LeetCode 2507: Smallest Value After Replacing With Sum of Prime Factors Solution

LeetCode 2507 Solution Explanation

Explanation

To solve this problem, we can repeatedly find the prime factors of the given number n, calculate their sum, and update n with this sum until n becomes a prime number. We can achieve this by iterating over numbers starting from 2 (the smallest prime number) and dividing n by the current number as long as it is a factor. If the current number is a factor, we add it to the sum and update n with the quotient. We repeat this process until n becomes a prime number.

Time Complexity

The time complexity of this solution is O(sqrt(n) log(log(n))) on average, where n is the given number.

Space Complexity

The space complexity of this solution is O(1) as we are using a constant amount of extra space.

LeetCode 2507 Solutions in Java, C++, Python

class Solution {
    public int smallestValue(int n) {
        while (!isPrime(n)) {
            int sum = 0;
            for (int i = 2; i <= Math.sqrt(n); i++) {
                while (n % i == 0) {
                    sum += i;
                    n /= i;
                }
            }
            if (n > 1) {
                sum += n;
            }
            n = sum;
        }
        return n;
    }

    private boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) return false;
        }
        return true;
    }
}

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