<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 func(int mask, vector<int> &ValidPermutations, map<int,int> &ValidPValues,vector<int>&dp){
        if(mask == 0){ // if all elements of numbers is covered
            return 0; 
        }
        if(dp[mask] != -1){
            return dp[mask]; // this is dp step
        }
        int ans = INT_MAX/2;
        for(auto it : ValidPermutations){
            if((mask & it) == it){ // wether we can take this permutation
                int ans1 = func(mask^it,ValidPermutations,ValidPValues,dp); // here we removed this mask such that other permutation can't take common elements
                ans = min(ans,ans1 + ValidPValues[it]);
            }
        }
        return dp[mask] = ans; // dp step
    }
    int minimumIncompatibility(vector<int>& nums, int k) {
        // first will check simple condition for not possible cases
        int n  = nums.size();
        if(n%k != 0){
            return -1;
        }
        int cnt = n/k;

        vector<int> ValidPermutations; // to keep track of valid permutations
        map<int,int> ValidPValues; // to keep track of valid permutation value

        // there can be 2^n permutations as every element has two options
        // now we are going to check which permutations are valid according to given codition
        for(int i = 1;i < (1<<n);i++){
            int count = 0;
            int mn = INT_MAX; // minimum element which is in this permutation
            int mx = INT_MIN; // maximum element which is in this permutation

            set<int> st; // to keep track of repeated elements
            bool isRepeat = false;
            for(int j = 0;j < n;j++){// itrate on all array elements
                int mask = 1 << j; // here we are checking this number is in mask or not
                if(mask & i){
                    if(st.find(nums[j])  != st.end()){// in this permutation there is element which is repeating
                        isRepeat = true;
                        break;
                    }
                    count ++;
                    if(count > cnt){
                        break;
                    }
                    mn = min(mn,nums[j]);
                    mx = max(mx,nums[j]);
                    st.insert(nums[j]);
                }
            }
            if(count != cnt){ // if it don't pass the count condition
                continue;
            }
            if(isRepeat == false){ // if pass the repatation condtion then will save permutation in validPermutations
                if(mx == INT_MIN || mn == INT_MAX){
                    continue;
                }
                ValidPermutations.push_back(i);
                ValidPValues[i] = mx-mn;
            }

        }
        vector<int>dp(1<<n,-1);
        int ans = func((1<<n) -1,ValidPermutations,ValidPValues,dp);
        if(ans >= INT_MAX/2){
            return -1;
        }
        return ans;

    }
};
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