<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 {
    int n, sz, dp[1 << 16];
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        sz = (n = nums.size()) / k;
        memset(dp, -1, 1 << n << 2), dp[0] = 0;
        solve(nums, (1 << n) - 1);
        return (dp[(1 << n) - 1] < 1e9)? dp[(1 << n) - 1]: -1;
    }

    int solve(vector<int>& nums, int mask) {
        if (~dp[mask]) return dp[mask];
        dp[mask] = pick(nums, mask, 0, sz, n - 1);
        return dp[mask];
    }

    int pick(vector<int>& nums, int mask, int used, int k, int i) {
        if (!k) return solve(nums, mask) + __lg(used) - __lg(used & -used);
        int res = 1e9;
        for (; ~i; i--) {
            if (~mask >> i & 1) continue;
            if (used >> nums[i] & 1) continue;
            res = min(res, pick(nums, mask ^ 1 << i, used | 1 << nums[i], k - 1, i - 1));
        }
        return res;
    }
};
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