# Count of the non-prime divisors of a given number

Given a number N, the task is to find the count of non-prime divisors of the given number N.

Examples:

Input: N = 8
Output:
Explanation:
Divisors of 8 are – {1, 2, 4, 8}
Non-Prime Divisors – {1, 4, 8}

Input: N = 20
Output:
Explanation:
Divisors of 20 are – {1, 2, 4, 5, 10, 20}
Non-Prime Divisors – {1, 4, 10, 20}

Approach: The key observation in the problem is that any number can be written as a product of its prime factors as

where K is the count of the prime factors of the given number. Using permutation and combination we can find that the total count of the
factors that are

Therefore, the count of the non-prime factors will be:

Below is the implementation of the above approach:

## C++

 // C++ program to find count of // non-prime divisors of given number   #include using namespace std;   // Function to factors of the given // number vector getFactorization(int x) {     int count = 0;     vector v;       // Loop to find the divisors of     // the number 2     while (x % 2 == 0) {         count++;         x = x / 2;     }     if (count != 0)         v.push_back(count);       // Loop to find the divisors of the     // given number upto SQRT(N)     for (int i = 3; i <= sqrt(x); i += 2) {         count = 0;         while (x % i == 0) {             count++;             x /= i;         }         if (count != 0)             v.push_back(count);     }       // Condition to check if the rest     // number is also a prime number     if (x > 1) {         v.push_back(1);     }     return v; }   // Function to find the non-prime // divisors of the given number int nonPrimeDivisors(int N) {     vector v = getFactorization(N);     int ret = 1;       // Loop to count the number of     // the total divisors of given number     for (int i = 0; i < v.size(); i++)         ret = ret * (v[i] + 1);       ret = ret - v.size();     return ret; }   // Driver Code int main() {     int N = 8;       // Function Call     cout << nonPrimeDivisors(N) << endl;     return 0; }

## Java

 // Java program to find // count of non-prime // divisors of given number import java.util.*; class GFG{   // Function to factors // of the given number static Vector getFactorization(int x) {   int count = 0;   Vector v = new Vector<>();     // Loop to find the   // divisors of the number 2   while (x % 2 == 0)   {     count++;     x = x / 2;   }       if (count != 0)     v.add(count);     // Loop to find the divisors   // of the given number upto SQRT(N)   for (int i = 3;            i <= Math.sqrt(x); i += 2)   {     count = 0;     while (x % i == 0)     {       count++;       x /= i;     }           if (count != 0)       v.add(count);   }     // Condition to check if   // the rest number is also   // a prime number   if (x > 1)   {     v.add(1);   }   return v; }   // Function to find the non-prime // divisors of the given number static int nonPrimeDivisors(int N) {   Vector v = getFactorization(N);   int ret = 1;     // Loop to count the number of   // the total divisors of given number   for (int i = 0; i < v.size(); i++)     ret = ret * (v.get(i) + 1);     ret = ret - v.size();   return ret; }   // Driver Code public static void main(String[] args) {   int N = 8;     // Function Call   System.out.println(nonPrimeDivisors(N)); } }   // This code is contributed by shikhasingrajput

## Python3

 # Python3 program to find count of # non-prime divisors of given number from math import sqrt   # Function to factors of the given # number def getFactorization(x):           count = 0     v = []       # Loop to find the divisors of     # the number 2     while (x % 2 == 0):         count += 1         x = x // 2       if (count != 0):         v.append(count)       # Loop to find the divisors of the     # given number upto SQRT(N)     for i in range(3, int(sqrt(x)) + 12):         count = 0                   while (x % i == 0):             count += 1             x //= i                       if (count != 0):             v.append(count)       # Condition to check if the rest     # number is also a prime number     if (x > 1):         v.append(1)               return v   # Function to find the non-prime # divisors of the given number def nonPrimeDivisors(N):           v = getFactorization(N)     ret = 1       # Loop to count the number of     # the total divisors of given number     for i in range(len(v)):         ret = ret * (v[i] + 1)     ret = ret - len(v)           return ret   # Driver Code if __name__ == '__main__':           N = 8       # Function Call     print(nonPrimeDivisors(N))   # This code is contributed by Samarth

## C#

 // C# program to find // count of non-prime // divisors of given number using System; using System.Collections.Generic; class GFG{   // Function to factors // of the given number static List getFactorization(int x) {   int count = 0;   List v = new List();     // Loop to find the   // divisors of the number 2   while (x % 2 == 0)   {     count++;     x = x / 2;   }       if (count != 0)     v.Add(count);     // Loop to find the divisors   // of the given number upto   // SQRT(N)   for (int i = 3;            i <= Math.Sqrt(x); i += 2)   {     count = 0;     while (x % i == 0)     {       count++;       x /= i;     }           if (count != 0)       v.Add(count);   }     // Condition to check if   // the rest number is also   // a prime number   if (x > 1)   {     v.Add(1);   }   return v; }   // Function to find the non-prime // divisors of the given number static int nonPrimeDivisors(int N) {   List v = getFactorization(N);   int ret = 1;     // Loop to count the number of   // the total divisors of given number   for (int i = 0; i < v.Count; i++)     ret = ret * (v[i] + 1);     ret = ret - v.Count;   return ret; }   // Driver Code public static void Main(String[] args) {   int N = 8;     // Function Call   Console.WriteLine(nonPrimeDivisors(N)); } }   // This code is contributed by gauravrajput1

## Javascript



Output

3

### Approach 2:

One approach is to check all the divisors of the given number and count the divisors that are not prime.

