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

<p>A subset&#39;s <strong>incompatibility</strong> is the difference between the maximum and minimum elements in that array.</p>

<p>Return <em>the <strong>minimum possible sum of incompatibilities</strong> of the </em><code>k</code> <em>subsets after distributing the array optimally, or return </em><code>-1</code><em> if it is not possible.</em></p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,4], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,3,8,1,3,1,2,2], k = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 16</code></li>
	<li><code>nums.length</code> is divisible by <code>k</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
</ul>

from typing import List
from functools import lru_cache

class Solution:
    def minimumIncompatibility(self, nums: List[int], k: int) -> int:
        n = len(nums)
        subset_size = n // k
        
        # Precompute all valid subsets and their incompatibilities
        all_subsets = []
        for mask in range(1 << n):
            if bin(mask).count('1') == subset_size:
                subset = [nums[i] for i in range(n) if (mask & (1 << i))]
                if len(set(subset)) == subset_size:  # Check if all elements are unique
                    incompatibility = max(subset) - min(subset)
                    all_subsets.append((mask, incompatibility))
        
        # Initialize dp array
        dp = [float('inf')] * (1 << n)
        dp[0] = 0
        
        # Fill dp array
        for mask in range(1 << n):
            if dp[mask] == float('inf'):
                continue
            for sub_mask, incompatibility in all_subsets:
                if mask & sub_mask == 0:  # Ensure sub_mask is disjoint with mask
                    new_mask = mask | sub_mask
                    dp[new_mask] = min(dp[new_mask], dp[mask] + incompatibility)
        
        # The answer is the minimum incompatibility sum using all elements
        result = dp[(1 << n) - 1]
        return result if result != float('inf') else -1

# Example usage:
# sol = Solution()
# print(sol.minimumIncompatibility([1,2,1,4], 2))  # Output: 4
# print(sol.minimumIncompatibility([6,3,8,1,3,1,2,2], 4))  # Output: 6
# print(sol.minimumIncompatibility([5,3,3,6,3,3], 3))  # Output: -1
17200