LeetCode 2507: Smallest Value After Replacing With Sum of Prime Factors Solution
Master LeetCode problem 2507 (Smallest Value After Replacing With Sum of Prime Factors), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2507. Smallest Value After Replacing With Sum of Prime Factors
Problem 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.
Solution Code
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;
}
}Try It Yourself
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What is the time complexity of LeetCode 2507 (Smallest Value After Replacing With Sum of Prime Factors)?
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