<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 {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        int n = nums.size();
        int subset_size = n / k;
        unordered_map<int, int> freq;
        for (int num : nums) {
            freq[num]++;
            if (freq[num] > k) {
                return -1;  // Impossible to distribute
            }
        }
        
        // Precompute valid subsets of size subset_size
        vector<pair<int, int>> validSubsets;  // Pair of (subset as bitmask, incompatibility)
        int N = nums.size();
        vector<int> indices(N);
        iota(indices.begin(), indices.end(), 0);  // Indices from 0 to N-1
        unordered_map<int, int> incompatibilityMap;
        
        // Map from bitmask to incompatibility
        // Generate all combinations of indices of size subset_size
        vector<vector<int>> combs = combinations(indices, subset_size);
        for (auto &comb : combs) {
            unordered_set<int> elements;
            int max_val = INT_MIN, min_val = INT_MAX;
            int bitmask = 0;
            bool valid = true;
            for (int idx : comb) {
                int val = nums[idx];
                if (elements.count(val)) {
                    valid = false;
                    break;
                }
                elements.insert(val);
                max_val = max(max_val, val);
                min_val = min(min_val, val);
                bitmask |= (1 << idx);
            }
            if (valid) {
                int incomp = max_val - min_val;
                validSubsets.push_back({bitmask, incomp});
            }
        }
        
        vector<int> dp(1 << N, INT_MAX);
        dp[0] = 0;
        for (int mask = 0; mask < (1 << N); ++mask) {
            if (dp[mask] == INT_MAX) continue;
            for (auto &p : validSubsets) {
                int subset = p.first;
                if ((mask & subset) == 0) {
                    int new_mask = mask | subset;
                    dp[new_mask] = min(dp[new_mask], dp[mask] + p.second);
                }
            }
        }
        int full_mask = (1 << N) - 1;
        return dp[full_mask] == INT_MAX ? -1 : dp[full_mask];
    }
    
    // Helper function to generate combinations of indices
    vector<vector<int>> combinations(vector<int>& nums, int k) {
        vector<vector<int>> result;
        vector<int> comb;
        generateCombinations(nums, k, 0, comb, result);
        return result;
    }
    
    void generateCombinations(vector<int>& nums, int k, int start, vector<int>& comb, vector<vector<int>>& result) {
        if (comb.size() == k) {
            result.push_back(comb);
            return;
        }
        for (int i = start; i < nums.size(); ++i) {
            comb.push_back(nums[i]);
            generateCombinations(nums, k, i + 1, comb, result);
            comb.pop_back();
        }
    }
};

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