<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>

/*
see if this solution could help in understanding time complexity and optimisation:
    https://leetcode.com/problems/partition-to-k-equal-sum-subsets/submissions/1390009398/


    TC analysis:
    worst case when no case omitted (no recursion tree prunning) :
    for each elemet : we have k places to be placed in, thus TC will be calsulated as:
    k*k*k*...n times : k^n

    Prun cases : 
    1. duplication of subset e.g. putting first element only in first subset makes sense,
        rest it will do duplication
    2. subset size constraint
    

    After prunning (ommitting duplicates and redundant cases) :
    think it like : we have k places, and we are checking for each place whether we can keep current element there or not
    
    if we can place element or not (selecting or not selecting) : gives 2^n
    as k places : k*(2^n)

    It's very important to understand complexity analysis

    Bitmasking is just way to convert recursion into iteration
*/

class Solution {
public:
    int targetSubsetSize;
    int fun(vector<int>& nums, int i, vector<vector<int>>& subsets) {
        if(i==nums.size()) {
            int sum=0;
            for(auto &subset: subsets) { //equal sizes
                sum+=(subset.back()-subset[0]);
            }
            return sum;
        }

        int ans=1e9;
        int emptySubsetExplored = false;
        for(auto &subset: subsets) {
            if(emptySubsetExplored) break;

            if(subset.size() < targetSubsetSize && binary_search(subset.begin(), subset.end(), nums[i])==false) {
                if(subset.size()==0) emptySubsetExplored = true;
                subset.push_back(nums[i]);
                ans=min(ans,fun(nums,i+1,subsets));
                subset.pop_back();
            }
        }

        return ans;
    }


    int minimumIncompatibility(vector<int>& nums, int k) {
        int n=nums.size();
        if(n%k != 0) return -1;

        targetSubsetSize = n/k;
        sort(nums.begin(), nums.end());
        vector<vector<int>> subsets(k);

        int ans = fun(nums,0,subsets);
        return ans==1e9 ? -1 : ans;
    }
};
9200