Algorithm:

Initialize a count variable to 0.
For each number i from 1 to N, check if i is a divisor of N.
If i is a divisor of N, check if it is prime.
If i is not prime, increment the count variable.
Return the count variable.

## C++

 #include #include // needed for sqrt() function   using namespace std;   // function to check if a number is prime bool is_prime(int n) {     if (n <= 1) { // 1 and numbers less than 1 are not prime         return false;     }     for (int i = 2; i <= sqrt(n); i++) { // check divisibility from 2 to the square root of n         if (n % i == 0) { // if n is divisible by i, then n is not a prime number             return false;         }     }     return true; // if no factors are found, then n is a prime number }   // function to count the number of non-prime divisors of a given integer int count_non_prime_divisors(int N) {     int count = 0;     for (int i = 1; i <= N; i++) {         if (N % i == 0) { // if i is a divisor of N             if (!is_prime(i)) { // if i is not a prime number, increment the count                 count += 1;             }         }     }     return count; }   // main function to test the count_non_prime_divisors function int main() {     int N = 20;     cout << "Number of non-prime divisors of " << N << " is " << count_non_prime_divisors(N) << endl;     return 0; }

## Java

 import java.util.*;   public class Main {     // helper method to check if an integer is prime   public static boolean isPrime(int n)   {           // check if n is less than or equal to 1, which is not a prime number     if (n <= 1) {       return false;     }           // check if n is divisible by any integer from 2 to the square root of n     for (int i = 2; i <= Math.sqrt(n); i++) {       if (n % i == 0) { // if it is divisible, then n is not prime         return false;       }     }           // if n is not divisible by any integer from 2     // to the square root of n, then n is prime     return true;   }     // helper method to count the number of non-prime divisors of an integer N   public static int countNonPrimeDivisors(int N)   {           int count = 0; // initialize the count of non-prime divisors to 0     // loop through all integers from 1 to N     for (int i = 1; i <= N; i++) {       if (N % i == 0) { // if i is a divisor of N         if (!isPrime(i)) { // if i is not prime           count += 1; // increment the count of non-prime divisors         }       }     }           // return the count of non-prime divisors of N     return count;   }     // main method to test the countNonPrimeDivisors method   public static void main(String[] args)   {     int N = 20; // test value of N     // print the number of non-prime divisors of     // N using the countNonPrimeDivisors method     System.out.println("Number of non-prime divisors of " + N + " is " + countNonPrimeDivisors(N));   } }

## Python3

 def count_non_prime_divisors(N):     count = 0     for i in range(1, N+1):         if N % i == 0:             if is_prime(i) == False:                 count += 1     return count   def is_prime(n):     if n <= 1:         return False     for i in range(2, int(n**0.5)+1):         if n % i == 0:             return False     return True   N = 20 print("Number of non-prime divisors of", N, "is", count_non_prime_divisors(N))

## C#

 using System;   class GFG {     // Function to check if a number is prime     static bool 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;     }       // Function to count the number of non-prime divisors   // of a given integer     static int CountNonPrimeDivisors(int N) {         int count = 0;         for (int i = 1; i <= N; i++) {             if (N % i == 0) {                 if (!IsPrime(i)) {                     count += 1;                 }             }         }         return count;     }       // Main function to test the CountNonPrimeDivisors function     static void Main() {         int N = 20;         Console.WriteLine("Number of non-prime divisors of " + N + " is " + CountNonPrimeDivisors(N));     } }

## Javascript

 // function to check if a number is prime function isPrime(n) {     if (n <= 1) { // 1 and numbers less than 1 are not prime         return false;     }     for (let i = 2; i <= Math.sqrt(n); i++) { // check divisibility from 2 to the square root of n         if (n % i === 0) { // if n is divisible by i, then n is not a prime number             return false;         }     }     return true; // if no factors are found, then n is a prime number }   // function to count the number of non-prime divisors of a given integer function countNonPrimeDivisors(N) {     let count = 0;     for (let i = 1; i <= N; i++) {         if (N % i === 0) { // if i is a divisor of N             if (!isPrime(i)) { // if i is not a prime number, increment the count                 count += 1;             }         }     }     return count; }   // main function to test the countNonPrimeDivisors function       let N = 20;     console.log(Number of non-prime divisors of ${N} is${countNonPrimeDivisors(N)});

Output

Number of non-prime divisors of 20 is 4

Time Complexity: O(N * sqrt(N)), as we need to check all the divisors of N and for each divisor, we check if it is prime, which takes O(sqrt(N)) time.
Auxiliary Space: O(1), as we are not using any extra space.

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