<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:
    int minimumIncompatibility(vector<int>& nums, int k) {
        int n = nums.size();
        int a = n/k;
        unordered_map<int, int> mp;
        for(int i=0; i<(1<<n); i++){
            if(__builtin_popcount(i)==a){
                set<int> st;
                for(int j = 0; j < n; j++){
                    if(i&(1<<j)){
                        if(st.find(nums[j])==st.end()){
                            st.insert(nums[j]);

                        }
                    }
                }
                if(st.size()==a){
                    auto it1 = st.end();
                    it1--;
                    auto it = st.begin();
                    mp[i] = (*it1 - *it);
                 }
            }
        }
        vector<int> dp((1<<n), INT_MAX);
        dp[0]= 0;
        for(auto it= mp.begin(); it!=mp.end(); it++){
            int mask = it->first; 

            for(int i=0; i<(1<<n); i++){
                if(dp[i]==INT_MAX){
                    continue;
                }
                if(!(i&(mask))){
                    if(it->second == INT_MAX){
                        break;
                    }
                    dp[i|mask] = min(dp[i]+it->second, dp[i|mask]);
                }
            }
        }
        if(dp[(1<<n)-1]==INT_MAX){
            return -1;
        }
        return dp[(1<<n)-1];

    }
};
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