Introduction

In this blog post, we will explore how to solve the LeetCode problem "2485" 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

Given a positive integer n, find the pivot integer x such that: The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively. Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input. Example 1: Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21. Example 2: Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1. Example 3: Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist. Constraints: 1 <= n <= 1000

Explanation

Here's an efficient solution to find the pivot integer:

• Mathematical Approach: Instead of iterating and calculating sums repeatedly, we can use the formula for the sum of an arithmetic series. The sum of numbers from 1 to x is x * (x + 1) / 2, and the sum from 1 to n is n * (n + 1) / 2. The sum from x to n can then be calculated as the difference between the sum from 1 to n and the sum from 1 to x-1. We're looking for an x where sum(1 to x) = sum(x to n).
• Iterative Search: Iterate through possible values of x from 1 to n. For each x, calculate the sum from 1 to x using the formula. Then, calculate the sum from x to n also using the formula and compare.

• Early Exit: As the problem states that there will be at most one pivot, we can immediately return once we find it.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def pivotInteger(n: int) -> int:
    """
    Finds the pivot integer x such that the sum of all elements between 1 and x inclusively
    equals the sum of all elements between x and n inclusively.

    Args:
        n: A positive integer.

    Returns:
        The pivot integer x. If no such integer exists, return -1.
    """

    total_sum = n * (n + 1) // 2

    for x in range(1, n + 1):
        sum_1_to_x = x * (x + 1) // 2
        sum_x_to_n = total_sum - (x - 1) * x // 2

        if sum_1_to_x == sum_x_to_n:
            return x

    return -1