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2507. Smallest Value After Replacing With Sum of Prime Factors

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.

 

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

Solutions

  • 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;
        }
    }
}

  • class Solution {
public:
    int smallestValue(int n) {
        while (1) {
            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;
        }
    }
};

  • 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


  • func smallestValue(n int) int {
	for {
		t, s := n, 0
		for i := 2; i <= n/i; i++ {
			for n%i == 0 {
				s += i
				n /= i
			}
		}
		if n > 1 {
			s += n
		}
		if s == t {
			return s
		}
		n = s
	}
}

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