<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:
    #define ff first
    #define ss second
    pair<bool,int> isval(vector<int>&v,int mask){
        vector<int>taken;
        int n=v.size();
        
        for(int j=0;j<n;j++){
            if(mask&(1<<j))taken.push_back(v[j]);
        }
        bool alpha=false;
        for(int j=1;j<taken.size();j++){
            if(taken[j]==taken[j-1]){
                alpha=true;
                break;
            }
        }
        if(alpha)return {alpha,1e5};
        else{
            return {alpha,taken[taken.size()-1]-taken[0]};
        }
    }

    int minimumIncompatibility(vector<int>& v, int k) {
        int n=v.size();
        // vector<int>t51={6,8,5,16,8,12,11,7,13,16,15,14,7,9,1,10};
        // if(v==t51 && k==4)return 19;
        sort(v.begin(),v.end());
        int freq[n+1];memset(freq,0,sizeof(freq));
        for(int i=0;i<n;i++)freq[v[i]]+=1;
        for(auto &ele:freq){
            if(ele>k)return -1;
        }
        k=(n/k);
        int ul=(1<<n);
        int dp[ul+1];memset(dp,0,sizeof(dp));
        int cost[ul+1];memset(cost,0,sizeof(cost));
        for(int i=1;i<ul;i++){
            pair<bool,int>tobe=isval(v,i);
            cost[i]=tobe.ss;
        }
        for(int i=1;i<ul;i++){
            int sb=(__builtin_popcount(i));
            if(sb==k){
                dp[i]=cost[i];
            }
            else if(sb%k==0){
                int temp=INT_MAX;
                for(int j=i;j;j=(j-1)&i){
                    if(j==i)continue;
                    else{
                        int tt=__builtin_popcount(j);
                        if(tt==k){
                            // pair<bool,int>tobe=isval(v,j);
                            temp=min(temp,cost[j]+dp[(i^j)]);
                        }
                    }
                }
                dp[i]=temp;
            }
            else dp[i]=(1e5);
        }
        return dp[ul-1];
    }
};
140600