    number_count = len(numbers)

    # Iterate through all possible numbers of groups
    for group_count in range(1, number_count + 1): 
        # Iterate through all possible combinations of groups
        for combination in generate_combinations(numbers, group_count): 

            is_valid_partition = True
            # Check each group to ensure it satisfies the difference requirement.

            for group in combination: 
                if len(group) > 0: 
                    maximum_value = max(group)
                    minimum_value = min(group)
                    if maximum_value - minimum_value > max_difference: 
                        is_valid_partition = False
                        break

            # If the current partition is valid, return true
            if is_valid_partition: 
                return True

    return False

def generate_combinations(numbers, group_count):
    if group_count == 1:
        yield [numbers]
        return

    if len(numbers) == 0 or group_count <= 0:
        yield []
        return

    for i in range(1, len(numbers) + 1):
        first_group = numbers[:i]
        for rest_groups in generate_combinations(numbers[i:], group_count - 1):
            yield [first_group] + rest_groups

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm explores all possible partitions of the input array. There are 2^(n-1) ways to partition an array of size n into subarrays. For each partition, the algorithm iterates through each subarray to find the maximum and minimum elements, which takes O(n) time in the worst case, as the sizes of the subarrays in a partition could add up to n. Therefore, the overall time complexity is O(n * 2^(n-1)), which simplifies to O(n * 2^n).

Space Complexity

O(N) – The provided brute force algorithm explores all possible partitions of the input array. In the worst-case scenario, to check if a given partition is valid, temporary lists representing each group might be created. If each number is in its own group, up to N temporary lists would be needed in memory at the same time, where N is the size of the input array. These lists store elements of the array for the validity check. Therefore, the algorithm could use O(N) auxiliary space.

Optimal Solution

Approach

To partition the array most efficiently, we sort the array first. Then, we create partitions greedily. The key is to add elements to a partition as long as the difference between the largest and smallest element in that partition remains less than or equal to K.

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

First, arrange the numbers in the array from smallest to largest.
Start a new group (partition) with the very first number.
Look at the next number in the sorted list. If this number, minus the smallest number in the current group, is no more than K, add it to the group.
Keep adding numbers to the current group using the same check (new number minus smallest number in the group must be no more than K).
If adding a new number would make the difference bigger than K, stop adding to this group. Start a new group with the new number.
Repeat steps 3-5 until you have gone through all the numbers.
The number of groups you created is the minimum number of partitions needed.

Code Implementation

def partition_array(numbers, max_difference):
    numbers.sort()
    partitions_count = 0
    start_index = 0

    # Iterate through the sorted array
    while start_index < len(numbers):
        partitions_count += 1
        current_partition_min = numbers[start_index]
        current_index = start_index + 1

        # Greedily add elements to the current partition
        while current_index < len(numbers):

            # Check if the new number exceeds max difference
            if numbers[current_index] - current_partition_min <= max_difference:
                current_index += 1
            else:
                break

        # Start a new partition from the next element
        start_index = current_index

    return partitions_count

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is sorting the array initially, which takes O(n log n) time, where n is the number of elements in the array. The subsequent greedy partitioning involves iterating through the sorted array once, which takes O(n) time. However, 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 algorithm's time complexity is O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in-place. Beyond the input array and the integer K (which doesn't scale with input size), the algorithm only uses a few integer variables to track the start of the current partition and iterate through the sorted array. The number of these variables does not depend on the size of the input array (N). Therefore, the auxiliary space used is constant, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list or throw an IllegalArgumentException as appropriate to signal invalid input.
Input array with only one element	Return a list containing the single element since a single element array is trivially partitioned.
Array with all identical elements and K is zero	The original array is already partitioned, so return a list containing the original array.
Array with all identical elements and K is non-zero	Return a list containing the original array as any partition will satisfy the condition.
Large input array with numbers close to Integer.MAX_VALUE and Integer.MIN_VALUE	Sorting the array might cause integer overflow when calculating differences, so use long to prevent overflow.
Array is already sorted and satisfies the condition	The algorithm should efficiently process this case, ideally without unnecessary operations, by returning the initial array as a single partition.
Array with negative numbers, zeros, and positive numbers	The solution needs to correctly handle the signed differences by taking absolute values or sorting to handle properly.
No possible partition exists given the input array and K	The algorithm should return the original array as a single partition indicating that no valid partition was found

Null or empty input array

How to Handle:

Return an empty list or throw an IllegalArgumentException as appropriate to signal invalid input.

Input array with only one element

How to Handle:

Return a list containing the single element since a single element array is trivially partitioned.

Array with all identical elements and K is zero

How to Handle:

The original array is already partitioned, so return a list containing the original array.

Array with all identical elements and K is non-zero

How to Handle:

Return a list containing the original array as any partition will satisfy the condition.

Large input array with numbers close to Integer.MAX_VALUE and Integer.MIN_VALUE

How to Handle:

Sorting the array might cause integer overflow when calculating differences, so use long to prevent overflow.

Array is already sorted and satisfies the condition

How to Handle:

The algorithm should efficiently process this case, ideally without unnecessary operations, by returning the initial array as a single partition.

Array with negative numbers, zeros, and positive numbers

How to Handle:

The solution needs to correctly handle the signed differences by taking absolute values or sorting to handle properly.

No possible partition exists given the input array and K

How to Handle:

The algorithm should return the original array as a single partition indicating that no valid partition was found