1681. Minimum Incompatibility

Description

Given an integer array nums and an integer k, the task is to distribute nums into k subsets of equal size such that no two equal elements are in the same subset. The incompatibility of a subset is the difference between its maximum and minimum elements. The goal is to find the minimum possible sum of incompatibilities across all k subsets. If it's impossible to distribute the array, return -1.

Solution: Preprocessing + State Compression + Dynamic Programming

This solution uses a combination of preprocessing, state compression, and dynamic programming to efficiently find the minimum incompatibility.

1. Preprocessing:

• Subset Incompatibility: The code first pre-calculates the incompatibility g[i] for each possible subset i. i is represented as a bitmask where the j-th bit is 1 if the j-th element of nums is included in the subset. The code checks if the subset has the correct size (m = n/k) and contains no duplicates. If valid, it calculates the incompatibility.

2. State Compression:

• State Representation: The dynamic programming uses a bitmask i to represent the state of the partitioning. Each bit in i corresponds to an element in nums. A bit set to 1 indicates the element is assigned to a subset; 0 indicates it's unassigned.

3. Dynamic Programming:

• DP Table: f[i] stores the minimum incompatibility sum for the state represented by i. f[0] = 0 (no elements assigned).

• Iteration: The code iterates through each state i. For each unassigned element, it explores all possible subsets of the unassigned elements (mask). If a valid subset is found (correct size, no duplicates), it updates the DP table:

  f[i | j] = min(f[i | j], f[i] + g[j])

  This means the minimum incompatibility for the new state (with subset j added) is the minimum of the current value and the sum of the incompatibility of j and the minimum incompatibility of the previous state i.

• Final Result: f[(1 << n) - 1] represents the minimum incompatibility when all elements are assigned. If this value is still infinity (meaning no valid partitioning was found), return -1; otherwise, return the value.

Time Complexity: O(3n) - The dominant factor is the three nested loops: the outer loop iterates through all possible states (2n), the inner loop explores subsets (mask), and within that loop it iterates through subsets j. While not exactly 3n, the complexity is exponential and closely related to it.

Space Complexity: O(2n) - The space used by the DP table f and the pre-calculated incompatibilities g is proportional to 2n.

Code (Python3)

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
 

The code in other languages (Java, C++, Go, TypeScript, C#) follows a similar structure and logic. The primary differences are in syntax and the handling of bit manipulation and data structures. The core DP algorithm remains consistent across all implementations.