<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 m = n/k, subset_in[(1<<n)];
        memset(subset_in, -1, sizeof(subset_in));

        for(int i=1;i<(1<<n);i++)
        {
            if(__builtin_popcount(i)!=m)
            {
                continue;
            }
            unordered_set<int> elements;
            int minimum = 20, maximum = 0;
            for(int j=0;j<n;j++)
            {
                if(i>>j&1)
                {
                    if(elements.count(nums[j]))
                    {
                        break;
                    }
                    elements.insert(nums[j]);
                    minimum = min(minimum, nums[j]);
                    maximum = max(maximum, nums[j]);
                }
            }
            if(elements.size()==m)
            {
                subset_in[i] = maximum - minimum;
            }
        }

        int dp[1<<n];
        memset(dp, 0x3f, sizeof(dp));
        dp[0] = 0;
        for(int i=0;i<(1<<n);i++)
        {
            if(dp[i]==0x3f3f3f3f)
            {
                continue;
            }
            int mask=0;
            unordered_set<int> used;
            for(int j=0;j<n;j++)
            {
                if(!(i>>j&1) && !(used.count(nums[j])))
                {
                    used.insert(nums[j]);
                    mask |= (1<<j);
                }
            }
            if(used.size()<m)
            {
                continue;
            }
            for(int j=mask;j;j=(j-1)&mask)
            {
                if(subset_in[j]!=-1)
                {
                    dp[i|j] = min(dp[i|j], dp[i]+subset_in[j]);
                }
            }
        }
        return (dp[(1<<n)-1]==0x3f3f3f3f)?-1:dp[(1<<n)-1];
    }
};
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