#1681

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Incompatibility

Optimize the sum of incompatibilities when distributing an array into subsets with unique elements.

array dynamic programming bit manipulation bitmask

Problem Statement

You are given an integer array nums and an integer k, and you need to distribute the array into k subsets of equal size. No subset should contain duplicate elements. The incompatibility of a subset is defined as the difference between the maximum and minimum elements in that subset.

Your task is to return the minimum possible sum of incompatibilities after optimally distributing the elements into the subsets, or return -1 if it's impossible to distribute the elements as required. The constraints guarantee that nums.length is divisible by k.

Examples

Example 1

Input: nums = [1,2,1,4], k = 2

Output: 4

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

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

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

Solution Approach

State Transition Dynamic Programming

Utilize dynamic programming to track possible states based on the elements already assigned to subsets. Using bitmasking, represent subsets and transition between states to find the minimal incompatibility sum. Each transition needs to respect the constraint that subsets have unique elements.

Backtracking with Memoization

Backtrack through the array, attempting to assign elements to subsets while ensuring no duplicates. Use memoization to store already explored states and avoid recalculating solutions for the same configuration, thus optimizing performance.

Bitmasking for Efficient Subset Assignment

Bitmasking allows efficient representation of subsets and enables quick checks to ensure no duplicate elements. By iterating over all bitmask combinations, you can explore all valid configurations while tracking incompatibilities.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend heavily on the approach used. A dynamic programming solution with bitmasking can run efficiently within the constraints. The overall complexity is determined by the number of states and the transitions between them, making it feasible for small arrays (up to 16 elements).

What Interviewers Usually Probe

The candidate must understand how to use dynamic programming to minimize incompatibilities across subsets.
Look for an understanding of bit manipulation, especially bitmasking for subset tracking.
Effective backtracking techniques and how to use memoization will signal the candidate's ability to optimize the solution.

Common Pitfalls or Variants

Common pitfalls

Forgetting to ensure that subsets contain no duplicate elements.
Using a brute-force approach without leveraging dynamic programming or bitmasking to optimize the solution.
Not accounting for impossible configurations where no valid distribution of the array exists.

Follow-up variants

Variation in the number of subsets k for testing different partitioning strategies.
Adjusting the input size to test scalability for different input lengths.
Modifying the condition of no duplicates within a subset to explore the trade-offs of optimization techniques.

FAQ

What is the minimum incompatibility problem in GhostInterview?

It is a problem where you need to distribute an array into subsets, minimizing the sum of incompatibilities between the max and min values of each subset.

How can dynamic programming help with the minimum incompatibility problem?

Dynamic programming can track possible configurations of subsets efficiently, minimizing incompatibilities by optimizing subset assignments.

What is the significance of bitmasking in the minimum incompatibility problem?

Bitmasking allows efficient representation of subsets, enabling quick validation of no-duplicate conditions while exploring optimal configurations.

How does backtracking play a role in solving the minimum incompatibility problem?

Backtracking helps explore possible subsets, while memoization ensures that the same states are not recomputed, making the process more efficient.

What is the time complexity of solving the minimum incompatibility problem?

The time complexity depends on the approach used, but with dynamic programming and bitmasking, it can be solved efficiently within the problem's constraints.

terminal

Solution

Solution 1: Preprocessing + State Compression + Dynamic Programming

Let's assume that the size of each subset after partitioning is $m$, so $m=\frac{n}{k}$, where $n$ is the length of the array.

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class Solution:
    def minimumIncompatibility(self, nums: List[int], k: int) -> int:
        n = len(nums)
        m = n // k
        g = [-1] * (1 << n)
        for i in range(1, 1 << n):
            if i.bit_count() != m:
                continue
            s = set()
            mi, mx = 20, 0
            for j, x in enumerate(nums):
                if i >> j & 1:
                    if x in s:
                        break
                    s.add(x)
                    mi = min(mi, x)
                    mx = max(mx, x)
            if len(s) == m:
                g[i] = mx - mi
        f = [inf] * (1 << n)
        f[0] = 0
        for i in range(1 << n):
            if f[i] == inf:
                continue
            s = set()
            mask = 0
            for j, x in enumerate(nums):
                if (i >> j & 1) == 0 and x not in s:
                    s.add(x)
                    mask |= 1 << j
            if len(s) < m:
                continue
            j = mask
            while j:
                if g[j] != -1:
                    f[i | j] = min(f[i | j], f[i] + g[j])
                j = (j - 1) & mask
        return f[-1] if f[-1] != inf else -1

Continue Practicing

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