# Refactorable number

Given an integer n. Check whether the number is refactorable or not. A refactorable number is an integer n that is divisible by count of all it’s divisors.

Example :

```Input:  n = 8
Output: yes

Explanation:
8 has 4 divisors: 1, 2, 4, 8
Since 8 is divisible by 4 therefore 8 is
refactorable number.

Input : n = 4
Output: no
```

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

This solution is pretty straightforward. The idea is to iterate from 1 to sqrt(n) and count all the divisors of a number. After that we just need to check whether the number n is divisible by it’s total count or not.

## C++

 `// C++ program to check whether number is  ` `// refactorable or not ` `#include ` ` `  `// Function to count all divisors ` `bool` `isRefactorableNumber(``int` `n) ` `{  ` `    ``// Initialize result ` `    ``int` `divCount = 0;  ` `     `  `    ``for` `(``int` `i = 1; i <= ``sqrt``(n); ++i) ` `    ``{ ` `        ``if` `(n % i==0) ` `        ``{ ` `            ``// If divisors are equal, count  ` `            ``// only one. ` `            ``if` `(n / i == i) ` `                ``++divCount; ` ` `  `            ``// Otherwise count both ` `            ``else`  `                ``divCount += 2; ` `        ``} ` `    ``} ` ` `  `    ``return` `n % divCount == 0; ` `} ` ` `  `//Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 8; ` `    ``if` `(isRefactorableNumber(n)) ` `        ``puts``(``"yes"``); ` `    ``else` `        ``puts``(``"no"``); ` ` `  `    ``n = 14; ` `    ``if` `(isRefactorableNumber(n)) ` `        ``puts``(``"yes"``); ` `    ``else` `        ``puts``(``"no"``); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check whether number is  ` `// refactorable or not ` ` `  `class` `GFG ` `{ ` `    ``// Function to count all divisors ` `    ``static` `boolean` `isRefactorableNumber(``int` `n) ` `    ``{ ` `        ``// Initialize result ` `        ``int` `divCount = ``0``;  ` `         `  `        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); ++i) ` `        ``{ ` `            ``if` `(n % i==``0``) ` `            ``{ ` `                ``// If divisors are equal, count  ` `                ``// only one. ` `                ``if` `(n / i == i) ` `                    ``++divCount; ` `                     `  `                ``// Otherwise count both ` `                ``else`  `                    ``divCount += ``2``; ` `            ``} ` `        ``} ` `        ``return` `n % divCount == ``0``; ` `    ``} ` `     `  `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = ``8``; ` `        ``if` `(isRefactorableNumber(n)) ` `            ``System.out.println(``"yes"``); ` `        ``else` `            ``System.out.println(``"no"``); ` `             `  `        ``n = ``14``; ` `        ``if` `(isRefactorableNumber(n)) ` `            ``System.out.println(``"yes"``); ` `        ``else` `            ``System.out.println(``"no"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Saket Kumar `

## Python

 `# Python program to check whether number is ` `# refactorable or not ` `import` `math ` ` `  `def` `isRefactorableNumber(n): ` ` `  `    ``# Initialize result ` `    ``divCount ``=` `0` ` `  `    ``for` `i ``in` `range``(``1``,``int``(math.sqrt(n))``+``1``): ` ` `  `        ``if` `n ``%` `i ``=``=` `0``: ` ` `  `            ``# If divisors are equal, count only one ` `            ``if` `n``/``i ``=``=` `i: ` `                ``divCount ``+``=` `1` ` `  `            ``else``:  ``# Otherwise count both ` `                ``divCount ``+``=` `2` ` `  `    ``return` `n ``%` `divCount ``=``=` `0` ` `  ` `  `# Driver Code ` `n ``=` `8` `if` `isRefactorableNumber(n): ` `    ``print` `"yes"` `else``: ` `    ``print` `"no"` ` `  `n ``=` `14` `if` `(isRefactorableNumber(n)): ` `    ``print` `"yes"` `else``: ` `    ``print` `"no"`

## C#

 `// C# program to check whether number is  ` `// refactorable or not ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function to count all divisors ` `    ``static` `bool` `isRefactorableNumber(``int` `n) ` `    ``{ ` `         `  `        ``// Initialize result ` `        ``int` `divCount = 0;  ` `         `  `        ``for` `(``int` `i = 1; i <= Math.Sqrt(n); ++i) ` `        ``{ ` `            ``if` `(n % i==0) ` `            ``{ ` `                ``// If divisors are equal, count  ` `                ``// only one. ` `                ``if` `(n / i == i) ` `                    ``++divCount; ` `                 `  `                ``// Otherwise count both ` `                ``else`  `                    ``divCount += 2; ` `            ``} ` `        ``} ` `        ``return` `n % divCount == 0; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 8; ` `        ``if` `(isRefactorableNumber(n)) ` `            ``Console.WriteLine(``"yes"``); ` `        ``else` `            ``Console.Write(``"no"``); ` `             `  `        ``n = 14; ` `        ``if` `(isRefactorableNumber(n)) ` `            ``Console.Write(``"yes"``); ` `        ``else` `            ``Console.Write(``"no"``); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

```Output:
yes
no
```

Time complexity: O(sqrt(n))
Auxiliary space: O(1)

1. There is no refactorable number which is perfect.
2. There is no three consecutive integers can all be refactorable.
3. Refactorable number have natural density zero.

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