<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:
    map<int,int>mp;
    int dp[1<<16];
    int recur(int mask){
        if(mask == 0){
            return 0 ;
        }
        if(dp[mask]!=-1){
            return dp[mask];
        }
        int ans = 1e6;
        for(auto &it : mp){
            int submask = it.first;
            int score = it.second;
            if( (submask & mask) == submask) {
                ans = min(ans,score + recur(mask^submask));
            }
        }
        return dp[mask] = ans;
    }
    int minimumIncompatibility(vector<int>& nums, int k){
        
        int n = nums.size();
        int l = n/k;
        for(int mask=1;mask<(1<<n);mask++){
            int q = __builtin_popcount(mask);
            if(q==l){
                bool flag = true;
                map<int,int>vis;
                vector<int>v;
                for(int i=0;i<n;i++){
                    if( (mask&(1<<i))){
                        if(vis.find(nums[i])==vis.end()){
                            vis[nums[i]]=1;
                            v.push_back(nums[i]) ;
                        }
                        else{
                            flag = false;
                            break;
                        }
                    }
                }
                if(flag){
                    sort(v.begin(),v.end());
                    int score = v.back()-v[0];
                    mp[mask] = score;
                }
            }
        }
        memset(dp,-1,sizeof(dp));
        int ans  = recur( (1<<n) -1);
        if(ans == 1e6){
            return -1;
        }
        return ans;
    }
};
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