No. Independent halves combine additively, trading one exponential exponent for two half-sized enumerations.

Every full choice is a pair of half choices. Listing both halves turns one giant search tree into two smaller catalogs that can be joined intelligently.

What the function promises: Return the minimum absolute difference between target and any subset sum. Empty subset is allowed.

from bisect import bisect_left

def closest_subset_difference(values, target):
    middle = len(values)//2
    def sums(part):
        result = [0]
        for value in part:
            result += [old+value for old in result]
        return result
    left = sums(values[:middle])
    right = sorted(sums(values[middle:]))
    best = abs(target)
    for subtotal in left:
        position = bisect_left(right, target-subtotal)
        for i in (position-1, position):
            if 0 <= i < len(right):
                best = min(best, abs(target-subtotal-right[i]))
    return best

assert closest_subset_difference([5,-7,3,5], 6) == 0
assert closest_subset_difference([], 9) == 9

Work and storage: O(2^(n/2) · n) generation in this simple list-building form, plus sorting and O(2^(n/2)) space.

When to recognize it: Independent halves combine additively, trading one exponential exponent for two half-sized enumerations.

Check before applying it: It still grows exponentially and needs substantial memory.

A sum-indexed DP or bitset can be pseudo-polynomial and much faster.