<p>You are given an integer array <code>nums</code> of <code>2 * n</code> integers. You need to partition <code>nums</code> into <strong>two</strong> arrays of length <code>n</code> to <strong>minimize the absolute difference</strong> of the <strong>sums</strong> of the arrays. To partition <code>nums</code>, put each element of <code>nums</code> into <strong>one</strong> of the two arrays.</p>

<p>Return <em>the <strong>minimum</strong> possible absolute difference</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="example-1" src="https://assets.leetcode.com/uploads/2021/10/02/ex1.png" style="width: 240px; height: 106px;" />
<pre>
<strong>Input:</strong> nums = [3,9,7,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> One optimal partition is: [3,9] and [7,3].
The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [-36,36]
<strong>Output:</strong> 72
<strong>Explanation:</strong> One optimal partition is: [-36] and [36].
The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="example-3" src="https://assets.leetcode.com/uploads/2021/10/02/ex3.png" style="width: 316px; height: 106px;" />
<pre>
<strong>Input:</strong> nums = [2,-1,0,4,-2,-9]
<strong>Output:</strong> 0
<strong>Explanation:</strong> One optimal partition is: [2,4,-9] and [-1,0,-2].
The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
	<li><code>nums.length == 2 * n</code></li>
	<li><code>-10<sup>7</sup> &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        n = len(nums)
        half = n // 2

        full_sum = sum(nums)
        half_sum = full_sum // 2

        # check if left and right part is optimal
        if sum(nums[: half]) == half_sum:
            # here and further
            # (full_sum - 2 * half_sum) is equivalent (full_sum % 2) and isn't equivalent (0)
            # because sum(nums) may be odd 
            return full_sum - 2 * half_sum
        if sum(nums[half :]) == half_sum:
            return full_sum - 2 * half_sum

        def get_subsets_sums(elems):
            result = defaultdict(set)
            result[0] = {0}
            for elem in elems:
                for cnt in reversed(list(result)):
                    result[cnt + 1].update(
                        value + elem
                        for value in result[cnt]
                    )
            return result

        left = get_subsets_sums(nums[: half])
        right = get_subsets_sums(nums[half :])

        def calc(left, right):
            # initial var result
            # find max of n-subset of nums, if sum of subset is equal or less than half_sum
            result = min(min(left[half]), min(right[half]))
            
            for left_cnt, left_sums in left.items():
                right_cnt = half - left_cnt
                right_sums = right[right_cnt]

                a = sorted(left_sums)
                b = sorted(right_sums)
                m = len(a)
                l = len(b)
                i = 0
                j = l - 1
                
                # use 2 pointers instead of a loop in a loop
                while i < m and j >= 0:
                    current = a[i] + b[j]
                    if current == half_sum:
                        return full_sum - 2 * current
                    if current > half_sum:
                        j -= 1
                    else:
                        result = max(result, current)
                        i += 1
            
            return full_sum - 2 * result

        return calc(left, right)
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