# Sphenic Number

A Sphenic Number is a positive integer n which is product of exactly three distinct primes. The first few sphenic numbers are 30, 42, 66, 70, 78, 102, 105, 110, 114, …
Given a number n, determine whether it is a Sphenic Number or not.

Examples:

```Input : 30
Output : Yes
Explanation : 30 is the smallest Sphenic number,
30 = 2 × 3 × 5
the product of the smallest three primes

Input : 60
Output : No
Explanation : 60 = 22 x 3 x 5
has exactly 3 prime factors but
is not a sphenic number

```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Sphenic number can be checked by first generating the Least prime factor of numbers till n.
Then we could simply divide the number by its least prime factor and then that number by its least prime factor, and so on and then check if the number has exactly 3 distinct prime factors.

Below is the C++ implementation of the idea.

```// C++ program to check whether a number is a
// Sphenic number or not
#include<bits/stdc++.h>
using namespace std;

const int MAX = 1000;

// Create a vector to store least primes.
// Initialize all entries as 0.
vector<int> least_pf(MAX, 0);

// This function fills values in least_pf[].
// such that the value of least_pf[i] stores
// smallest prime factor of i.
// This function is based on sieve of Eratosthenes
void leastPrimeFactor(int n)
{
// Least prime factor for 1 is 1
least_pf[1] = 1;

// Store least prime factor for all other
// numbers.
for (int i = 2; i <= n; i++)
{
// least_pf[i] == 0 means i is prime
if (least_pf[i] == 0)
{
// Initializing prime number as its own
// least prime factor.
least_pf[i] = i;

// Mark 'i' as a divisor for all its
// multiples that are not already marked
for (int j = 2*i; j <= n; j += i)
if (least_pf[j] == 0)
least_pf[j] = i;
}
}
}

// Function returns true if n is a sphenic number and
// No otherwise
bool isSphenic(int n)
{
// Stores three prime factors of n. We have at-most
// three elements in s.
set<int> s;

// Keep finding least prime factors until n becomes 1
while (n > 1)
{
// Find least prime factor of current value of n.
int lpf = least_pf[n];

// We store current size of s to check if a prime
// factor repeats
int init_size = s.size();

// Insert least prime factor of current value of n
s.insert(lpf);

// If either lpf repeats or number of lpfs becomes
// more than 3, then return false.
if (s.size() == init_size || s.size() > 3)

// same prime divides the
// number more than once
return false;

// Divide n by lpf
n /= lpf;
}

// Return true if size of set is 3
return (s.size() == 3);
}

// Driver program to test above functions
int main()
{
leastPrimeFactor(MAX);
for (int i=1; i<100; i++)
if (isSphenic(i))
cout << i << " ";
return 0;
}
```

Output:

```Yes
```

Time Complexity: O(nlog(n))
Auxiliary Space: O(n)

References:
1. OEIS
2. https://en.wikipedia.org/wiki/Sphenic_number

This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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