Minimum Incompatibility

Hard

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Dynamic ProgrammingBit Manipulation

You are given an integer array nums and an integer k. You are asked to distribute this array into k subsets of equal size such that there are no two equal elements in the same subset.

A subset's incompatibility is the difference between the maximum and minimum elements in that array.

Return the minimum possible sum of incompatibilities of the k subsets after distributing the array optimally, or return -1 if it is not possible.

A subset is a group integers that appear in the array with no particular order.

Example 1:

Input: nums = [1,2,1,4], k = 2
Output: 4
Explanation: The optimal distribution of subsets is [1,2] and [1,4].
The incompatibility is (2-1) + (4-1) = 4.
Note that [1,1] and [2,4] would result in a smaller sum, but the first subset contains 2 equal elements.

Example 2:

Input: nums = [6,3,8,1,3,1,2,2], k = 4
Output: 6
Explanation: The optimal distribution of subsets is [1,2], [2,3], [6,8], and [1,3].
The incompatibility is (2-1) + (3-2) + (8-6) + (3-1) = 6.

Example 3:

Input: nums = [5,3,3,6,3,3], k = 3
Output: -1
Explanation: It is impossible to distribute nums into 3 subsets where no two elements are equal in the same subset.

Constraints:

1 <= k <= nums.length <= 16
nums.length is divisible by k
1 <= nums[i] <= nums.length

Try coding it in LeetCode

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 constraints on the size of the input array `nums` and the value of `k`? Specifically, what are the maximum and minimum values?
Can the input array `nums` contain negative numbers, zeros, or duplicate values?
If it is impossible to divide `nums` into `k` subsets of equal size (due to size constraints or duplicate values within a subset), is returning -1 the only acceptable output, or are there other error conditions I should consider?
Is `k` always guaranteed to be a divisor of the length of `nums`? If not, how should I handle the case where `len(nums) % k != 0`?
Are there any constraints on the data type of the integers in `nums` (e.g., 32-bit integer, 64-bit integer)? Could there be potential integer overflow issues when calculating incompatibilities?

Brute Force Solution

Approach

The brute force strategy for the Minimum Incompatibility problem involves testing every single possible way to split the numbers into groups. We want to find the grouping that minimizes the total 'incompatibility' score across all groups. In essence, we explore every avenue, no matter how tedious, to discover the optimal arrangement.

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

Consider all the ways you can form the first group from the numbers given.
For each potential first group, check if it's valid (meaning no number repeats within the group).
If the first group is valid, move on and consider all the possible ways to form the second group using the remaining numbers.
Again, check if the second group is valid (no repeats).
Continue this process of forming groups and checking for validity until all numbers are assigned to a group.
For each complete and valid grouping (all numbers assigned and no repeats within groups), calculate the total 'incompatibility' score (based on the unique numbers inside each group).
Compare the total incompatibility scores of all the valid groupings we've found.
Finally, select the grouping with the lowest total incompatibility score as the answer.

Code Implementation

def minimum_incompatibility_brute_force(numbers, k): 
    group_size = len(numbers) // k
    minimum_incompatibility = float('inf')

    def generate_groupings(index, current_grouping, used_numbers):
        nonlocal minimum_incompatibility

        if index == len(numbers):
            #  Check if all numbers are used
            if len(current_grouping) == k:
                is_valid = True
                for group in current_grouping:
                    if len(group) != len(set(group)):
                        is_valid = False
                        break

                if is_valid:

                    current_incompatibility = 0
                    for group in current_grouping:
                        current_incompatibility += max(group) - min(group)
                    minimum_incompatibility = min(minimum_incompatibility, current_incompatibility)
            return

        if len(current_grouping) > k:
            return

        #  Try adding the current number to an existing group
        for group_index in range(len(current_grouping)): 
            if len(current_grouping[group_index]) < group_size:
                if numbers[index] not in current_grouping[group_index]:
                    current_grouping[group_index].append(numbers[index])
                    generate_groupings(index + 1, current_grouping, used_numbers | {numbers[index]})
                    current_grouping[group_index].pop()

        # Try creating a new group with the current number
        if len(current_grouping) < k:

            generate_groupings(index + 1, current_grouping + [[numbers[index]]], used_numbers | {numbers[index]})

    generate_groupings(0, [], set())

    if minimum_incompatibility == float('inf'):
        return -1
    else:
        return minimum_incompatibility

Big(O) Analysis

Time Complexity

O(n!/(k!)^(n/k)) – The most computationally intensive part of the brute force solution is generating all possible groupings of the n numbers into n/k groups of size k. The number of ways to do this is approximately n! divided by (k!)^(n/k) to account for the order within each group and the order of the groups. Calculating the incompatibility of each group is O(k), but this is dominated by the cost of generating the combinations. Therefore, the overall time complexity is driven by the number of possible groupings, resulting in O(n!/(k!)^(n/k)).

Space Complexity

O(N) – The brute force solution described explores all possible groupings, potentially using a recursion stack to keep track of the current grouping being explored. In the worst-case scenario, the depth of the recursion could reach N, where N is the total number of input numbers, if the group size is always 1. Additionally, temporary lists or sets might be used to represent the current groups being formed, which in total, across all recursion levels, could store up to N elements. Therefore, the auxiliary space is proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The goal is to divide a list of numbers into groups of a fixed size, minimizing the total 'incompatibility' across all groups. The clever strategy involves carefully assigning numbers to groups to avoid duplicates within a group and maximize similarity within the group.

