<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) {
        sort(nums.begin(), nums.end());

        int n = nums.size();
        int cap = n / k;

        const int INF = 2e9;
        vector<int> cost(1 << n, INF), dp(1 << n, INF);
        vector<int> gr;
        for (int i = 0; i < (1 << n); i++) {
            int sz = __builtin_popcount(i);
            if (sz != cap) continue;
            vector<bool> cnt(n + 1);
            bool ok = true;
            int mx = -1, mn = n + 1;
            for (int j = 0; j < n; j++) {
                if (!((i >> j) & 1)) continue;
                if (cnt[nums[j]]) { ok = false; break; }
                cnt[nums[j]] = true;
                mx = max(mx, nums[j]), mn = min(mn, nums[j]);
            }
            if (!ok) continue;
            cost[i] = mx - mn;
            dp[i] = mx - mn;
            gr.push_back(i);
        }

        for (int i = 1; i < (1 << n); i++) {
            int sz = __builtin_popcount(i);
            if (sz == cap) continue;
            if (sz % cap != 0) continue;
            for (auto c : gr) {
                if ((c | i) != i) continue;
                dp[i] = min(dp[i], dp[i ^ c] + cost[c]);
            }
        }

        if (dp[(1 << n) - 1] == INF) return -1;
        return dp[(1 << n) - 1];
    }
};
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