Check if a number has prime count of divisors

Given an integer N, the task is to check if the count of divisors of N is prime or not.

Examples:

Input: N = 13 
Output: Yes 
The divisor count is 2 (1 and 13) which is prime.

Input: N = 8 
Output: No 
The divisors are 1, 2, 4 and 8. 

Approach: Please read this article to find the count of divisors of a number. So find the maximum value of i for every prime divisor p such that N % (pⁱ) = 0. So the count of divisors gets multiplied by (i + 1). The count of divisors will be (i¹ + 1) * (i² + 1) * … * (i^k + 1). 
It can now be seen that there can only be one prime divisor for the maximum i and if N % pⁱ = 0 then (i + 1) should be prime. The primality can be checked in sqrt(n) time and the prime factors can also be found in sqrt(n) time. So the overall time complexity will be O(sqrt(n)).

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true
// if n is prime
bool Prime(int n)
{
    // There is no prime
    // less than 2
    if (n < 2)
        return false;

    // Run a loop from 2 to sqrt(n)
    for (int i = 2; i <= sqrt(n); i++)

        // If there is any factor
        if (n % i == 0)
            return false;

    return true;
}

// Function that returns true if n
// has a prime count of divisors
bool primeCountDivisors(int n)
{
    if (n < 2)
        return false;

    // Find the prime factors
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0) {

            // Find the maximum value of i for every
            // prime divisor p such that n % (p^i) == 0
            long a = n, c = 0;
            while (a % i == 0) {
                a /= i;
                c++;
            }

            // If c+1 is a prime number and a = 1
            if (a == 1 && Prime(c + 1))
                return true;

            // The number cannot have two factors
            // to have count of divisors prime
            else
                return false;
        }

    // Else the number is prime so
    // it has only two divisors
    return true;
}

// Driver code
int main()
{
    int n = 13;

    if (primeCountDivisors(n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function that returns true
    // if n is prime
    static boolean Prime(int n)
    {
        // There is no prime
        // less than 2
        if (n < 2)
            return false;

        // Run a loop from 2 to sqrt(n)
        for (int i = 2; i <= (int)Math.sqrt(n); i++)

            // If there is any factor
            if (n % i == 0)
                return false;
        return true;
    }

    // Function that returns true if n
    // has a prime count of divisors
    static boolean primeCountDivisors(int n)
    {
        if (n < 2)
            return false;

        // Find the prime factors
        for (int i = 2; i <= (int)Math.sqrt(n); i++)
            if (n % i == 0)
            {

                // Find the maximum value of i for every
                // prime divisor p such that n % (p^i) == 0
                long a = n, c = 0;
                while (a % i == 0)
                {
                    a /= i;
                    c++;
                }

                // If c+1 is a prime number and a = 1
                if (a == 1 && Prime((int)c + 1) == true)
                    return true;

                // The number cannot have two factors
                // to have count of divisors prime
                else
                    return false;
            }

        // Else the number is prime so
        // it has only two divisors
        return true;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 13;

        if (primeCountDivisors(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import sqrt

# Function that returns true
# if n is prime
def Prime(n) :

    # There is no prime
    # less than 2
    if (n < 2) :
        return False;

    # Run a loop from 2 to sqrt(n)
    for i in range(2, int(sqrt(n)) + 1) :

        # If there is any factor
        if (n % i == 0) :
            return False;

    return True;

# Function that returns true if n
# has a prime count of divisors
def primeCountDivisors(n) :

    if (n < 2) :
        return False;

    # Find the prime factors
    for i in range(2, int(sqrt(n)) + 1) :
        if (n % i == 0) :

            # Find the maximum value of i for every
            # prime divisor p such that n % (p^i) == 0
            a = n; c = 0;
            while (a % i == 0) :
                a //= i;
                c += 1;

            # If c + 1 is a prime number and a = 1
            if (a == 1 and Prime(c + 1)) :
                return True;

            # The number cannot have two factors
            # to have count of divisors prime
            else :
                return False;

    # Else the number is prime so
    # it has only two divisors
    return True;

# Driver code
if __name__ == "__main__" :

    n = 13;

    if (primeCountDivisors(n)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function that returns true
    // if n is prime
    static bool Prime(int n)
    {

        // There is no prime
        // less than 2
        if (n < 2)
            return false;

        // Run a loop from 2 to sqrt(n)
        for (int i = 2; i <= (int)Math.Sqrt(n); i++)

            // If there is any factor
            if (n % i == 0)
                return false;
        return true;
    }

    // Function that returns true if n
    // has a prime count of divisors
    static bool primeCountDivisors(int n)
    {
        if (n < 2)
            return false;

        // Find the prime factors
        for (int i = 2; i <= (int)Math.Sqrt(n); i++)
            if (n % i == 0)
            {

                // Find the maximum value of i for every
                // prime divisor p such that n % (p^i) == 0
                long a = n, c = 0;
                while (a % i == 0)
                {
                    a /= i;
                    c++;
                }

                // If c+1 is a prime number and a = 1
                if (a == 1 && Prime((int)c + 1) == true)
                    return true;

                // The number cannot have two factors
                // to have count of divisors prime
                else
                    return false;
            }

        // Else the number is prime so
        // it has only two divisors
        return true;
    }

    // Driver code
    public static void Main()
    {
        int n = 13;

        if (primeCountDivisors(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of the approach

// Function that returns true
// if n is prime
function Prime(n)
{

    // There is no prime
    // less than 2
    if (n < 2)
        return false;

    // Run a loop from 2 to sqrt(n)
    for(var i = 2; i <= Math.sqrt(n); i++)

        // If there is any factor
        if (n % i == 0)
            return false;

    return true;
}

// Function that returns true if n
// has a prime count of divisors
function primeCountDivisors( n)
{
    if (n < 2)
        return false;

    // Find the prime factors
    for(var i = 2; i <= Math.sqrt(n); i++)
        if (n % i == 0)
        {

            // Find the maximum value of i for every
            // prime divisor p such that n % (p^i) == 0
            var a = n, c = 0;

            while (a % i == 0)
            {
                a /= i;
                c++;
            }

            // If c+1 is a prime number and a = 1
            if (a == 1 && Prime(c + 1))
                return true;

            // The number cannot have two factors
            // to have count of divisors prime
            else
                return false;
        }

    // Else the number is prime so
    // it has only two divisors
    return true;
}

// Driver rcode
n = 13;

if (primeCountDivisors(n))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by SoumikMondal

</script>

Output:

Yes

Time Complexity: O(sqrt(n)), as we are using a loop to traverse sqrt (n) times. Where n is the integer given as input.

Auxiliary Space: O(1), as we are not using any extra space.