<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:
    const int inf = 0x3f3f3f3f;
    int minimumIncompatibility(vector<int>& nums, int k) {
      int n = nums.size();
      int sz = n/k; // size of every group
      sort(nums.begin(), nums.end());
      vector<vector<int>> dp(1<<n, vector<int>(n, inf));
      for (int i = 0; i < n; ++i) { dp[0][i] = 0; }  
      for (int m = 0; m < (1<<n)-1; ++m) {
        int cnt = __builtin_popcount(m);
        for (int i = 0; i < n; ++i) {  
          if (dp[m][i] == inf) { continue; }
          if (cnt%sz == 0) {
            int j = __builtin_ffs(~m)-1;
            dp[m|(1<<j)][j] = min(dp[m|(1<<j)][j], dp[m][i]); 
          } else {
            for (int j = i+1; j < n; ++j) { 
              if (m&(1<<j) || nums[j] == nums[i]) { continue; }
              dp[m|(1<<j)][j] = min(dp[m|(1<<j)][j], dp[m][i]+nums[j]-nums[i]);
            }
          }
        }
      }
      return dp[(1<<n)-1][n-1] == inf? -1: dp[(1<<n)-1][n-1];  
    }
};

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