# Square free semiprimes in a given range using C++ STL

Given two integers L and R (L < = R). The task is to find all square free semiprimes in the range L to R (both inclusive).

Examples:

Input : L = 1, R = 10
Output : 2
4, 6, 9, 10 are semi primes. But 6, 10 are square free semi primes.

Input : L = 10, R = 20
Output : 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisites: Sieve of Eratosthenes, upper and lower bound

Understanding :
Semiprimes are numbers of the form where p and q are primes, not necessarily distinct. All semiprime has only 4 factors where p and q are the only two prime factors and .

Naive Approach:
Precompute all prime number upto . Find all combinations of two prime p and q such that is between L and R. Iterating through all combinations of prime would give a time complexity of . This solution, however, will not work for large L and R values.

Time Complexity: O(N^2)

Efficient Approach:
Precompute all prime number up to . We can divide the problem of finding two primes p and q into a simpler form.
As we can say that . Similarly as we can say that .
Now the problem is reduced to finding the count of q such that for all p.

Here, we can use binary search for finding upper_bound of from list of prime numbers and subtract it from index of lower_bound of from list of prime numbers to find the count of all q in range to for the given p. Repeating the above step for all prime p will give the answer for given range L to R

Below is the implementation of the above approach :

 // CPP program to find square free semiprimes in the range  #include  using namespace std;     #define N 78498     void Sieve(int pre[])  {      // Max size of prime array      int MX = 1e6;         // Array of prime of size 1000001.      // will be marked true for prime, false otherwise      // i     = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...]      // prime = {0, 0, 1, 1, 0, 1, 0, 1, 0, 0, ...}      bool prime[MX + 1];         // i -> keeps track of index of prime       // (current integer)      // idx -> keeps track of index of pre      int i = 2;      int idx = 0;         // Initialize all entries to true      memset(prime, true, sizeof(prime));         // For all i from 2 to MX iterate      for (i = 2; i <= MX; i++)      {          // If i is prime          if (prime[i])           {              // Set idx'th index of pre as i and               // increment idx after setting the value              pre[idx++] = i;                 // For all multiple of i from 2*i to MX              // mark them as false i.e. not prime              for (int j = (i << 1); j <= MX; j += i)                  prime[j] = false;          }      }  }     // Function to find square free semi primes in the range  int semiPrimeCount(int L, int R)  {         // Array of prime integer      // will contain first 78498 primes      // idx = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...]      // pre = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...}      int pre[N];         // Prepare pre array      Sieve(pre);         // Result of current query      int res = 0;         // Offset index from start      int j = 1;         // For all prime integers p in pre       // from 2 to sqrt(R) iterate      for (auto p : pre) {          // ub_num     = count of all multiple of p less           //             than R => p = R/p          //          // ub     = iterator of first element greater           //         than ub_num in pre          //          int ub_num = R / p;          auto ub = upper_bound(pre + j, pre + N, ub_num);             // lb_num     = count of all multiple of p           //             less than L => p = L/p          // If L is divisible by p increment p by           // 1 => p = p+1 = p+(L%p>0)          //          // lb     = iterator of first element greater           //         than or equal lb_num in pre                 int lb_num = (L / p) + ((L % p) > 0);          auto lb = lower_bound(pre + j, pre + N, lb_num);             // Add the difference between ub and lb          res += ub - lb;             // Increment j          j++;             // If prime p is greater than or equal to sqrt(R)          if (p * p >= R)              break;      }      return res;  }     // Driver code  int main()  {      int L = 10, R = 20;             // Function call      cout << semiPrimeCount(L, R);         return 0;  }

Output:

3


Time Complexity: O(N*logN)

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