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, 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
nand compute their sum. - Iteration: Continue the process until
nconverges to a stable value (i.e.,ndoesn't change after the sum of its prime factors is calculated). Return Stable Value: The stable value is the smallest value
nwill take on.Runtime Complexity: O(n log n), Storage Complexity: O(1)
Code
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