<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();
        if (k == n) {
            return 0;
        }

        int sz = n / k;

        int pre[1<<n];
        int E[n+1];
        for (int mask = 0; mask < 1<<n; mask++) {
            pre[mask] = INT_MAX;
            int b = __builtin_popcount(mask);
            if (b != sz) {
                continue;
            }
            int mi = INT_MAX;
            int ma = INT_MIN;
            bool ok = true;
            memset(E, 0, sizeof(E));
            for (int i = 0; i < n; i++) {
                if (1<<i & mask) {
                    if (E[nums[i]]) {
                        ok = false;
                    }
                    E[nums[i]] = 1;
                    mi = min(mi, nums[i]);
                    ma = max(ma, nums[i]);
                }
            }
            if (ok) {
                pre[mask] = ma - mi;
            }
        }

        int dp[1<<n];
        dp[0] = 0;
        for (int mask = 1; mask < 1<<n; mask++) {
            int b = __builtin_popcount(mask);
            if (b % sz != 0) {
                continue;
            }
            if (b == sz) {
                dp[mask] = pre[mask];
                continue;
            }
            dp[mask] = INT_MAX;
            for (int submask = mask; submask; submask = (submask-1) & mask) {
                int sb = __builtin_popcount(submask);
                if (sb != (b - sz)) {
                    continue;
                }
                if (dp[submask] == INT_MAX) {
                    continue;
                }
                if (pre[mask - submask] == INT_MAX) {
                    continue;
                }
                dp[mask] = min(dp[mask], dp[submask] + pre[mask - submask]);
            }
        }

        return dp[(1<<n)-1] == INT_MAX ? -1 : dp[(1<<n)-1];
    }
};
9500