<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 dfs(int i, int n, vector<int>&nums, int x, vector<vector<int>>&subsets, int k, int&ans, int currsum){
        if(currsum>ans) return;
        if(i==n){
            ans=min(ans,currsum);
            return;
        }
        set<vector<int>> vis;
        for(int j=0;j<k;j++){
            if(vis.count(subsets[j])) continue;
            vis.insert(subsets[j]);
            if(subsets[j].empty()){
                //first number
                subsets[j].push_back(nums[i]);
                dfs(i+1,n,nums,x,subsets,k,ans,currsum);
                subsets[j].pop_back();
            }else{
                int last=subsets[j].back();
                if(nums[i]<=last||subsets[j].size()==x){
                    continue;
                }
                last=subsets[j].back();
                int diff=nums[i]-last;
                subsets[j].push_back(nums[i]);
                dfs(i+1,n,nums,x,subsets,k,ans,currsum+diff);
                subsets[j].pop_back();
            }
        }
    }
    int minimumIncompatibility(vector<int>& nums, int k) {
        int n=nums.size();
        vector<int> mp(n+1);
        int maxcnt=0;
        for(int i:nums){
            mp[i]++;
            maxcnt=max(maxcnt,mp[i]);
        }
        if(maxcnt > k) return -1;
        int x=n/k;
        sort(nums.begin(),nums.end());
        vector<vector<int>> subsets(k);
        int ans=INT_MAX;
        dfs(0,n,nums,x,subsets,k,ans,0);
        return ans==INT_MAX?-1:ans;
    }
};
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