Given a positive integer N, check if it can be expressed as a sum of two semi-primes or not.
Semi-primes A number is said to be a semi-prime if it can be expressed as product of two primes number ( not necessarily distinct ).
Semi-primes in the range 1 -100 are:
4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.
Examples:
Input : N = 30
Output: YES
Explanation: 30 can be expressed as '15 + 15'
15 is semi-primes as 15 is a product of two primes 3 and 5.
Input : N = 45
Output : YES
Explanation: 45 can be expressed as '35 + 10'
35 and 10 are also semi-primes as it can a be expressed
as product of two primes:
35 = 5 * 7
10 = 2 * 5
Prerequisite:
A Simple Solution is to traverse form i=1 to 
and check if i and N-i are semi-primes or not. If yes, print i and n-i.
An Efficient Solution is to pre-compute semi-primes in an array up to the given range and then traverse the semi-prime array and check if n-arr[i] is semi-prime or not. As, arr[i] is already a semi-prime if n-arr[i] is also a semi-prime, then n can be expressed as sum of two semi-primes.
Below is the implementation of above approach:
C++
// CPP Code to check if an integer // can be expressed as sum of // two semi-primes #include <bits/stdc++.h> using namespace std; #define MAX 1000000 vector<int> arr; bool sprime[MAX]; // Utility function to compute // semi-primes in a range void computeSemiPrime() { memset(sprime, false, sizeof(sprime)); for (int i = 2; i < MAX; i++) { int cnt = 0; int num = i; for (int j = 2; cnt < 2 && j * j <= num; ++j) { while (num % j == 0) { num /= j, ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2) { sprime[i] = true; arr.push_back(i); } } } // Utility function to check // if a number sum of two // semi-primes bool checkSemiPrime(int n) { int i = 0; while (arr[i] <= n / 2) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then we a number can be expressed as // sum of two semi-primes if (sprime[n - arr[i]]) { return true; } i++; } return false; } // Driver code int main() { computeSemiPrime(); int n = 30; if (checkSemiPrime(n)) cout << "YES"; else cout << "NO"; return 0; } |
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Java
// Java Code to check if an integer // can be expressed as sum of // two semi-primes import java.util.*; class GFG { static final int MAX = 1000000; static Vector<Integer> arr = new Vector<>(); static boolean[] sprime = new boolean[MAX]; // Utility function to compute // semi-primes in a range static void computeSemiPrime() { for (int i = 0; i < MAX; i++) sprime[i] = false; for (int i = 2; i < MAX; i++) { int cnt = 0; int num = i; for (int j = 2; cnt < 2 && j * j <= num; ++j) { while (num % j == 0) { num /= j; ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2) { sprime[i] = true; arr.add(i); } } } // Utility function to check // if a number is sum of two // semi-primes static boolean checkSemiPrime(int n) { int i = 0; while (arr.get(i) <= n / 2) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then a number can be expressed as // sum of two semi-primes if (sprime[n - arr.get(i)]) { return true; } i++; } return false; } // Driver code public static void main(String[] args) { computeSemiPrime(); int n = 30; if (checkSemiPrime(n)) System.out.println("YES"); else System.out.println("NO"); } } |
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Python3
# Python3 Code to check if an integer can # be expressed as sum of two semi-primes MAX = 10000 arr = [] sprime = [False] * (MAX) # Utility function to compute # semi-primes in a range def computeSemiPrime(): for i in range(2, MAX): cnt, num, j = 0, i, 2 while cnt < 2 and j * j <= num: while num % j == 0: num /= j # Increment count of prime numbers cnt += 1 j += 1 # If number is greater than 1, add it # to the count variable as it indicates # the number remain is prime number if num > 1: cnt += 1 # if count is equal to '2' then # number is semi-prime if cnt == 2: sprime[i] = True arr.append(i) # Utility function to check # if a number sum of two # semi-primes def checkSemiPrime(n): i = 0 while arr[i] <= n // 2: # arr[i] is already a semi-prime # if n-arr[i] is also a semi-prime # then a number can be expressed as # sum of two semi-primes if sprime[n - arr[i]] == True: return True i += 1 return False # Driver code if __name__ == "__main__": computeSemiPrime() n = 30 if checkSemiPrime(n) == True: print("YES") else: print("NO") # This code is contributed by # Rituraj Jain |
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C#
// C# Code to check if an integer // can be expressed as sum of // two semi-primes using System.Collections.Generic; class GFG { static int MAX = 1000000; static List<int> arr = new List<int>(); static bool[] sprime = new bool[MAX]; // Utility function to compute // semi-primes in a range static void computeSemiPrime() { for (int i = 0; i < MAX; i++) sprime[i] = false; for (int i = 2; i < MAX; i++) { int cnt = 0; int num = i; for (int j = 2; cnt < 2 && j * j <= num; ++j) { while (num % j == 0) { num /= j; ++cnt; // Increment count // of prime numbers } } // If number is greater than 1, add it to // the count variable as it indicates the // number remain is prime number if (num > 1) ++cnt; // if count is equal to '2' then // number is semi-prime if (cnt == 2) { sprime[i] = true; arr.Add(i); } } } // Utility function to check // if a number is sum of two // semi-primes static bool checkSemiPrime(int n) { int i = 0; while (arr[i] <= n / 2) { // arr[i] is already a semi-prime // if n-arr[i] is also a semi-prime // then a number can be expressed as // sum of two semi-primes if (sprime[n - arr[i]]) { return true; } i++; } return false; } // Driver code public static void Main() { computeSemiPrime(); int n = 30; if (checkSemiPrime(n)) System.Console.WriteLine("YES"); else System.Console.WriteLine("NO"); } } // This code is contributed by mits |
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Output:
YES
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