#2507

Medium

LeetCode Problem Workspace

Smallest Value After Replacing With Sum of Prime Factors

Replace a number with the sum of its prime factors until it stabilizes, and return the smallest value.

Problem Statement

You are given a positive integer n. Continuously replace n with the sum of its prime factors. Return the smallest value n will take on.

Each replacement process involves breaking down the number into its prime factors, summing them, and replacing n with the result. The process continues until the value of n cannot be reduced further, which will occur when n becomes a prime number.

Examples

Example 1

Input: n = 15

Output: 5

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

Initially, n = 3. 3 is the smallest value n will take on.

Constraints

2 <= n <= 105

Solution Approach

Prime Factorization

Start by finding the prime factorization of the number n. This can be done using trial division up to the square root of n, or using efficient algorithms like the Sieve of Eratosthenes to precompute primes. The key is breaking n into its prime factors and summing them.

Iterative Replacement

After each prime factorization, replace n with the sum of its prime factors. This replacement is repeated until n stabilizes, meaning it can no longer be reduced further because it is a prime number. Keep track of each replacement step.

Optimization Considerations

To avoid redundant calculations, precompute a list of primes and utilize memoization for faster factorization. Using efficient algorithms to reduce the number of steps is essential for handling large inputs up to 10^5.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity primarily depends on the factorization process. Each factorization step takes O(sqrt(n)) for trial division, and the number of iterations depends on the prime factorization chain. Using an optimized sieve and memoization can improve efficiency for large n. The space complexity depends on the storage of prime numbers and any memoization structures used.

What Interviewers Usually Probe

Candidate understands the process of prime factorization and iterative reduction.
Candidate shows ability to optimize factorization and apply efficient algorithms.
Candidate demonstrates good handling of large input sizes, such as n up to 10^5.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle edge cases where n is already prime.
Failing to optimize factorization for large numbers, leading to time complexity issues.
Not recognizing that the final result is always a prime number.

Follow-up variants

Implement the solution using a sieve of Eratosthenes to precompute primes.
Modify the solution to track the number of iterations required to reach the smallest value.
Add additional constraints where n could be a very large number, requiring more efficient algorithms.

FAQ

What is the smallest value n will take on?

The smallest value is the prime number that n will eventually reduce to after all prime factorizations are summed.

How do I factorize a number efficiently?

You can use trial division up to sqrt(n) or precompute primes with a sieve for more efficient factorization.

What is the time complexity of this problem?

The time complexity depends on the factorization step, typically O(sqrt(n)) for each iteration. Optimizations can reduce redundant calculations.

How does the iterative process work?

After factorizing n and summing its prime factors, n is replaced with the sum and the process repeats until n stabilizes.

What pattern is this problem related to?

This problem combines number theory with a simulation pattern, specifically focusing on prime factorization and iterative replacement.

terminal

Solution

Solution 1: Brute Force Simulation

According to the problem statement, we can perform a process of prime factorization, i.e., continuously decompose a number into its prime factors until it can no longer be decomposed. During the process, add the prime factors each time they are decomposed, and perform this recursively or iteratively.

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class Solution:
    def smallestValue(self, n: int) -> int:
        while 1:
            t, s, i = n, 0, 2
            while i <= n // i:
                while n % i == 0:
                    n //= i
                    s += i
                i += 1
            if n > 1:
                s += n
            if s == t:
                return t
            n = s

Continue Practicing

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