2507. Smallest Value After Replacing With Sum of Prime Factors

Problem

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

Solution (Java)

class Solution {
    public int smallestValue(int n) {
        if(isPrime(n)) return n; // otherwise it will run forever
        int sum = getPrimeFactorSum(n);
        if(sum == n) return n;  // otherwise it will run forever
        return smallestValue(sum);
    }

    public boolean isPrime(int n) { // to check if number is prime
        if(n == 2) return true;
        for(int i = 2; i < Math.sqrt(n) + 1; i++) {
            if(n % i == 0) return false;
        }
        return true;
    }

    public int getFirstPrimeFactor(int n) { // to get first prime number
        if(isPrime(n)) return n;
        for(int i = 2; i < n; i++) {
            if(n % i == 0) return i;
        }
        return n;
    }

    public int getPrimeFactorSum(int n) { //sum of prime factors of a number
        int sum = 0;
        while(!isPrime(n)) {
            int m = getFirstPrimeFactor(n);
            n /= m;
            sum += m;
        }
        sum += n;
        return sum;
    }

}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).