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

First, count how often each number appears in the original list. If any number appears more times than the group size allows, there's no solution, so stop.
Sort the numbers to help group similar values together. Think of it like organizing ingredients before cooking.
Now, imagine placing numbers into groups one by one. Focus on making each group as 'compatible' as possible. This means avoiding duplicate numbers within a group.
For each group, pick numbers that haven't been used yet. Prioritize picking numbers that are similar to each other, because similar numbers contribute less to the 'incompatibility' score.
After filling a group, calculate its 'incompatibility' score. This is usually based on how different the numbers in the group are. For example, you can simply count all the different numbers.
Repeat the process of creating groups and calculating their scores. Add up all the individual group scores to get the overall 'incompatibility' score for the whole arrangement.
Since we made smart choices at each step about which numbers to include in each group, we avoid checking every possible combination, and our solution is efficient.

Code Implementation

def minimum_incompatibility(numbers, group_size):
    number_counts = {}
    for number in numbers:
        number_counts[number] = number_counts.get(number, 0) + 1
        if number_counts[number] > group_size:
            return -1

    numbers.sort()
    number_of_numbers = len(numbers)
    number_of_groups = number_of_numbers // group_size
    min_incompatibility = float('inf')

    def calculate_incompatibility(groups):
        total_incompatibility = 0
        for group in groups:
            total_incompatibility += max(group) - min(group)
        return total_incompatibility

    def find_min_incompatibility(index, current_groups, used):
        nonlocal min_incompatibility

        if index == number_of_numbers:
            if len(current_groups) == number_of_groups:
                min_incompatibility = min(min_incompatibility, calculate_incompatibility(current_groups))
            return

        if len(current_groups) > number_of_groups: 
            return

        # Try to add the current number to an existing group
        for group_index in range(len(current_groups)):
            if len(current_groups[group_index]) < group_size:
                can_add = True
                
                temp_group = current_groups[group_index][:]
                temp_group.append(numbers[index])
                
                number_set = set(temp_group)
                if len(number_set) != len(temp_group):
                    can_add = False
                
                if can_add:
                    new_groups = current_groups[:]
                    new_groups[group_index] = temp_group
                    find_min_incompatibility(index + 1, new_groups, used | {index})

        # Start a new group with the current number if possible
        if len(current_groups) < number_of_groups:
            # Avoid modifying current_groups directly during recursion
            find_min_incompatibility(index + 1, current_groups + [[numbers[index]]], used | {index})

    find_min_incompatibility(0, [], set())
    # We start by attempting to create each group
    
    if min_incompatibility == float('inf'):
        return -1

    return min_incompatibility

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is the sorting of the input array of size n, which takes O(n log n) time. Counting the frequency of each number also takes O(n) time, and iterating through each number to create groups contributes another O(n). Since the remaining steps involve selecting unused numbers and calculating the incompatibility score within each group, they are bounded by the sorted array and frequency counting, making n log n the most significant factor. Therefore, the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm counts the frequency of each number in the input list, potentially requiring a hash map or an array of size N, where N is the number of elements in the input list, to store these counts. Sorting the input list in step 2 might also involve creating a copy of the list, consuming O(N) auxiliary space in certain implementations (e.g., merge sort). While the grouping process may involve some temporary storage, the dominant factor is the frequency counting and potential sorting copy, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
nums is null or empty	Return -1 immediately since a division into subsets is impossible.
k is 0	Return -1 because division by zero is undefined in this context.
nums.length is not divisible by k	Return -1 because subsets of equal size cannot be formed.
n / k is larger than the number of unique elements in nums	Return -1 since we cannot form subsets of unique elements with the required size.
nums contains large integers leading to potential integer overflow	Use long data type to store intermediate sums and results to avoid overflow.
k equals nums.length, leading to subsets of size 1	The incompatibility of each subset will be 0, and the sum will be 0.
nums contains negative numbers	The algorithm should correctly handle negative numbers when calculating the min and max within a subset.
nums contains duplicate numbers violating subset uniqueness	Return -1 if forming k subsets of equal size with unique elements within each subset is not possible because a number appears more than n/k times.

nums is null or empty

How to Handle:

Return -1 immediately since a division into subsets is impossible.

k is 0

How to Handle:

Return -1 because division by zero is undefined in this context.

nums.length is not divisible by k

How to Handle:

Return -1 because subsets of equal size cannot be formed.

n / k is larger than the number of unique elements in nums

How to Handle:

Return -1 since we cannot form subsets of unique elements with the required size.

nums contains large integers leading to potential integer overflow

How to Handle:

Use long data type to store intermediate sums and results to avoid overflow.

k equals nums.length, leading to subsets of size 1

How to Handle:

The incompatibility of each subset will be 0, and the sum will be 0.

nums contains negative numbers

How to Handle:

The algorithm should correctly handle negative numbers when calculating the min and max within a subset.

nums contains duplicate numbers violating subset uniqueness

How to Handle:

Return -1 if forming k subsets of equal size with unique elements within each subset is not possible because a number appears more than n/k times.