Smallest Value After Replacing With Sum of Prime Factors - Video Solutions
2507. Smallest Value After Replacing With Sum of Prime Factors | Weekly Contest 324 | LeetCode 2507
Smallest Value After Replacing With Sum of Prime Factors - Video Solution
Watch 10 video solutions for Smallest Value After Replacing With Sum of Prime Factors, a medium level problem involving Math, Simulation, Number Theory. This walkthrough by Bro Coders has 1,144 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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
nmultiple times, it should be included in the sum as many times as it dividesn.
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
Approach Overview
Problem Overview: You start with an integer n. Replace it with the sum of its prime factors (counted with multiplicity). Repeat this operation until the value no longer changes. The task is to return the smallest value reached after the process stabilizes.
Approach 1: Iterative Prime Factor Sum Replacement (O(k * sqrt(n)) time, O(1) space)
This approach directly simulates the transformation process. For a given value x, compute the sum of its prime factors using standard prime factorization: iterate from 2 to sqrt(x), repeatedly dividing whenever a factor is found. If the remaining value is greater than 1, it is also a prime factor and gets added to the sum. Replace x with this sum and continue until the new value equals the previous one. Each iteration reduces the number or keeps it the same, so the process converges quickly. This method uses only a few variables and no additional data structures, making it efficient and easy to implement using concepts from number theory and simulation.
Approach 2: Recursive Prime Factor Sum Replacement (O(k * sqrt(n)) time, O(log n) space)
The recursive version expresses the same process using function calls. Compute the prime factor sum for the current number. If the sum equals the number, return it because the value has stabilized. Otherwise, call the function again with the new sum. The recursion naturally models the repeated replacement process. The main work still happens in the prime factorization step, which iterates up to sqrt(n). The recursion depth is small because values shrink quickly, but it still uses call stack space. This approach highlights the mathematical structure of the problem and fits well when practicing recursion with math-heavy problems.
Recommended for interviews: The iterative simulation is usually preferred. It avoids recursion overhead and keeps the logic explicit: compute prime factors, update the value, repeat until stable. Showing the brute simulation first demonstrates understanding of the process, while implementing efficient prime factorization shows practical problem-solving skill expected in technical interviews.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Prime Factor Sum Replacement | O(k * sqrt(n)) | O(1) | Best general solution; minimal memory usage and clear control flow |
| Recursive Prime Factor Sum Replacement | O(k * sqrt(n)) | O(log n) | Useful for demonstrating recursive problem modeling or practicing recursion |