<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) {
    const int n = nums.size(); 
    const int c = n / k;
    int dp[1 << 16][16];
    memset(dp, 0x7f, sizeof(dp));
    for (int i = 0; i < n; ++i) dp[1 << i][i] = 0;
    for (int s = 0; s < 1 << n; ++s)
      for (int i = 0; i < n; ++i) {
        if ((s & (1 << i)) == 0) continue;
        for (int j = 0; j < n; ++j) {
          if ((s & (1 << j))) continue;
          const int t = s | (1 << j);
          if (__builtin_popcount(s) % c == 0) {
            dp[t][j] = min(dp[t][j], dp[s][i]);
          } else if (nums[j] > nums[i]) {           
            dp[t][j] = min(dp[t][j], 
                           dp[s][i] + nums[j] - nums[i]);            
          }
        }        
    }
    int ans = *min_element(begin(dp[(1 << n) - 1]), 
                           end(dp[(1 << n) - 1]));
    return ans > 1e9 ? - 1 : ans;
  }
};
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