<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(), m = n/k;
        // bmap contains the value of all acceptable subsets
        const vector<int> submap = getMap(nums, m);
        vector<int> dp(1<<n, -1);
        dp[0] = 0;
        for (int b = 0; b < (1<<n); b++)
        {
            if (__builtin_popcount(b)%m != 0) continue;//b is not a combination of some subsets (maybe 1, 2, ..., k subsets)
            //iterating thru all b's submasks, which is acceptable
            //ex: b=1011001 -> sub = 1011000, 1010000, 1000000, 0000000
            for (int sub = b; sub; sub = (sub-1)&b)
            {
                if (submap[sub] == -1) continue; // if submask hasnt caculate -> not acceptable
                if (dp[b-sub] == -1) continue; // if this submask is oke, but it cant combine with other subset to from b
                if (dp[b] == -1) dp[b] = dp[b-sub] + submap[sub];
                else dp[b] = min(dp[b], dp[b-sub] + submap[sub]);
            }
        }
        return dp.back();
    }
private:
    // Generates a vetor v where:
    //   v[b] = the incompatibility of bitmask b, if b represents a valid subset of size m
    //   v[b] = -1, if b is not a valid subset of size m
    vector<int> getMap(const vector<int>& nums, int m)
    {
        int n = nums.size();
        vector<int> ans(1<<n, -1);
        for (int mask = 0; mask < (1<<n); mask++)
        {
            // mask is a subset
            // numbers of elements != m -> not a subset
            if (__builtin_popcount(mask) != m) continue;
            int flag = 1;
            int mX = 0, mI = 17, vis = 0;
            // caculate the value of this subset
            // find max and min element
            for (int i = 0; i < n; i++)
            {
                // if this subset doesnot contain this index -> continue;
                if ((mask & (1<<i)) == 0) continue;
                if (vis & (1<<nums[i])) // if an elements appear twice -> this subset not acceptable
                {
                    flag = 0;
                    break;
                }
                vis |= (1<<nums[i]);
                mX = max(mX, nums[i]);
                mI = min(mI, nums[i]);
            }
            // if this subset not ok -> dont caculate its value
            if (!flag) continue; 
            ans[mask] = abs(mX - mI);
        }
        return ans;
    }
};
16400