Sphenic Number

Last Updated : 11 Aug, 2026

Given a positive integer n, return true if n is a Sphenic Number; otherwise, return false. A number is called a Sphenic Number if it is the product of exactly three distinct prime numbers.

Examples: 

Input: n = 30
Output: true
Explanation: 30 = 2 * 3 * 5 so n is product of 3 distinct prime numbers.

Input: n = 60
Output: false
Explanation: 60 = 2 * 2 * 3 * 5 so n is product of 4 prime numbers.

Try It Yourself
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Table of Content

[Naive Approach] Check Every Number as a Prime Factor - O(n√n) Time and O(1) Space

The idea is to check every number from 2 to n and identify the prime numbers that divide n. Count the prime factors and make sure each one occurs only once. If exactly three distinct prime factors are found, then n is a Sphenic number.

Working of Approach:

  • Traverse all numbers from 2 to n.
  • For every divisor, check whether it is prime.
  • If it is prime, count it and divide it from n.
  • If the same prime divides n again, return false.
  • Finally, return true if exactly three distinct prime factors are found.
C++ 
#include <iostream>
using namespace std;

// Check whether a number is prime.
bool isPrime(int x)
{
    if (x < 2)
        return false;

    // Check all possible divisors.
    for (int i = 2; i * i <= x; i++)
    {
        if (x % i == 0)
            return false;
    }

    return true;
}

bool isSphenicNo(int n)
{
    int cnt = 0;

    // Check every number as a possible prime factor.
    for (int i = 2; i <= n; i++)
    {
        if (n % i == 0 && isPrime(i))
        {
            cnt++;
            n /= i;

            // A repeated prime factor makes it non-sphenic.
            if (n % i == 0)
                return false;

            // More than three distinct prime factors.
            if (cnt > 3)
                return false;
        }
    }

    // A Sphenic number has exactly three distinct prime factors.
    return cnt == 3;
}

int main()
{

    int n = 30;

    cout << boolalpha << isSphenicNo(n) << endl;

    return 0;
}
Java 
public class GFG {
    // Check whether a number is prime.
    public static boolean isPrime(int x)
    {
        if (x < 2)
            return false;

        // Check all possible divisors.
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0)
                return false;
        }

        return true;
    }

    public static boolean isSphenicNo(int n)
    {
        int cnt = 0;

        // Check every number as a possible prime factor.
        for (int i = 2; i <= n; i++) {
            if (n % i == 0 && isPrime(i)) {
                cnt++;
                n /= i;

                // A repeated prime factor makes it
                // non-sphenic.
                if (n % i == 0)
                    return false;

                // More than three distinct prime factors.
                if (cnt > 3)
                    return false;
            }
        }

        // A Sphenic number has exactly three distinct prime
        // factors.
        return cnt == 3;
    }

    public static void main(String[] args)
    {
        int n = 30;

        System.out.println(isSphenicNo(n));
    }
}
Python 
def isPrime(x):
    if x < 2:
        return False

    # Check all possible divisors.
    for i in range(2, int(x**0.5) + 1):
        if x % i == 0:
            return False

    return True

def isSphenicNo(n):
    cnt = 0

    # Check every number as a possible prime factor.
    for i in range(2, n + 1):
        if n % i == 0 and isPrime(i):
            cnt += 1
            n //= i

            # A repeated prime factor makes it non-sphenic.
            if n % i == 0:
                return False

            # More than three distinct prime factors.
            if cnt > 3:
                return False

    # A Sphenic number has exactly three distinct prime factors.
    return cnt == 3

if __name__ == "__main__":
    n = 30

    print(isSphenicNo(n))
C# 
using System;

public class GFG {
    // Check whether a number is prime.
    public static bool isPrime(int x)
    {
        if (x < 2)
            return false;

        // Check all possible divisors.
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0)
                return false;
        }

        return true;
    }

    public static bool isSphenicNo(int n)
    {
        int cnt = 0;

        // Check every number as a possible prime factor.
        for (int i = 2; i <= n; i++) {
            if (n % i == 0 && isPrime(i)) {
                cnt++;
                n /= i;

                // A repeated prime factor makes it
                // non-sphenic.
                if (n % i == 0)
                    return false;

                // More than three distinct prime factors.
                if (cnt > 3)
                    return false;
            }
        }

        // A Sphenic number has exactly three distinct prime
        // factors.
        return cnt == 3;
    }

    public static void Main()
    {
        int n = 30;

        Console.WriteLine(isSphenicNo(n));
    }
}
JavaScript 
function isPrime(x)
{
    if (x < 2)
        return false;

