Minimum Absolute Difference of Subsequence Sum

Problem Description

You are given an array of integers nums and an integer goal. The task is to find a subsequence of nums such that the absolute difference between the sum of the subsequence and goal is minimized. Return this minimum absolute difference.

Naive Approach (Brute Force)

The most straightforward approach is to generate all possible subsequences of the input array nums, calculate their sums, and find the minimum absolute difference between these sums and the goal. This involves iterating through all possible combinations of elements in nums.

Code (Python)

def min_abs_difference_brute_force(nums, goal):
    min_diff = float('inf')
    n = len(nums)

    for i in range(1 << n):  # Iterate through all possible subsequences
        current_sum = 0
        for j in range(n):
            if (i >> j) & 1:  # If j-th bit is set, include nums[j] in the subsequence
                current_sum += nums[j]

        min_diff = min(min_diff, abs(current_sum - goal))

    return min_diff

Time Complexity

The brute force approach has a time complexity of O(2ⁿ * n), where n is the length of nums. This is because there are 2ⁿ possible subsequences, and for each subsequence, we need to compute the sum, which takes O(n) time.

Space Complexity

The space complexity is O(1) as we are only using a constant amount of extra space.

Optimal Approach (Meet in the Middle)

Since the constraint 1 <= nums.length <= 40, the brute force approach will result in TLE. A more efficient approach is to use the "Meet in the Middle" technique. We split the array into two halves, generate all possible sums for each half, and then use a two-pointer approach to find the closest sum to the goal.

Steps

1. Split the array: Divide nums into two halves, left and right.
2. Generate sums for each half: Create two sets, left_sums and right_sums, containing all possible subsequence sums for each half. Use recursion for this.
3. Sort one of the sum sets: Sort right_sums to enable efficient searching.
4. Iterate and find closest sums: Iterate through left_sums and for each sum, use binary search in right_sums to find the closest sum that minimizes the absolute difference with the goal.

Code (Python)

def min_abs_difference(nums, goal):
    def generate_sums(arr):
        sums = {0}  # Initialize with 0 for empty subsequence
        for num in arr:
            new_sums = set()
            for s in sums:
                new_sums.add(s + num)
            sums.update(new_sums)
        return sorted(list(sums))

    n = len(nums)
    left = nums[:n // 2]
    right = nums[n // 2:]

    left_sums = generate_sums(left)
    right_sums = generate_sums(right)

    right_sums.sort()

    min_diff = float('inf')

    for left_sum in left_sums:
        remaining = goal - left_sum
        
        # Binary search to find the closest value in right_sums
        l, r = 0, len(right_sums) - 1
        while l <= r:
            mid = (l + r) // 2
            min_diff = min(min_diff, abs(left_sum + right_sums[mid] - goal))
            if right_sums[mid] < remaining:
                l = mid + 1
            elif right_sums[mid] > remaining:
                r = mid - 1
            else:
                return 0  # Exact match found
        
        # Check the values at indices l and r after the loop finishes
        if l < len(right_sums):
            min_diff = min(min_diff, abs(left_sum + right_sums[l] - goal))
        if r >= 0:
            min_diff = min(min_diff, abs(left_sum + right_sums[r] - goal))

    return min_diff

Time Complexity

The optimal approach has a time complexity of O(2^n/2 + 2^n/2 log(2^n/2)), where n is the length of nums. This simplifies to O(2^n/2 + n * 2^{n/2 - 1}) since generating the sums for each half takes O(2^n/2) and sorting takes O(2^n/2 log(2^n/2)). The binary search contributes O(n * 2^n/2) to find the best combination, so, overall, it can be simplified to O(n * 2^n/2).

Space Complexity

The space complexity is O(2^n/2) to store the sums for each half.

Edge Cases

• Empty input array: The function should handle the case when the input array is empty. Although the constraints specify at least one element, handling this edge case makes the solution more robust.
• Goal is achievable: The function should return 0 if the goal is achievable by some subsequence sum.
• Large differences: The function should handle cases where the difference between the subsequence sum and the goal is very large.

By using the meet in the middle approach, we significantly reduce the time complexity compared to the brute-force method, making it feasible for larger input arrays.