Find the total number of composite factor for a given number

Given an integer N, the task is to find the total number of composite factors of N. Composite factors of a number are the factors which are not prime.

Examples:

Input: N = 24
Output: 5
1, 2, 3, 4, 6, 8, 12 and 24 are the factors of 24.
Out of which only 4, 6, 8, 12 and 24 are composites.

Input: N = 100
Output: 6

Approach:

  • Find all the factors of N and store it in a variable totalFactors
  • Find all the prime factors of N and store it in a variable primeFactors
  • Now, total composite factors will be totalFactors – primeFactors – 1 (1 is subtracted because 1 is neither prime nor composite).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of prime factors of n
int composite_factors(int n)
{
    int count = 0;
    int i, j;
  
    // Initialise array with 0
    int a[n + 1] = { 0 };
    for (i = 1; i <= n; ++i) {
        if (n % i == 0) {
  
            // Stored i value into an array
            a[i] = i;
        }
    }
  
    // Every non-zero value at a[i] denotes
    // that i is a factor of n
    for (i = 2; i <= n; i++) {
        j = 2;
        int p = 1;
  
        // Find if i is prime
        while (j < a[i]) {
            if (a[i] % j == 0) {
                p = 0;
                break;
            }
            j++;
        }
  
        // If i is a factor of n
        // and i is not prime
        if (p == 0 && a[i] != 0) {
            count++;
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    int n = 100;
  
    cout << composite_factors(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class Gfg 
  
// Function to return the count
// of prime factors of n
public static int composite_factors(int n)
{
    int count = 0;
    int i, j;
  
    // Initialise array with 0
    int[] a=new int[n+1];
    for( i = 0; i < n; i++)
    {
        a[i]=0;
    }
    for (i = 1; i <= n; ++i) 
    {
        if (n % i == 0)
        {
  
            // Stored i value into an array
            a[i] = i;
        }
    }
  
    // Every non-zero value at a[i] denotes
    // that i is a factor of n
    for (i = 2; i <= n; i++) 
    {
        j = 2;
        int p = 1;
  
        // Find if i is prime
        while (j < a[i]) 
        {
            if (a[i] % j == 0
            {
                p = 0;
                break;
            }
            j++;
        }
  
        // If i is a factor of n
        // and i is not prime
        if (p == 0 && a[i] != 0
        {
            count++;
        }
      
}
    return count;
}
  
  
// Driver code
public static void main(String[] args) 
    int n = 100;
      
    System.out.println(composite_factors(n));
  
}
}
  
// This code is contributed by nidhi16bcs2007

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Python3

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# Python3 implementation of the approach 
  
# Function to return the count 
# of prime factors of n 
def composite_factors(n) : 
  
    count = 0;
      
    # Initialise array with 0
    a = [0]*(n + 1) ;
      
    for i in range(1, n + 1) : 
        if (n % i == 0) :
  
            # Stored i value into an array 
            a[i] = i;
  
    # Every non-zero value at a[i] denotes 
    # that i is a factor of n 
    for i in range(2,n + 1) :
        j = 2
        p = 1
  
        # Find if i is prime 
        while (j < a[i]) :
            if (a[i] % j == 0) :
                p = 0
                break
                  
            j += 1
  
  
        # If i is a factor of n 
        # and i is not prime 
        if (p == 0 and a[i] != 0) :
            count += 1
  
    return count; 
  
  
# Driver code 
if __name__ == "__main__" :
  
    n = 100
  
    print(composite_factors(n)); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the count
// of prime factors of n
static int composite_factors(int n)
{
    int count = 0;
    int i, j;
  
    // Initialise array with 0
    int[] a = new int[n + 1];
    for( i = 0; i < n; i++)
    {
        a[i]=0;
    }
    for (i = 1; i <= n; ++i)
    {
        if (n % i == 0)
        {
  
            // Stored i value into an array
            a[i] = i;
        }
    }
  
    // Every non-zero value at a[i] denotes
    // that i is a factor of n
    for (i = 2; i <= n; i++)
    {
        j = 2;
        int p = 1;
  
        // Find if i is prime
        while (j < a[i])
        {
            if (a[i] % j == 0)
            {
                p = 0;
                break;
            }
            j+=1;
        }
  
        // If i is a factor of n
        // and i is not prime
        if (p == 0 && a[i] != 0)
        {
            count += 1;
        }
  
}
    return count;
}
  
  
// Driver code
public static void Main()
{
    int n = 100;
  
    Console.WriteLine(composite_factors(n));
}
}
  
// This code is contributed by mohit kumar 29

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Output:

6


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