<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 groupSize = n/k;

        unordered_map<int, int> allGroups; // all groups of size = groupSize
        unordered_map<int, int> memo;

        sort(nums.begin(), nums.end());
        generateAllGroups(allGroups, groupSize, nums);

        return findMinScore(0, k, (1 << n) - 1, allGroups, memo);
    }

    int findMinScore(int i, int groups, int remain, unordered_map<int, int> &allGroups, unordered_map<int, int>& memo) {
        if (i == groups) return 0;
        if (memo.contains(remain)) return memo[remain];
        
        int minScore = INT_MAX;

        for (auto &g : allGroups) {
            if ((g.first & remain) == g.first) {
                remain -= g.first;

                int score = findMinScore(i+1, groups, remain, allGroups, memo);

                if (score != -1) {
                    minScore = min(minScore, g.second + score);
                }

                remain += g.first;
            }
        }

        memo[remain] = minScore == INT_MAX ? -1 : minScore;
    
        return memo[remain];
    }

    void generateAllGroups(unordered_map<int, int> &allGroups, int groupSize, vector<int>& nums) {
        int n = nums.size();

        for (int i = 0; i < (1 << n); ++i) {
            int bits = 0;
            int minVal = INT_MAX;
            int maxVal = INT_MIN;

            for (int j = 0, k = -1; j < n; ++j) {
                if (i & (1 << j)) {
                    if (k != -1 && nums[j] == nums[k]) { // same number, invalid group
                        bits = 0;
                        break;
                    }

                    bits++;
                    minVal = min(minVal, nums[j]);
                    maxVal = max(maxVal, nums[j]);
                    k = j;
                }
            }

            if (bits == groupSize) {
                allGroups[i] = maxVal - minVal;
            }
        }
    }
};
30900