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Minimum Incompatibility

Difficulty: Hard

Problem Description

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.

Key Insights

  • Each subset must have a unique set of elements (no duplicates).
  • The incompatibility is defined as the difference between the maximum and minimum values in a subset.
  • The total number of elements in nums is divisible by k, which allows for equal-sized subsets.
  • The problem can be approached using dynamic programming and bit manipulation to explore different subset combinations effectively.

Space and Time Complexity

Time Complexity: O(2^n * n), where n is the length of the nums array. This arises from exploring all possible subsets using bitmasking. Space Complexity: O(2^n), for storing the dynamic programming states.

Solution

The solution involves using a dynamic programming approach with bitmasking to keep track of which elements have been assigned to subsets. We'll maintain a DP array where each entry corresponds to a bitmask representing the elements used so far. The algorithm iterates through each possible subset, calculates its incompatibility, and updates the DP state for the next subsets. If we can successfully fill k subsets without violating the uniqueness constraint, we return the minimum sum of incompatibilities.

Code Solutions

def minimumIncompatibility(nums, k):
    from collections import defaultdict

    n = len(nums)
    if n % k != 0:
        return -1
    
    size = n // k
    nums.sort()
    
    dp = defaultdict(lambda: float('inf'))
    dp[0] = 0
    
    for mask in range(1 << n):
        if bin(mask).count('1') % size != 0:
            continue
        
        for submask in range(1 << n):
            if mask & submask != submask:  # submask must be part of mask
                continue
            
            elements = [nums[i] for i in range(n) if (submask & (1 << i)) != 0]
            if len(elements) != len(set(elements)):  # Check for duplicates
                continue
            
            incompatibility = max(elements) - min(elements)
            new_mask = mask ^ submask
            dp[mask] = min(dp[mask], dp[new_mask] + incompatibility)
    
    return dp[(1 << n) - 1] if dp[(1 << n) - 1] != float('inf') else -1
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