Partition Array Into Two Arrays to Minimize Sum Difference

Naive Approach (Brute Force)

The most straightforward approach is to generate all possible partitions of the nums array into two subarrays of size n. For each partition, calculate the absolute difference between the sums of the two subarrays, and keep track of the minimum difference found so far.

This can be implemented using recursion or bit manipulation to generate all possible subsets. Each element can either belong to the first or the second subarray.

Code (Python):

import itertools

def min_abs_difference_brute_force(nums):
    n = len(nums) // 2
    min_diff = float('inf')
    for i in range(1 << (2 * n)):
        group1 = []
        group2 = []
        for j in range(2 * n):
            if (i >> j) & 1:
                group1.append(nums[j])
            else:
                group2.append(nums[j])
        if len(group1) == n:
            min_diff = min(min_diff, abs(sum(group1) - sum(group2)))
    return min_diff

Time Complexity: O(2²ⁿ * n), where n is half the length of the input array. We iterate through all 2²ⁿ possible subsets and calculate the sum of each subset, taking O(n) time.

Space Complexity: O(n) to store the two subarrays.

Optimal Solution (Meet in the Middle)

Since the constraint 1 <= n <= 15, a brute-force approach will have a time complexity of 2^30 in the worst case. To optimize, we can use the meet-in-the-middle technique.

1. Split the nums array into two halves, each of size n. Let's call them left and right.
2. Generate all possible subset sums for each half. For each possible number of elements k (0 to n), store all subset sums of size k for both left and right in separate lists.
3. Iterate through all possible subset sizes k from 0 to n.
   • For each k, iterate through all possible subset sums sum_left in the left list of size k.
   • The goal is to find a sum_right in the right list of size n - k such that sum_left + sum_right is as close as possible to total_sum / 2. Use binary search to find the closest sum_right.
   • Update the minimum absolute difference with the new calculated difference.

Code (Python):

import bisect

def min_abs_difference(nums):
    n = len(nums) // 2
    total_sum = sum(nums)

    def subset_sums(arr):
        sums = [[] for _ in range(n + 1)]
        for mask in range(1 << n):
            count = 0
            subset_sum = 0
            for i in range(n):
                if (mask >> i) & 1:
                    subset_sum += arr[i]
                    count += 1
            sums[count].append(subset_sum)
        for i in range(n + 1):
            sums[i].sort()
        return sums

    left = nums[:n]
    right = nums[n:]

    left_sums = subset_sums(left)
    right_sums = subset_sums(right)

    min_diff = float('inf')

    for k in range(n + 1):
        for sum_left in left_sums[k]:
            remaining_sum = total_sum // 2 - sum_left
            right_arr = right_sums[n - k]
            idx = bisect.bisect_left(right_arr, remaining_sum)

            if idx < len(right_arr):
                min_diff = min(min_diff, abs(total_sum - 2 * (sum_left + right_arr[idx])))
            if idx > 0:
                min_diff = min(min_diff, abs(total_sum - 2 * (sum_left + right_arr[idx - 1])))

    return min_diff

Time Complexity: O(n * 2ⁿ + n² * log(2ⁿ)). Generating subsets takes O(n * 2ⁿ). Sorting the subset sums takes O(2ⁿlog(2ⁿ)) which becomes O(n*2ⁿ). The binary search part takes O(n²*log(2ⁿ))

Space Complexity: O(n * 2ⁿ) to store the subset sums for both halves.

Edge Cases

• Empty Input: If the input array is empty, the problem statement specifies that 1 <= n <= 15 and nums.length == 2 * n. Thus, empty input will never occur.
• Negative Numbers: The presence of negative numbers is handled correctly as the algorithm considers both positive and negative sums.
• Large Numbers: The absolute values of numbers could be up to 10⁷. The sum of n such numbers could be as large as 15 * 10^7 = 1.5 * 10^8. This is within the range of int in Python, and will not overflow.