<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:
        ctrNums = Counter(nums)
        if any(ctrNums[num] > k for num in ctrNums):
            return -1
        bags, res, sz = [[] for _ in range(k)], inf, len(nums)
        maxNumsperBag, numbers = sz // k, sorted(ctrNums.keys())
        nums.sort()
        curSize, curBag, gaps, gapsSum = 0, [], defaultdict(int), defaultdict(int)
        cntUniqueNums = len(numbers)
        for i in range(1, cntUniqueNums):
            gaps[numbers[i]] = numbers[i] - numbers[i - 1]
        for i in range(sz - 1, -1, -1):
            curBag.append(gaps[nums[i]])
            curBag.sort()
            curSize += gaps[nums[i]]
            gapsSum[i, 0] = curSize
            for empty in range(1, min(k + 1, len(curBag))):
                gapsSum[i, empty] = gapsSum[i, empty - 1] - curBag[-empty]
        
        def dfs(i: int, curRes: int, emptyBags: int) -> None:
            nonlocal res
            if i == sz:
                res = min(res, curRes)
                return
            if curRes + gapsSum[i, emptyBags] >= res: 
                return
            for j in range(k):
                if (j > 0 and bags[j] == bags[j - 1]) or (bags[j] and bags[j][-1] == nums[i]) or (len(bags[j])) == maxNumsperBag:
                    continue
                else:
                    bags[j].append(nums[i])
                    if len(bags[j]) == 1:
                        dfs(i + 1, curRes, emptyBags - 1)
                    else:
                        dfs(i + 1, curRes + bags[j][-1] - bags[j][-2], emptyBags)
                    bags[j].pop()
        
        dfs(0, 0, k)
        return res
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