<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>

#include <vector>
#include <unordered_map>
#include <algorithm>
#include <climits>

class Solution {
public:
    int minimumIncompatibility(std::vector<int>& nums, int k) {
        int n = nums.size();
        int subsetSize = n / k;

        std::unordered_map<int, int> freq;
        for (int num : nums) {
            freq[num]++;
            if (freq[num] > k) {
                return -1;
            }
        }

        int totalSubsets = 1 << n; 
        std::vector<int> subsetIncompatibility(totalSubsets, INT_MAX);
        std::vector<int> dp(totalSubsets, INT_MAX);
        dp[0] = 0;

        for (int subset = 0; subset < totalSubsets; subset++) {
            std::vector<int> elements;
            for (int i = 0; i < n; i++) {
                if (subset & (1 << i)) {
                    elements.push_back(nums[i]);
                }
            }
            if (elements.size() == subsetSize) {
                std::sort(elements.begin(), elements.end());
                bool valid = true;
                for (int i = 1; i < elements.size(); i++) {
                    if (elements[i] == elements[i - 1]) {
                        valid = false;
                        break;
                    }
                }
                if (valid) {
                    subsetIncompatibility[subset] = elements.back() - elements.front();
                }
            }
        }

        for (int subset = 0; subset < totalSubsets; subset++) {
            if (__builtin_popcount(subset) % subsetSize == 0) {
                for (int smallerSubset = subset; smallerSubset; smallerSubset = (smallerSubset - 1) & subset) {
                    if (subsetIncompatibility[smallerSubset] != INT_MAX && dp[subset ^ smallerSubset] != INT_MAX) {
                        
                        if (dp[subset ^ smallerSubset] <= INT_MAX - subsetIncompatibility[smallerSubset]) {
                            dp[subset] = std::min(dp[subset], dp[subset ^ smallerSubset] + subsetIncompatibility[smallerSubset]);
                        }
                    }
                }
            }
        }

        return dp[totalSubsets - 1] == INT_MAX ? -1 : dp[totalSubsets - 1];
    }
};

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