<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 minIncompatibility = INT_MAX;

    void backtrack(vector<int>& nums, vector<int>& subsetMax, vector<int>& subsetMin, vector<int>& subsetSize, int index, int subsetCount, int k) {
        if (index == nums.size()) {
            int currSum = 0;
            for (int i = 0; i < k; i++) {
                currSum += (subsetMax[i] - subsetMin[i]);
            }
            minIncompatibility = min(minIncompatibility, currSum);
            return;
        }
        
        for (int i = 0; i < k; i++) {
            if (subsetSize[i] == 0) {
                subsetMax[i] = nums[index];
                subsetMin[i] = nums[index];
                subsetSize[i]++;
                backtrack(nums, subsetMax, subsetMin, subsetSize, index + 1, subsetCount + 1, k);
                subsetSize[i]--;
                break; // Avoid placing the same element in an empty subset
            } else if (subsetSize[i] < nums.size() / k && nums[index] != subsetMax[i] && nums[index] != subsetMin[i]) {
                int originalMax = subsetMax[i];
                int originalMin = subsetMin[i];
                subsetMax[i] = max(subsetMax[i], nums[index]);
                subsetMin[i] = min(subsetMin[i], nums[index]);
                subsetSize[i]++;
                backtrack(nums, subsetMax, subsetMin, subsetSize, index + 1, subsetCount, k);
                subsetSize[i]--;
                subsetMax[i] = originalMax;
                subsetMin[i] = originalMin;
            }
        }
    }

    int minimumIncompatibility(vector<int>& nums, int k) {
        int n = nums.size();
        if (n % k != 0) return -1; // Cannot divide into equal subsets

        sort(nums.begin(), nums.end());

        vector<int> subsetMax(k, INT_MIN);
        vector<int> subsetMin(k, INT_MAX);
        vector<int> subsetSize(k, 0);

        backtrack(nums, subsetMax, subsetMin, subsetSize, 0, 0, k);

        return minIncompatibility == INT_MAX ? -1 : minIncompatibility;
    }
};

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