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Check whether a number has exactly three distinct factors or not
  • Difficulty Level : Easy
  • Last Updated : 09 Apr, 2021
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Given a positive integer n(1 <= n <= 1018). Check whether a number has exactly three distinct factors or not. Print “Yes” if it has otherwise “No“.

Examples : 

Input : 9
Output: Yes
Explanation
Number 9 has exactly three factors:
1, 3, 9, hence answer is 'Yes'

Input  : 10
Output : No

Simple approach is to count factors by generating all divisors of a number by using this approach, after that check whether the count of all factors are equal to ‘3’ or not. Time complexity of this approach is O(sqrt(n)). 

Better approach is to use Number theory. According to property of perfect square, “Every perfect square(x2) always have only odd numbers of factors“. 

If the square root of given number(say x2) is prime(after conforming that number is perfect square) then it must have exactly three distinct factors i.e., 



  1. A number 1 of course. 
  2. Square root of a number i.e., x(prime number). 
  3. Number itself i.e., x2

Below is the implementation of above approach: 

C++




// C++ program to check whether number
// has exactly three distinct factors
// or not
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to check whether a
// number is prime or not
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check whether given number
// has three distinct factors or not
bool isThreeDisctFactors(long long n)
{
    // Find square root of number
    int sq = (int)sqrt(n);
 
    // Check whether number is perfect
    // square or not
    if (1LL * sq * sq != n)
        return false;
 
    // If number is perfect square, check
    // whether square root is prime or
    // not
    return isPrime(sq) ? true : false;
}
 
// Driver program
int main()
{
    long long num = 9;
    if (isThreeDisctFactors(num))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    num = 15;
    if (isThreeDisctFactors(num))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    num = 12397923568441;
    if (isThreeDisctFactors(num))
        cout << "Yes\n";
    else
        cout << "No\n";
 
    return 0;
}

Java




// Java program to check whether number
// has exactly three distinct factors
// or not
public class GFG {
 
 
// Utility function to check whether a
// number is prime or not
static boolean isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check whether given number
// has three distinct factors or not
static boolean isThreeDisctFactors(long n)
{
    // Find square root of number
    int sq = (int)Math.sqrt(n);
 
    // Check whether number is perfect
    // square or not
    if (1L * sq * sq != n)
        return false;
 
    // If number is perfect square, check
    // whether square root is prime or
    // not
    return isPrime(sq) ? true : false;
}
 
// Driver program
    public static void main(String[] args) {
        long num = 9;
    if (isThreeDisctFactors(num))
        System.out.println("Yes");
    else
        System.out.println("No");
 
    num = 15;
    if (isThreeDisctFactors(num))
        System.out.println("Yes");
    else
        System.out.println("No");
 
    num = 12397923568441L;
    if (isThreeDisctFactors(num))
        System.out.println("Yes");
    else
        System.out.println("No");
    }
}

Python3




# Python 3 program to check whether number
# has exactly three distinct factors
# or not
 
from math import sqrt
# Utility function to check whether a
# number is prime or not
def isPrime(n):
    # Corner cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
 
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False
     
    k= int(sqrt(n))+1
    for i in range(5,k,6):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
 
    return True
 
# Function to check whether given number
# has three distinct factors or not
def isThreeDisctFactors(n):
    # Find square root of number
    sq = int(sqrt(n))
 
    # Check whether number is perfect
    # square or not
    if (1 * sq * sq != n):
        return False
 
    # If number is perfect square, check
    # whether square root is prime or
    # not
    if (isPrime(sq)):
        return True
    else:
        return False
 
# Driver program
if __name__ == '__main__':
    num = 9
    if (isThreeDisctFactors(num)):
        print("Yes")
    else:
        print("No")
 
    num = 15
    if (isThreeDisctFactors(num)):
        print("Yes")
    else:
        print("No")
 
    num = 12397923568441
    if (isThreeDisctFactors(num)):
        print("Yes")
    else:
        print("No")
 
# This code is contributd by
# Surendra_Gangwar

C#




// C# program to check whether number
// has exactly three distinct factors
// or not
using System;
 
public class GFG {
 
    // Utility function to check whether
    // a number is prime or not
    static bool isPrime(int n)
    {
 
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can
        // skip middle five numbers in
        // below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // Function to check whether given number
    // has three distinct factors or not
    static bool isThreeDisctFactors(long n)
    {
 
        // Find square root of number
        int sq = (int)Math.Sqrt(n);
 
        // Check whether number is perfect
        // square or not
        if (1LL * sq * sq != n)
            return false;
 
        // If number is perfect square, check
        // whether square root is prime or
        // not
        return isPrime(sq) ? true : false;
    }
 
    // Driver program
    static public void Main()
    {
        long num = 9;
        if (isThreeDisctFactors(num))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
 
        num = 15;
        if (isThreeDisctFactors(num))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
 
        num = 12397923568441;
        if (isThreeDisctFactors(num))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This Code is contributed by vt_m.

PHP




<?php
// PHP program to check whether
// number has exactly three
// distinct factors or not
 
// Utility function to check
// whether a number is prime
// or not
function isPrime($n)
{
    // Corner cases
    if ($n <= 1)
        return false;
    if ($n <= 3)
        return true;
 
    // This is checked so that
    // we can skip middle five
    // numbers in below loop
    if ($n % 2 == 0 || $n % 3 == 0)
        return false;
 
    for ($i = 5; $i * $i <= $n;
                   $i = $i + 6)
        if ($n % $i == 0 ||
            $n % ($i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check
// whether given number
// has three distinct
// factors or not
function isThreeDisctFactors($n)
{
    // Find square root of number
    $sq = sqrt($n);
 
    // Check whether number is
    // perfect square or not
    if ($sq * $sq != $n)
        return false;
 
    // If number is perfect square,
    // check whether square root is
    // prime or not
    return isPrime($sq) ? true : false;
}
 
// Driver Code
$num = 9;
if (isThreeDisctFactors($num))
    echo "Yes\n";
else
    echo "No\n";
 
$num = 15;
if (isThreeDisctFactors($num))
    echo "Yes\n";
else
    echo "No\n";
 
$num = 12397923568441;
if (isThreeDisctFactors($num))
    echo "Yes\n";
else
    echo "No\n";
 
// This code is contributed by ak_t
?>

Javascript




<script>
 
// Javascript program to check whether
// number has exactly three distinct
// factors or not
 
// Utility function to check whether a
// number is prime or not
function isPrime(n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for(let i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check whether given number
// has three distinct factors or not
function isThreeDisctFactors(n)
{
     
    // Find square root of number
    let sq = parseInt(Math.sqrt(n));
 
    // Check whether number is perfect
    // square or not
    if (sq * sq != n)
        return false;
 
    // If number is perfect square, check
    // whether square root is prime or
    // not
    return isPrime(sq) ? true : false;
}
 
// Driver code
let num = 9;
if (isThreeDisctFactors(num))
    document.write("Yes<br>");
else
    document.write("No<br>");
 
num = 15;
if (isThreeDisctFactors(num))
    document.write("Yes<br>");
else
    document.write("No<br>");
 
num = 12397923568441;
if (isThreeDisctFactors(num))
    document.write("Yes<br>");
else
    document.write("No<br>");
     
// This code is contributed by souravmahato348 
 
</script>

Output : 

Yes
No
No

Time complexity : O(n1/4
Auxiliary space: O(1)

This article is contributed by Shubham Bansal. 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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