<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 gs=n/k;        
        if (gs==1) return 0;
        
        sort(nums.begin(), nums.end());
        
        vector<vector<int>> dp(1<<n, vector<int>(n, INT_MAX));
        for (int i=0; i<n; i++) dp[1<<i][i]=0;
        
        for (int mask=1, limit=1<<n; mask<limit; mask++) {
            int ones=__builtin_popcount(mask);
            if (ones==1) continue;
            
            if (ones%gs==1) {
                for (int i=0; i<n; i++) {
                    if ((mask&(1<<i))==0) continue;
                    for (int j=0; j<n; j++) {
                        if (i!=j && (mask&(1<<j))) {
                            int prev=mask^(1<<i);
                            if (dp[prev][j]!=INT_MAX) {
                                dp[mask][i]=min(dp[mask][i], dp[prev][j]);
                            }
                        }
                    }
                }
            }
            else {
                for (int i=0; i<n; i++) {
                    if ((mask&(1<<i))==0) continue;
                    for (int j=0; j<i; j++) {
                        if (nums[j]!=nums[i] && (mask&(1<<j))) {
                            int prev=mask^(1<<i);
                            if (dp[prev][j]!=INT_MAX) {
                                dp[mask][i]=dp[prev][j]+nums[i]-nums[j];
                            }
                        }
                    }
                }
            }
        }
        
        int ans=INT_MAX;
        for (int i=0; i<n; i++) {
            ans=min(ans, dp.back()[i]);
        }
        
        return (ans==INT_MAX)?-1: ans;
    }
};
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