2035 Partition Array Into Two Arrays to Minimize Sum Difference

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

Return the minimum possible absolute difference.

Example 1:

arr1: 3  9
nums: 3  9  7  3
arr2:       7  3

Input: nums = [3,9,7,3]
Output: 2
Explanation: 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.

Example 2:

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

Example 3:

arr1: 2         4      -9
nums: 2  -1  0  4  -2  -9
arr2:    -1  0     -2

Input: nums = [2,-1,0,4,-2,-9]
Output: 0
Explanation: 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.

Constraints:

• 1 <= n <= 15
• nums.length == 2 * n
• -107 <= nums[i] <= 107

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class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        h = len(nums) // 2
        left_half, right_half = nums[:h], nums[h:]
        ans = abs(sum(left_half) - sum(right_half))
        total = sum(nums)
        half = total // 2
        
        for k in range(1, h // 2 + 1):
            # select k from left half, and h-k from right half
            left = [sum(comb) for comb in combinations(left_half, k)]
            right = [sum(comb) for comb in combinations(right_half, h - k)]
            right.sort()

            for l in left:
                r = half - l
                p = bisect.bisect_left(right, r) 
                for pp in [p, p - 1]:
                    if 0 <= pp < len(right):
                        left_ans_sum = l + right[pp]
                        right_ans_sum = total - left_ans_sum
                        diff = abs(left_ans_sum - right_ans_sum)
                        ans = min(ans, diff) 
        return ans