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++ 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 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 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# 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 |
<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.
#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;
} |
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));
}
} |
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))
|
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));
}
} |
// 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.