Introduction

In this blog post, we will explore how to solve the LeetCode problem "1982" 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 n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2n subset sums of the unknown array (in no particular order). Return the array ans of length n representing the unknown array. If multiple answers exist, return any of them. An array sub is a subset of an array arr if sub can be obtained from arr by deleting some (possibly zero or all) elements of arr. The sum of the elements in sub is one possible subset sum of arr. The sum of an empty array is considered to be 0. Note: Test cases are generated such that there will always be at least one correct answer. Example 1: Input: n = 3, sums = [-3,-2,-1,0,0,1,2,3] Output: [1,2,-3] Explanation: [1,2,-3] is able to achieve the given subset sums: - []: sum is 0 - [1]: sum is 1 - [2]: sum is 2 - [1,2]: sum is 3 - [-3]: sum is -3 - [1,-3]: sum is -2 - [2,-3]: sum is -1 - [1,2,-3]: sum is 0 Note that any permutation of [1,2,-3] and also any permutation of [-1,-2,3] will also be accepted. Example 2: Input: n = 2, sums = [0,0,0,0] Output: [0,0] Explanation: The only correct answer is [0,0]. Example 3: Input: n = 4, sums = [0,0,5,5,4,-1,4,9,9,-1,4,3,4,8,3,8] Output: [0,-1,4,5] Explanation: [0,-1,4,5] is able to achieve the given subset sums. Constraints: 1 <= n <= 15 sums.length == 2n -104 <= sums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Core Idea: The algorithm works by iteratively finding one element of the unknown array at a time. It leverages the fact that the difference between the largest and smallest subset sum in sums will always be an element of the original array. After finding such an element, the algorithm reconstructs the remaining subset sums and proceeds recursively until the complete array is recovered.

• Subset Reconstruction: After identifying an element, the sums array is split into two groups based on whether the element was included or excluded from each subset sum. This effectively reduces the problem size for the next iteration.

• Base Case: When n = 0, it means we have found all the elements and return an empty array.

• Complexity:

  • Runtime: O(n * 2ⁿ). The recursion depth is 'n', and in each level, we iterate through sums array which has a size of 2ⁿ .
  • Storage: O(2ⁿ) due to the space occupied by the sums array and the call stack of the recursive calls (which can go up to n levels deep, with each level potentially storing the sums array).

Code

    def recoverArray(n: int, sums: list[int]) -> list[int]:
    sums.sort()
    if n == 0:
        return []

    diff = sums[-1] - sums[0]
    pos = []
    neg = []
    count = 0
    used = [False] * len(sums)

    for i in range(len(sums)):
        if used[i]:
            continue
        pos.append(sums[i])
        used[i] = True
        for j in range(i + 1, len(sums)):
            if not used[j] and sums[j] == sums[i] + diff:
                neg.append(sums[j])
                used[j] = True
                break

    if pos.count(0) > 0:
        return [diff] + recoverArray(n - 1, pos)
    else:
        return [-diff] + recoverArray(n - 1, neg)