<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) // 2
        total_sum = sum(nums)
        target = total_sum // 2

        # Function to generate all possible sums for a subset of half the array
        def generate_sums(subset):
            subset_sums = defaultdict(list)
            m = len(subset)
            for k in range(m + 1):
                for comb in combinations(subset, k):
                    subset_sums[k].append(sum(comb))
            return subset_sums

        # Split the array into two halves
        first_half, second_half = nums[:n], nums[n:]

        # Generate all possible sums for both halves
        first_sums = generate_sums(first_half)
        second_sums = generate_sums(second_half)

        # Sort all sum lists to make binary search possible
        for k in second_sums:
            second_sums[k].sort()

        min_diff = float('inf')

        # For each possible sum in the first half, find the best matching sum in the second half
        for k in range(n + 1):
            for s1 in first_sums[k]:
                # We need to find s2 such that s1 + s2 is closest to target
                s2_needed = target - s1
                # Search in the (n-k) sums in the second half
                if n - k in second_sums:
                    s2_list = second_sums[n - k]

                    # Binary search to find the closest value to s2_needed
                    idx = self.binary_search(s2_list, s2_needed)

                    # Check the current candidate
                    candidate_sum = s1 + s2_list[idx]
                    min_diff = min(min_diff, abs(total_sum - 2 * candidate_sum))

                    # Check the neighbor in binary search (if it exists)
                    if idx + 1 < len(s2_list):
                        candidate_sum = s1 + s2_list[idx + 1]
                        min_diff = min(min_diff, abs(total_sum - 2 * candidate_sum))

        return min_diff

    # Binary search helper function to find the closest sum
    def binary_search(self, arr, target):
        lo, hi = 0, len(arr) - 1
        while lo < hi:
            mid = (lo + hi) // 2
            if arr[mid] < target:
                lo = mid + 1
            else:
                hi = mid
        return lo
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