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++
// 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;
} |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of prime factors of n function composite_factors(n)
{ var count = 0;
var i, j;
// Initialise array with 0
var a = Array(n + 1).fill(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;
var 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 var n = 100;
document.write(composite_factors(n));
</script> |
Output:
6
Time Complexity: O(n*val) where n is the given number and val is the largest factor of n.
Auxiliary Space: O(n)