We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Smallest Value After Replacing With Sum of Prime Factors

Difficulty: Medium

Problem Description

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.

Key Insights

  • The problem requires continuously transforming the number n into the sum of its prime factors until it stabilizes at a minimum value.
  • Prime factorization is essential to identify and sum the factors correctly.
  • The process may result in a cycle where n eventually becomes a prime number, which is the smallest possible value.

Space and Time Complexity

Time Complexity: O(sqrt(n) * log(n)), where the log(n) accounts for the number of prime factors. Space Complexity: O(1), as we are using a constant amount of space for calculations.

Solution

To solve the problem, we utilize the concept of prime factorization. The approach involves:

  1. Finding the prime factors of n and summing them.
  2. Replacing n with this sum and repeating the process until n reaches a stable value.
  3. Since we are dealing with prime factors, we can employ trial division up to the square root of n to find the factors efficiently.

The algorithm can be implemented using a loop that continues until n remains unchanged after summing its prime factors.

Code Solutions

def smallest_value(n: int) -> int:
    def sum_of_prime_factors(x):
        total_sum = 0
        factor = 2
        while factor * factor <= x:
            while (x % factor) == 0:
                total_sum += factor
                x //= factor
            factor += 1
        if x > 1:
            total_sum += x
        return total_sum

    while True:
        new_n = sum_of_prime_factors(n)
        if new_n == n:
            break
        n = new_n
    return n
← Back to All Questions