Square Free Number

Given a number, check if it is square-free or not. A number is said to be square-free if no prime factor divides it more than once, i.e., the largest power of a prime factor that divides n is one. First few square-free numbers are 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, …
Examples: 

Input: n = 10
Output: Yes
Explanation: 10 can be factorized as 2*5. Since no prime factor appears more than once, it is a square free number.

Input:  n = 20
Output: No
Explanation: 20 can be factorized as 2 * 2 * 5. Since prime factor appears more than once, it is not a square free number. 

The idea is simple, one by one find all prime factors. For every prime factor, we check if its square also divides n. If yes, then we return false. Finally, if we do not find a prime factor that is divisible more than once, we return false.

C++

// C++ Program to print
// all prime factors

# include <bits/stdc++.h>
using namespace std;

// Returns true if n is a square free
// number, else returns false.
bool isSquareFree(int n)
{
    if (n % 2 == 0)
       n = n/2;

    // If 2 again divides n, then n is
    // not a square free number.
    if (n % 2 == 0)
       return false;

    // n must be odd at this point.  So we can
    // skip one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i+2)
    {
        // Check if i is a prime factor
        if (n % i == 0)
        {
           n = n/i;

           // If i again divides, then
           // n is not square free
           if (n % i == 0)
               return false;
        }
    }

    return true;
}

// Driver Code
int main()
{
    int n = 10;
    if (isSquareFree(n))
       cout << "Yes";
    else
       cout << "No";
    return 0;
}

Java

// Java Program to print
// all prime factors

class GFG {

    // Returns true if n is a square free
    // number, else returns false.
    static boolean isSquareFree(int n)
    {
        if (n % 2 == 0)
        n = n / 2;

        // If 2 again divides n, then n is
        // not a square free number.
        if (n % 2 == 0)
        return false;

        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i = i + 2)
        {
            // Check if i is a prime factor
            if (n % i == 0)
            {
                n = n / i;

                // If i again divides, then
                // n is not square free
                if (n % i == 0)
                    return false;
            }
        }

        return true;
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 10;
        if (isSquareFree(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by prerna saini.

Python3

# Python3 Program to print
# all prime factors
from math import sqrt

# Returns true if n is
# a square free number,
# else returns false.
def isSquareFree(n):

    if n % 2 == 0:
        n = n / 2

    # If 2 again divides n,
    # then n is not a square
    # free number.
    if n % 2 == 0:
        return False

    # n must be odd at this
    # point. So we can skip
    # one element
    # (Note i = i + 2)
    for i in range(3, int(sqrt(n) + 1)):

        # Check if i is a prime
        # factor
        if n % i == 0:
            n = n / i

        # If i again divides, then
        # n is not square free
        if n % i == 0:
            return False
    return True

# Driver program
n = 10

if isSquareFree(n):
    print ("Yes")
else:
    print ("No")

# This code is contributed by Shreyanshi Arun.

C#

// C# Program to print
// all prime factors
using System;

class GFG {

    // Returns true if n is a square free
    // number, else returns false.
    static bool isSquareFree(int n)
    {
        if (n % 2 == 0)
        n = n / 2;

        // If 2 again divides n, then n is
        // not a square free number.
        if (n % 2 == 0)
        return false;

        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.Sqrt(n); i = i + 2)
        {
            // Check if i is a prime factor
            if (n % i == 0)
            {
                n = n / i;

                // If i again divides, then
                // n is not square free
                if (n % i == 0)
                    return false;
            }
        }

        return true;
    }

    // Driver program
    public static void Main()
    {
        int n = 10;
        if (isSquareFree(n))
        Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP Program to print
// all prime factors

// Returns true if n is a square free
// number, else returns false.
function isSquareFree($n)
{
    if ($n % 2 == 0)
        $n = $n / 2;

    // If 2 again divides n, then n is
    // not a square free number.
    if ($n % 2 == 0)
        return false;

    // n must be odd at this
    // point. So we can skip
    // one element (Note i = i +2)
    for ($i = 3; $i <= sqrt($n);
                   $i = $i + 2)
    {

        // Check if i is a prime factor
        if ($n % $i == 0)
        {
            $n = $n / $i;

            // If i again divides, then
            // n is not square free
            if ($n % $i == 0)
                return false;
        }
    }

    return true;
}

// Driver Code
$n = 10;
if (isSquareFree($n))
    echo("Yes");
else
    echo("No");

// This code is contributed by Ajit.
?>

Javascript

<script>

// JavaScript Program to print
// all prime factors

    // Returns true if n is a square free
    // number, else returns false.
    function isSquareFree(n)
    {
        if (n % 2 == 0)
        n = n / 2;

        // If 2 again divides n, then n is
        // not a square free number.
        if (n % 2 == 0)
        return false;

        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (let i = 3; i <= Math.sqrt(n); i = i + 2)
        {
            // Check if i is a prime factor
            if (n % i == 0)
            {
                n = n / i;

                // If i again divides, then
                // n is not square free
                if (n % i == 0)
                    return false;
            }
        }

        return true;
    }

// Driver code


       let n = 10;
        if (isSquareFree(n))
            document.write("Yes");
        else
            document.write("No");

</script>

Output

Yes

Time Complexity: O(sqrt(N)), In the worst case when the number is a perfect square, then there will be sqrt(n)/2 iterations.
Auxiliary Space: O(1)