CountSemiprimes

Count the semiprime numbers in the given range [a..b]


Solution:

Solution to Codility's Count the semiprime problem which is from the Codility Lesson 11: Sieve of Eratosthenes and, is solved in Java 8 with 100% performance and correctness scores. The goal here is to count the semiprime numbers in the given range [a..b] You can find the question of this CountSemiprimes problem in the Codility website.


package Codility.Lesson11;

import java.util.*;

public class CountSemiprimes {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int n = 26;
		int[] a1 = { 1, 4, 16 };
		int[] a2 = { 26, 10, 20 };
		int result2[] = solution(n, a1, a2);
		System.out.println(Arrays.toString(result2));
	}

	public static int[] solution(int N, int[] P, int[] Q) {

		// main idea:
		// using "sieve of Eratosthenes"
		// https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

		boolean[] primeArray = new boolean[N + 1]; // note: plus one for "0"

		// initial settting (sieve of Eratosthenes)
		Arrays.fill(primeArray, true); // initial setting: all primes
		primeArray[0] = false; // not prime
		primeArray[1] = false; // not prime
		int sqrtN = (int) Math.sqrt(N);
		// sieve of Eratosthenes
		for (int i = 2; i < sqrtN; i++) {
			int j = i + i;
			for (j = j; j <= N; j = j + i) {
				primeArray[j] = false; // not prime
			}
		}

		// store all primes in "List"
		List<Integer> primeList = new ArrayList<>();
		for (int i = 2; i <= N; i++) {
			if (primeArray[i] == true) {
				primeList.add(i); // "i" is prime
			}
		}

		// find "semiprimes"
		boolean[] semiprimeArray = new boolean[N + 1]; // note: plus one for "0"
		Arrays.fill(semiprimeArray, false); // initial setting: all "not" semiprimes
		long semiprimeTemp; // using "long" (be careful)
		// for "primeList.size()"
		for (int i = 0; i < primeList.size(); i++) {
			for (int j = i; j < primeList.size(); j++) {
				semiprimeTemp = (long) primeList.get(i) * (long) primeList.get(j); // semiprimes
				if (semiprimeTemp > N) {
					break;
				} else {
					semiprimeArray[(int) semiprimeTemp] = true; // semiprimes
				}
			}
		}

		// compute "cumulative Count of semiprimes"
		int[] semiprimeCumulateCount = new int[N + 1]; // note: plus one for "0"
		for (int i = 1; i <= N; i++) {
			semiprimeCumulateCount[i] = semiprimeCumulateCount[i - 1]; // cumulative
			if (semiprimeArray[i] == true) {
				semiprimeCumulateCount[i]++; // semiprimes
			}
		}

		// compute "results" (for each query)
		int numQuery = Q.length;
		int[] result = new int[numQuery];
		for (int i = 0; i < numQuery; i++) {
			result[i] = semiprimeCumulateCount[Q[i]] - semiprimeCumulateCount[P[i] - 1]; // note: "P[i]-1" (not
																							// included)
		}
		return result;
	}
}

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