<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 nn;
    vector<int> dp;

    int solve(int mk, vector<vector<int>> &mp) {
        if (mk == (1 << nn) - 1) return 0; 
        if (dp[mk] != -1) return dp[mk];  

        int ans = 1e8;
        for (int i = 0; i < mp.size(); i++) {
            if ((mk & mp[i][0]) == 0) {
                ans = min(ans, mp[i][1] + solve(mk | mp[i][0], mp));
            }
        }

        return dp[mk] = ans;  
    }

    int minimumIncompatibility(vector<int>& nums, int k) {
        vector<vector<int>> mp;

        int n = nums.size();
        nn = n;
        int r = n / k;  
                for (int i = 0; i < (1 << n); i++) {
            if (__builtin_popcount(i) == r) {
                set<int> st;
                int mi = n + 10, ma = -1;

                for (int b = 0; b < n; b++) {
                    if ((i & (1 << b)) != 0) {
                        mi = min(mi, nums[b]);
                        ma = max(ma, nums[b]);
                        st.insert(nums[b]);
                    }
                }

               
                if (st.size() == r) {
                    mp.push_back({i, ma - mi}); 
                }
            }
        }

        dp.resize(1 << n, -1);  
        int ans = solve(0, mp);

        return ans == 1e8 ? -1 : ans;
    }
};

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