<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>

int k;
int dp[16][1<<16];
int rec(int i, int mask, vector<int>& nums) {//O(k.2^n.nCk)
    if (i==k) return 0;

    int &ans=dp[i][mask];
    if (ans!=-1) return ans;

    ans=1e9;
    int nottaken=(~mask)&((1<<nums.size())-1);
    for (int s=nottaken; s; s=(s-1)&nottaken) {
        if (__builtin_popcount(s)==(nums.size()/k)) {
            int min_=1e9, max_=-1e9;
            int val=0, consider=1;
            for (int _=0;_<nums.size();_++) {
                if ((s>>_)&1) {
                    if ((val>>nums[_])&1) {consider=0; break;}
                    val|=(1<<nums[_]);
                    min_=min(min_, nums[_]);
                    max_=max(max_, nums[_]);
                }
            }
            if (consider) ans=min(ans, rec(i+1, mask|s,nums)+(max_-min_));
        }
    }
    return ans;
}

class Solution {
public:
int minimumIncompatibility(vector<int>& nums, int k_) {
    k=k_;
    memset(dp,-1,sizeof(dp));
    int ans=rec(0,0,nums);
    return ans!=1e9?ans:-1;
}
};
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