    // Check all possible divisors.
    for (let i = 2; i * i <= x; i++) {
        if (x % i === 0)
            return false;
    }

    return true;
}

function isSphenicNo(n)
{
    let cnt = 0;

    // Check every number as a possible prime factor.
    for (let i = 2; i <= n; i++) {
        if (n % i === 0 && isPrime(i)) {
            cnt++;
            n /= i;

            // A repeated prime factor makes it non-sphenic.
            if (n % i === 0)
                return false;

            // More than three distinct prime factors.
            if (cnt > 3)
                return false;
        }
    }

    // A Sphenic number has exactly three distinct prime
    // factors.
    return cnt === 3;
}

// Driver Code
let n = 30;
console.log(isSphenicNo(n));

Output
true

[Expected Approach] Prime Factorization Using Trial Division - O(√n) Time and O(1) Space

The idea is to factorize n using trial division up to √n, count its distinct prime factors, and immediately return false if any factor occurs more than once. If exactly three distinct prime factors are found, n is a Sphenic number.

Working of Approach:

  • Start checking factors from 2 up to √n.
  • If i divides n, count it as a distinct prime factor.
  • Divide i from n and check whether it divides n again.
  • If it occurs again, return false.
  • Count any remaining prime factor and check whether the count is exactly 3.

Let us understand with an example:
Input: n = 30

  • Initially, cnt = 0 and n = 30.
  • i = 2: 30 % 2 == 0, so cnt = 1 and n = 15; 2 does not repeat.
  • i = 3: 15 % 3 == 0, so cnt = 2 and n = 5; 3 does not repeat.
  • The loop stops because 4 * 4 > 5; since n > 1, count 5 as the third prime factor, so cnt = 3.
  • Finally, cnt == 3, so the function returns true.
C++ 
#include <iostream>
using namespace std;

bool isSphenicNo(int n)
{
    int cnt = 0;

    // Count distinct prime factors and ensure none repeats.
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            cnt++;
            n /= i;

            // A repeated prime factor makes the number non-sphenic.
            if (n % i == 0)
            {
                return false;
            }
        }
    }

    // Count the remaining prime factor if present.
    if (n > 1)
    {
        cnt++;
    }

    return cnt == 3;
}

int main()
{
    int n = 30;

    cout << boolalpha << isSphenicNo(n) << endl;

    return 0;
}
Java 
import java.util.Scanner;

public class GFG {
    public static boolean isSphenicNo(int n)
    {
        int cnt = 0;

        // Count distinct prime factors and ensure none
        // repeats.
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                cnt++;
                n /= i;

                // A repeated prime factor makes the number
                // non-sphenic.
                if (n % i == 0) {
                    return false;
                }
            }
        }

        // Count the remaining prime factor if present.
        if (n > 1) {
            cnt++;
        }

        return cnt == 3;
    }

    public static void main(String[] args)
    {
        int n = 30;

        System.out.println(isSphenicNo(n));
    }
}
Python 
def isSphenicNo(n):
    cnt = 0

    # Count distinct prime factors and ensure none repeats.
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            cnt += 1
            n //= i

            # A repeated prime factor makes the number non-sphenic.
            if n % i == 0:
                return False

    # Count the remaining prime factor if present.
    if n > 1:
        cnt += 1

    return cnt == 3

if __name__ == "__main__":
    n = 30

    print(isSphenicNo(n))
C# 
using System;

public class GFG {
    public static bool isSphenicNo(int n)
    {
        int cnt = 0;

        // Count distinct prime factors and ensure none
        // repeats.
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                cnt++;
                n /= i;

                // A repeated prime factor makes the number
                // non-sphenic.
                if (n % i == 0) {
                    return false;
                }
            }
        }

        // Count the remaining prime factor if present.
        if (n > 1) {
            cnt++;
        }

        return cnt == 3;
    }

    public static void Main()
    {
        int n = 30;

        Console.WriteLine(isSphenicNo(n));
    }
}
JavaScript 
function isSphenicNo(n)
{
    let cnt = 0;

    // Count distinct prime factors and ensure none repeats.
    for (let i = 2; i * i <= n; i++) {
        if (n % i === 0) {
            cnt++;
            n /= i;

            // A repeated prime factor makes the number
            // non-sphenic.
            if (n % i === 0) {
                return false;
            }
        }
    }

    // Count the remaining prime factor if present.
    if (n > 1) {
        cnt++;
    }

    return cnt === 3;
}

// Driver Code
const n = 30;
console.log(isSphenicNo(n));

Output
true
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