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Closest Subsequence Sum

Difficulty: Hard

Problem Description

You are given an integer array nums and an integer goal. You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal). Return the minimum possible value of abs(sum - goal).

Key Insights

  • A subsequence can be formed by removing some elements from the original array, making the problem combinatorial in nature.
  • The maximum length of nums is 40, allowing for potential use of combinatorial methods or dynamic programming.
  • The problem can be split into two halves, enabling an approach utilizing the "meet-in-the-middle" strategy to reduce the overall complexity.
  • The challenge is to efficiently calculate potential sums from subsequences and find the closest one to the given goal.

Space and Time Complexity

Time Complexity: O(2^(n/2) * log(2^(n/2))) where n is the length of nums. This accounts for generating sums for half the array and then searching for the closest sum in the other half. Space Complexity: O(2^(n/2)) to store all possible sums from the first half of the array.

Solution

The solution employs a meet-in-the-middle strategy, which involves dividing the array into two halves. For each half, we compute all possible sums of its subsequences and store these sums in a list. After generating these sums, we sort one of the lists. For each sum from the first half, we perform a binary search on the sorted list of sums from the second half to find the sum that, when added to the current sum, is closest to the goal. The absolute difference is then calculated and tracked to find the minimum difference.

Code Solutions

def minAbsDifference(nums, goal):
    from itertools import combinations
    
    def all_sums(arr):
        sums = set()
        for i in range(len(arr) + 1):
            for combo in combinations(arr, i):
                sums.add(sum(combo))
        return sums

    n = len(nums)
    left_sums = all_sums(nums[:n // 2])
    right_sums = sorted(all_sums(nums[n // 2:]))
    
    min_diff = float('inf')
    
    for left in left_sums:
        target = goal - left
        # Binary search for the closest sum in right_sums
        lo, hi = 0, len(right_sums) - 1
        while lo <= hi:
            mid = (lo + hi) // 2
            current_sum = right_sums[mid]
            min_diff = min(min_diff, abs(current_sum + left - goal))
            if current_sum < target:
                lo = mid + 1
            else:
                hi = mid - 1
    
    return min_diff
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