Introduction

In this blog post, we will explore how to solve the LeetCode problem "1755" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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). Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array. Example 1: Input: nums = [5,-7,3,5], goal = 6 Output: 0 Explanation: Choose the whole array as a subsequence, with a sum of 6. This is equal to the goal, so the absolute difference is 0. Example 2: Input: nums = [7,-9,15,-2], goal = -5 Output: 1 Explanation: Choose the subsequence [7,-9,-2], with a sum of -4. The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum. Example 3: Input: nums = [1,2,3], goal = -7 Output: 7 Constraints: 1 <= nums.length <= 40 -107 <= nums[i] <= 107 -109 <= goal <= 109

Explanation

Here's a breakdown of the approach and the Python code:

• Divide and Conquer: Split the input array nums into two halves.
• Generate Subsequence Sums: Generate all possible subsequence sums for each half.

• Two Pointer Optimization: Sort the sums of the second half and use a two-pointer approach to find the closest sum to goal - sum1 for each sum1 in the first half's sums.

• Runtime Complexity: O(N * 2^(N/2) * log(2^(N/2))) where N is the length of nums. Specifically, generating subsequence sums takes O(2^(N/2)), sorting takes O(2^(N/2) * log(2^(N/2)), and the two-pointer search takes O(2^(N/2) * log(2^(N/2))). Storage Complexity: O(2^(N/2)) to store the subsequence sums.

Code

    def minAbsDifference(nums, goal):
    def get_sums(arr):
        sums = {0}
        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 = get_sums(left)
    right_sums = get_sums(right)

    min_diff = float('inf')
    j = len(right_sums) - 1

    for left_sum in left_sums:
        target = goal - left_sum

        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] < target:
                l = mid + 1
            elif right_sums[mid] > target:
                r = mid - 1
            else:
                return 0


    return min_diff