<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 {
    int incompat(vector<int>& nums, int group) {
        int n=nums.size();
        int mn=INT_MAX, mx=INT_MIN;
        unordered_set<int> us;
        for (int i=0; i<n; i++) {
            if (group&(1<<i)) {
                if (us.count(nums[i])) return -1;
                us.insert(nums[i]);
                mn=min(mn, nums[i]);
                mx=max(mx, nums[i]);
            }
        }
        return mx-mn;
    }
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        int n=nums.size();
        int gs=n/k;        

        if (gs==1) return 0;
        if (k==1) return incompat(nums, (1<<n)-1);
        
        vector<int> dp1(1<<n, -1);
        for (int i=1, m=1<<n; i<m; i++) {
            if (__builtin_popcount(i)==gs) {
                dp1[i]=incompat(nums, i);
            }
        }
        
        vector<int> dp2(1<<n, -1);
        dp2[0]=0;
        
        for (int i=1, m=1<<n; i<m; i++) {
            if (__builtin_popcount(i)%gs==0) {
                for (int group=i; group; group=(group-1)&i) {                    
                    if (dp1[group]==-1) continue;
                    if (dp2[i-group]==-1) continue;
                    if (dp2[i]==-1) dp2[i]=dp1[group]+dp2[i-group];
                    else dp2[i]=min(dp2[i], dp1[group]+dp2[i-group]);
                }
            }
        }
        
        return dp2[(1<<n)-1];
    }
};
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