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: 3 
Explanation:
Divisors of 8 are – {1, 2, 4, 8} 
Non-Prime Divisors – {1, 4, 8}

Input: N = 20 
Output: 4 
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 <bits/stdc++.h>
using namespace std;

// Function to factors of the given
// number
vector<int> getFactorization(int x)
{
    int count = 0;
    vector<int> 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<int> 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<Integer> getFactorization(int x)
{
  int count = 0;
  Vector<Integer> 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<Integer> 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<int> getFactorization(int x)
{
  int count = 0;
  List<int> v = new List<int>();

  // 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<int> 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

<script>

// Javascript program to find
// count of non-prime
// divisors of given number

// Function to factors
// of the given number
function getFactorization(x)
{
      let count = 0;
      let v = [];

      // Loop to find the
      // divisors of the number 2
      while (x % 2 == 0)
      {
        count++;
        x = Math.floor(x / 2);
      }

      if (count != 0)
        v.push(count);

      // Loop to find the divisors
      // of the given number upto
      // SQRT(N)
      for (let i = 3;
        i <= Math.floor(Math.sqrt(x)); i += 2)
      {
        count = 0;
        while (x % i == 0)
        {
              count++;
              x = Math.floor(x / i);
        }

        if (count != 0)
          v.push(count);
      }

      // Condition to check if
      // the rest number is also
      // a prime number
      if (x > 1)
      {
        v.push(1);
      }
      return v;
}

// Function to find the non-prime
// divisors of the given number
function nonPrimeDivisors(N)
{
      let v = getFactorization(N);
      let ret = 1;

      // Loop to count the number of
      // the total divisors of given number
      for (let i = 0; i < v.length; i++)
        ret = ret * (v[i] + 1);

      ret = ret - v.length;
      return ret;
}

// Driver Code

    let N = 8;

  // Function Call
  document.write(nonPrimeDivisors(N));

</script>

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 <iostream>
#include <cmath> // 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.