# Solving Leetcode Interviews in Seconds with AI: Consecutive Numbers Sum


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```