<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 {
private:
    int diff[(1<<16)], ok[(1<<16)] = {}, bits[(1<<16)] = {};
    int dp[(1<<16)][16][23];
    int m = 0;
public:
    void init(vector<int>& nums, int k) {
        for (int mask = 1; mask < (1 << nums.size()); mask++) {
            int save = 0;
            int minb = INT_MAX;
            int maxb = INT_MIN;
            ok[mask] = (__builtin_popcount(mask) == nums.size()/k); 
            for (int i = 0; i < nums.size(); i++) {
                if (mask & (1<<i)) {
                    if (save & (1<<nums[i])) {ok[mask] = false; i = nums.size() + 9;  continue;}
                    save |= (1<<nums[i]); 
                    minb = min(minb, nums[i]);
                    maxb = max(maxb, nums[i]);
                }
            }
            if (ok[mask]) {
                bits[m] = mask;
                diff[m] = maxb-minb;
                m++;
            }
        }
    }

    int magic(int mask, int i, int spot, int k, int n) {
        if (i == k) return spot;
        if (dp[mask][i][spot] != -1) return dp[mask][i][spot];
        int ans = INT_MAX;
        for (int j = 0; j < m; j++) {
            if ((mask&bits[j]) == 0) {
                if (spot + diff[j] <= 22) ans = min(ans, magic(mask|bits[j], i+1, spot + diff[j], k, n));
            }
        }
        return dp[mask][i][spot] = ans;
    }
    int minimumIncompatibility(vector<int>& nums, int k) {
        if (nums.size() % k != 0) return -1;
        init(nums, k);
        memset(dp,-1,sizeof(dp));
        int ans = magic(0, 0, 0, k, nums.size());
        return ans == INT_MAX ? -1 : ans;
    }
};
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