Introduction

In this blog post, we will explore how to solve the LeetCode problem "2178" 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 finalSum. Split it into a sum of a maximum number of unique positive even integers. For example, given finalSum = 12, the following splits are valid (unique positive even integers summing up to finalSum): (12), (2 + 10), (2 + 4 + 6), and (4 + 8). Among them, (2 + 4 + 6) contains the maximum number of integers. Note that finalSum cannot be split into (2 + 2 + 4 + 4) as all the numbers should be unique. Return a list of integers that represent a valid split containing a maximum number of integers. If no valid split exists for finalSum, return an empty list. You may return the integers in any order. Example 1: Input: finalSum = 12 Output: [2,4,6] Explanation: The following are valid splits: (12), (2 + 10), (2 + 4 + 6), and (4 + 8). (2 + 4 + 6) has the maximum number of integers, which is 3. Thus, we return [2,4,6]. Note that [2,6,4], [6,2,4], etc. are also accepted. Example 2: Input: finalSum = 7 Output: [] Explanation: There are no valid splits for the given finalSum. Thus, we return an empty array. Example 3: Input: finalSum = 28 Output: [6,8,2,12] Explanation: The following are valid splits: (2 + 26), (6 + 8 + 2 + 12), and (4 + 24). (6 + 8 + 2 + 12) has the maximum number of integers, which is 4. Thus, we return [6,8,2,12]. Note that [10,2,4,12], [6,2,4,16], etc. are also accepted. Constraints: 1 <= finalSum <= 1010

Explanation

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

• High-Level Approach:

  • If finalSum is odd, it cannot be split into even numbers, so return an empty list.
  • Start with the smallest even number (2) and keep adding the next available even number to the list as long as the sum of even numbers is less than finalSum.
  • If the sum equals finalSum, return the list. Otherwise, adjust the last element in the list to accommodate the remaining difference.

• Complexity:

  • Runtime Complexity: O(sqrt(finalSum)) because, in the worst-case scenario, we iterate up to the square root of finalSum to generate the sequence.
  • Storage Complexity: O(sqrt(finalSum)) to store the list of even numbers.

Code

    def maximumEvenSplit(finalSum: int) -> list[int]:
    """
    Splits an integer into a sum of a maximum number of unique positive even integers.

    Args:
        finalSum: The integer to split.

    Returns:
        A list of integers that represent a valid split containing a maximum number of integers.
        If no valid split exists for finalSum, return an empty list.
    """
    if finalSum % 2 != 0:
        return []

    result = []
    current_sum = 0
    num = 2

    while current_sum + num <= finalSum:
        result.append(num)
        current_sum += num
        num += 2

    if current_sum < finalSum:
        result[-1] += (finalSum - current_sum)

    return result