# Square Free Number

• Difficulty Level : Medium
• Last Updated : 05 Aug, 2022

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 ``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

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## Javascript

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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)

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