Solution Explanation

This problem asks to find the minimum absolute difference between the sum of a subsequence of the input array nums and a given goal. The constraints suggest that a brute-force approach (iterating through all possible subsequences) would be too slow. The efficient solutions leverage a divide-and-conquer strategy combined with binary search.

Core Idea:

1. Divide: Split the input array nums into two halves.
2. Conquer: Generate all possible subsequence sums for each half recursively. This is done using a Depth-First Search (DFS) function. Each subsequence sum is stored in a set (or list in Python).
3. Combine: For each sum (l) from the left half's set, find the closest sum (r) in the sorted right half's set such that l + r is closest to the goal. Binary search is used to efficiently find this closest sum.
4. Result: The minimum absolute difference is the minimum difference found across all pairs of sums from the left and right halves.

Time Complexity Analysis:

• DFS: The time complexity of the DFS function is O(2n/2), where n is the length of nums. This is because each element can either be included or excluded in a subsequence, resulting in 2n/2 possibilities for each half.
• Binary Search: The binary search within the sorted right half takes O(log(2n/2)) = O(n/2) time in the worst case.
• Combining: The combination step iterates through the left half's set (size approximately 2n/2) and performs a binary search for each element. Therefore, the time complexity of this step is O(2n/2 * n/2).

Overall Time Complexity: The overall time complexity is dominated by the DFS and combining steps, which are both approximately O(n * 2n/2). This is significantly better than the brute-force O(n * 2n) approach.

Space Complexity Analysis:

The space complexity is dominated by the sets (or lists) used to store the subsequence sums. Each set can contain approximately 2n/2 elements. Therefore, the space complexity is O(2n/2).

Code Explanation (Python):

The Python code efficiently implements this strategy:

class Solution:
    def minAbsDifference(self, nums: List[int], goal: int) -> int:
        n = len(nums)
        left = set()
        right = set()
 
        self.getSubSeqSum(0, 0, nums[: n // 2], left)
        self.getSubSeqSum(0, 0, nums[n // 2 :], right)
 
        result = inf
        right = sorted(right)
        rl = len(right)
 
        for l in left:
            remaining = goal - l
            idx = bisect_left(right, remaining)
 
            if idx < rl:
                result = min(result, abs(remaining - right[idx]))
 
            if idx > 0:
                result = min(result, abs(remaining - right[idx - 1]))
 
        return result
 
    def getSubSeqSum(self, i: int, curr: int, arr: List[int], result: Set[int]):
        if i == len(arr):
            result.add(curr)
            return
 
        self.getSubSeqSum(i + 1, curr, arr, result)
        self.getSubSeqSum(i + 1, curr + arr[i], arr, result)
 

The getSubSeqSum function recursively generates all subsequence sums for a given array portion. The main function then combines the results from both halves using binary search to find the closest sum to the goal. bisect_left from the bisect module efficiently finds the insertion point for a value in a sorted list.

The other languages (Java, C++, Go) implement the same algorithm with minor syntactic differences. They use similar data structures and techniques to achieve the same time and space complexity.