<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 freqCount[20] = { 0 };
        for (auto x : nums)
            freqCount[x]++;
        
        numsSize_ = nums.size();
        for (int i = 0; i < numsSize_; ++i)
        {
            if (freqCount[i] > k)
                return -1;
        }

        nums_ = nums;
        k_ = k;
        std::sort(nums_.begin(), nums_.end());

        visited_ = 0;
        result_ = INT_MAX;

        visited_ |= (1 << 0);
        dfs(0, 1, nums_[0], nums_[0], 0);
        return result_;
    }

private:
    void dfs(int index, int pickedCount, int groupMin, int groupMax, int currScore)
    {
        if (pickedCount == (numsSize_ / k_))
        {
            if (visited_ == ((1 << numsSize_) - 1))
                result_ = std::min(result_, currScore + groupMax - groupMin);
            else
            {
                int j = 0;
                for (j = 0; j < numsSize_; ++j)
                {
                    if ((visited_ >> j) & 0b01)
                        continue;
                    break;
                }
                
                visited_ |= (1 << j);
                dfs(j, 1, nums_[j], nums_[j], currScore + groupMax - groupMin);
                visited_ ^= (1 << j);
            }

            return;
        }

        int last = -1;
        for (int i = index + 1; i < numsSize_; ++i)
        {
            if ((visited_ >> i) & 0b01)
                continue;

            if (nums_[i] == nums_[index])
                continue;

            if (nums_[i] == last)
                continue;

            last = nums_[i];
            visited_ |= (1 << i);

            dfs(i, pickedCount + 1, groupMin, nums_[i], currScore);
            visited_ ^= (1 << i);
        }
    }

private:
    std::vector<int> nums_;
    int numsSize_;
    int k_;

    int visited_ = 0;
    int result_ = INT_MAX;
};

// 0 1 2 3
// 1 1 2 4
8800