<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 ans = 1e9;
    bool validate(vector<set<int>> &dp){
        int tempans = 0;    
        for(int i = 0;i<dp.size();i++){
            if(!dp[i].empty())
                tempans += abs(*dp[i].begin() - *dp[i].rbegin());
            if(tempans > ans){
                return false;
            }
        }
        return true;
    }
    void solve(vector<set<int>> &dp,vector<int> &nums,int size,int curr){
        if(curr >= nums.size()){
            int tempans = 0;
            
            for(int i = 0;i<dp.size();i++){
                tempans += abs(*dp[i].begin() - *dp[i].rbegin());
            }
            ans = min(ans,tempans);
            return;
        }
        // cout << "curr " << curr << " sz " << size << " dps " << dp.size() <<  endl;
        set<set<int>> vis;
        for(int i = 0;i<dp.size();i++){
            if(dp[i].size() < size && dp[i].find(nums[curr]) == dp[i].end() && vis.find(dp[i]) == vis.end()
            && validate(dp) ){
                vis.insert(dp[i]);
                dp[i].insert(nums[curr]);
                solve(dp,nums,size,curr+1);
                dp[i].erase(nums[curr]);
            }
        }
    }
    int minimumIncompatibility(vector<int>& nums, int k) {
        
        vector<int> freq(17,0);
        int n = nums.size();
        for(int i = 0;i<n;i++){
            freq[nums[i]]++;
            // cout << freq[nums[i]] << " ";
            if(freq[nums[i]] > (k)){
                return -1;
            }
        }
        if(k == 1){
            return 0;
        }
        sort(nums.begin(),nums.end());
        
        // vector<set<int>> dp;
        std::vector<std::set<int>> dp(k);
        solve(dp,nums,n/k,0);
        return ans;
    }
};
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