<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:
    void f(vector<int> &nums, int size, int cur, int count, int sum, int l, int h, vector<int> &vis, int &n, int &k, int &res) {
        if (count == size) {
            int j = 0;
            while (j < n && vis[j] == 1) {
                j++;
            }
            if (j == n) {
                res = min(res, sum+h-l);
            } else {
                vis[j] = 1;
                f(nums, size, j, 1, sum+h-l, nums[j], nums[j], vis, n, k, res);
                vis[j] = 0;
            }
        } else {
            int last = -1;
            for (int i = cur+1; i < n; i++) {
                if (vis[i] == 1) continue;
                if (nums[i] == nums[cur]) continue;
                if (i > cur+1 && nums[i] == last) continue;
                vis[i] = 1;
                f(nums, size, i, count+1, sum, l, nums[i], vis, n, k, res);
                last = nums[i];
                vis[i] = 0;
            }
        }
    }
    int minimumIncompatibility(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        for (auto &i : nums) {
            mp[i]++;
            if (mp[i] > k) return -1;
        }
        int res = INT_MAX;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<int> vis(n, 0);
        if (n % k != 0) return -1;
        vis[0] = 1;
        f(nums, n/k, 0, 1, 0, nums[0], nums[0], vis, n, k, res);
        return res;
    }
};
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