LeetCode #2507 — MEDIUM

Smallest Value After Replacing With Sum of Prime Factors

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

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

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

15

Example 2

3

Related Problems

  • Happy Number (happy-number)
  • 2 Keys Keyboard (2-keys-keyboard)
  • Count Ways to Make Array With Product (count-ways-to-make-array-with-product)
  • Distinct Prime Factors of Product of Array (distinct-prime-factors-of-product-of-array)
  • Minimum Division Operations to Make Array Non Decreasing (minimum-division-operations-to-make-array-non-decreasing)
Step 02

Core Insight

What unlocks the optimal approach

  • Every time you replace n, it will become smaller until it is a prime number, where it will keep the same value each time you replace it.
  • n decreases logarithmically, allowing you to simulate the process.
  • To find the prime factors, iterate through all numbers less than n from least to greatest and find the maximum number of times each number divides n.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2507: Smallest Value After Replacing With Sum of Prime Factors
class Solution {
    public int smallestValue(int n) {
        while (true) {
            int t = n, s = 0;
            for (int i = 2; i <= n / i; ++i) {
                while (n % i == 0) {
                    s += i;
                    n /= i;
                }
            }
            if (n > 1) {
                s += n;
            }
            if (s == t) {
                return s;
            }
            n = s;
        }
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

ITERATIVE
O(n) time
O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT
O(log n) time
O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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