# Solving Leetcode Interviews in Seconds with AI: Smallest Value After Replacing With Sum of Prime Factors


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2507" 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
	> You are given a positive integer n. Continuously replace n with the sum of its prime factors.  Note that if a prime factor divides n multiple times, it should be included in the sum as many times as it divides n.  Return the smallest value n will take on.   Example 1:  Input: n = 15 Output: 5 Explanation: Initially, n = 15. 15 = 3 * 5, so replace n with 3 + 5 = 8. 8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6. 6 = 2 * 3, so replace n with 2 + 3 = 5. 5 is the smallest value n will take on.  Example 2:  Input: n = 3 Output: 3 Explanation: Initially, n = 3. 3 is the smallest value n will take on.    Constraints:  2 <= n <= 105  

	# Explanation
	Here's the breakdown of the solution:

*   **Prime Factorization:** Repeatedly find the prime factors of `n` and compute their sum.
*   **Iteration:** Continue the process until `n` converges to a stable value (i.e., `n` doesn't change after the sum of its prime factors is calculated).
*   **Return Stable Value:** The stable value is the smallest value `n` will take on.

*   **Runtime Complexity:** O(n log n), **Storage Complexity:** O(1)

	
	# Code
	```python
	def smallestValue(n):
    """
    Continuously replaces n with the sum of its prime factors until n becomes stable.

    Args:
        n: A positive integer.

    Returns:
        The smallest value n will take on.
    """

    while True:
        sum_of_prime_factors = 0
        temp_n = n
        i = 2
        while i * i <= temp_n:
            while temp_n % i == 0:
                sum_of_prime_factors += i
                temp_n //= i
            i += 1

        if temp_n > 1:
            sum_of_prime_factors += temp_n

        if sum_of_prime_factors == n:
            return n
        else:
            n = sum_of_prime_factors
	```