<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 m = n / k;
        int bitsets = 1 << n;

        auto mp = map_gen(nums, m);
        vector<int> dp(bitsets, INT_MAX);
        dp[0] = 0;

        for (int i = 0; i < bitsets; i++) {
            if (__builtin_popcount(i) % m != 0) continue;
            for (int submask = i; submask; submask = (submask - 1) & i) {
                if (mp[submask] == INT_MAX) continue;
                if ((submask & i) != submask) continue;
                if (dp[i - submask] == INT_MAX) continue;
                dp[i] = min(dp[i], dp[i - submask] + mp[submask]);
            }
        }

        int ret = dp.back();
        return ret == INT_MAX ? -1 : ret;
    }

private:
    vector<int> map_gen(auto& nums, int m) {
        int n = nums.size();
        vector<int> map(1 << n, INT_MAX);
        for (int bitset = 0; bitset < (1 << n); bitset++) {
            if (__builtin_popcount(bitset) != m) continue;

            int vis = 0;
            bool is_good = true;
            int mx = INT_MIN, mn = INT_MAX;
            for (int i = 0; i < n; i++) {
                if ((bitset & (1 << i)) == 0) continue;
                else if (vis & (1 << nums[i])) {
                    is_good = false;
                    break;
                }

                vis |= (1 << nums[i]);
                mx = max(mx, nums[i]);
                mn = min(mn, nums[i]);
            }

            if (!is_good) continue;

            map[bitset] = mx - mn;
        }

        return map;
    }
};
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