<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>

class Solution:
    def minimumIncompatibility(self, nums: List[int], k: int) -> int:
        c = Counter(nums)
        for n, v in c.items(): 
            if v > k: return -1
        nums.sort()

        def min_max(selection):
            mi, ma = inf, -inf
            used = set()
            for k in range(len(nums)):
                if selection & (1 << k) == 0: continue
                if nums[k] in used: return -inf
                used.add(nums[k])
                mi = min(nums[k], mi)
                ma = max(nums[k], ma)
            return ma - mi

        parts = []
        for idx in range(2**len(nums)):
            if idx.bit_count() == len(nums) // k: 
                out = min_max(idx)
                if out == -inf: continue
                parts.append([idx, out])
        
        @cache
        def go(selected):
            partition = selected.bit_count() // k
            if partition == len(nums) // k: return 0
            best = inf
            for p in parts:
                if selected & p[0] != 0: continue
                choice = p[1] + go(p[0] | selected)
                best = min(best, choice)
            return best
        return go(0)

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