<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>



const int INF = 1e9;

class Solution {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        int n=nums.size();
        int required_size=n/k;
        sort(nums.begin(),nums.end());

        // pre calculate the range for each subset i.e. (maxi-mini) for each subset of nums;
        vector<int>range(1<<n,INT_MAX);

        // note that there should not be any duplicates
        for(int mask=1;mask<(1<<n) ; mask++){
            int mini=-1,maxi=-1;
            bool duplicate=0;
            for(int i=0;i<n && !duplicate ; i++){
                if(mask &(1<<i)){
                    if(mini == -1){
                        mini=nums[i];
                    }
                    if(maxi==nums[i])duplicate=1;

                    maxi=nums[i] ; // since nums is sorted;
                }
            }
            if(!duplicate) range[mask]=maxi-mini;
        }
        vector<long long int>dp(1<<n,INT_MAX);
        // dp[mask]=maximum compatibility of an array if only the element corresponding to which bit in the mask=1 are taken in  the subset.
        // we have to return dp[(1<<n) -1] if ans exist else -1;

        dp[0]=0;
        // iterate over each subset of nums
        for(int mask=1;mask<(1<<n);mask++){

            // iterate over each submask for current mask/subset;
            
            for(int submask=mask;submask ; submask=(submask-1)&mask){
                // chk that submask has right number of element
                if(__builtin_popcount(submask)==required_size){
                    
                    //(mask ^ submask) represents those elements in mask that are not in submask
                    // we have already calculated dp[mask^submask] earlier since mask^submask<mask;
                    dp[mask]=min(dp[mask],range[submask]+dp[mask^submask]);

                }
            }

        }
        return dp.back()>=INT_MAX ? -1:dp.back();
    }
};
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