Introduction

In this blog post, we will explore how to solve the LeetCode problem "829" 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 an integer n, return the number of ways you can write n as the sum of consecutive positive integers. Example 1: Input: n = 5 Output: 2 Explanation: 5 = 2 + 3 Example 2: Input: n = 9 Output: 3 Explanation: 9 = 4 + 5 = 2 + 3 + 4 Example 3: Input: n = 15 Output: 4 Explanation: 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 Constraints: 1 <= n <= 109

Explanation

Here's an efficient solution to the problem:

• Mathematical Insight: The core idea is to transform the problem into finding odd factors of n. If n can be represented as a sum of k consecutive integers starting from a + 1, then n = (a + 1) + (a + 2) + ... + (a + k) = k * a + k * (k + 1) / 2. Rearranging, we get 2n = k * (2a + k + 1). Since a must be a non-negative integer, 2a + k + 1 > k. Also, one of k and 2a + k + 1 must be odd and the other even. Furthermore, 2a + k + 1 = (2n / k) and since 2a + k + 1 > k, it follows that k < sqrt(2n). Therefore, we only need to iterate for odd factors k.

• Odd Factor Counting: We iterate through odd numbers k starting from 1. For each k, we check if n is divisible by k. If it is, we determine if (2n / k - k - 1) is an even, non-negative integer. This implies that a is a non-negative integer.

• Optimization: We only check odd factors of n, reducing the number of iterations significantly. We can stop iterating when k > sqrt(2n).

• Runtime: O(sqrt(n)), Space: O(1)

Code

    def consecutiveNumbersSum(n: int) -> int:
    """
    Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.
    """
    count = 0
    i = 1
    while i * i <= 2 * n:
        if n % i == 0:
            if (2 * n / i - i - 1) % 2 == 0 and (2 * n / i - i - 1) >= 0:
                count += 1
        i += 2

    return count