# 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., 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
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
20 can be factorized as 2 * 2 * 5.
Since prime factor appears more than
once, it is not a square free number.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is simple, we 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 which 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 program to test above function */` `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

 ` `

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 .

My Personal Notes arrow_drop_up Maths is the language of nature

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