<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 ;
    int k ;
    vector<bool>visited ;
    int ret = INT_MAX ;
    
    void dfs(vector<int>&nums, int cur, int count, int low, int high, int sum){
        if(count == n/k){
            int i = 0;  
            while(i < n && visited[i])
                i++ ;
            if(i == n){
                ret = min(ret , sum + high - low) ;
                return ;
            }
            else{
                visited[i] = true ;
                dfs(nums, i, 1, nums[i], nums[i], sum + high - low) ;
                visited[i] = false ;
            }
        }
        else{
            int last = -1 ;
            for(int i = cur+1; i < n; i++){
                if(visited[i])
                    continue ;
                if(nums[i] == nums[cur])
                    continue ;
                if(nums[i] == last)
                    continue ;
                visited[i] = true ;
                dfs(nums, i, count+1, low, nums[i], sum) ;
                last = nums[i] ;
                visited[i] = false ;
            }
        }
    }
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        this->n = nums.size() ;
        this->k = k ;
        visited.resize(n) ;
        sort(nums.begin(), nums.end() ) ;
        int cn = 1 ;
        
        for(int i = 1; i < n; i++){
            if(nums[i] == nums[i-1]){
                cn++ ;
                if(cn > k)
                    return -1 ;
            }
            else
                cn = 1 ;
        }
        if(cn > k)
            return -1 ;
        visited[0] = true ;
        dfs(nums, 0, 1, nums[0], nums[0], 0) ;
        return ret ;
    }
};
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