# Java Challenge - Semi-Prime Numbers

This community-built FAQ covers the “Semi-Prime Numbers” code challenge in Java. You can find that challenge here, or pick any challenge you like from our list.

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I made separate functions to check whether a number is prime, find the next prime after a number, and to check whether a number is semi-prime (by checking if it has 2 prime factors using part of the prime factorization algorithm).
I only checked whether number is prime by checking if an integer up to the square root of the number is a factor.
My code is not the most efficient at finding prime numbers.

my code
``````public class SemiPrimeNumbers {
public static void main(String[] args) {
System.out.println(semiPrimeCount(10));
}

public static boolean isPrime(int x) {
if (x < 0) {
x = x * -1;
}
if (x < 2) {
return false;
}
else if (x < 4) {
return true;
}
double sqrt = Math.ceil(Math.sqrt(x));
for (int i = 2; i <= sqrt; i++) {
if ((x % i) == 0) {
return false;
}
}
return true;
}

public static int getNextPrime(int p) {
int j = p + 1;
while (!isPrime(j)) {
j++;
}
return j;
}

public static boolean isSemiPrime(int x) {
if (x < 0) {
x = x * -1;
}
if (x < 4) {
return false;
}
int primeFactorsCount = 0;
int p = 2;
while (x >= 2) {
// if prime p is a factor, factor p out of x, and increment counter
while ((x % p) == 0) {
x = x / p;
primeFactorsCount++;
if (primeFactorsCount > 2) {
return false;
}
}
p = getNextPrime(p);
}
return (primeFactorsCount == 2);
}

public static int semiPrimeCount(int n) {
int count = 0;
if (n < 4) {
return 0;
}
for (int i = 4; i < n; i++) {
if (isSemiPrime(i)) {
count++;
}
}
return count;
}
}
``````
1 Like