# Find permutation of n which is divisible by 3 but not divisible by 6

• Last Updated : 22 Apr, 2021

Given an integer . The task is to find another integer which is permutation of n, divisible by 3 but not divisible by 6. Given that n is divisible by 6. If no such permutation is possible print -1.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

```Input: n = 336
Output: 363

Input: n = 48
Output: -1```

For a number to be divisible by 6, it must be divisible by 3 as well as 2, means every even integer divisible by 3 is divisible by 6. So, an integer which is divisible by 3 but not 6 is odd integer divisible by 3.
So, if integer n contains any odd integer then there exists a permutation which is divisible by 3 but not 6, else no such permutation exist.

Algorithm:

1. let LEN is length of integer (i.e. ceil(log10(n))).
2. iterate over LEN and check whether n is even or odd.
• if n is odd return n
• else right – rotate n once. and continue.
3. if LEN is over return -1

Below is the implementation of the above approach:

## C++

 `// C++ program to find permutation of n``// which is divisible by 3 but not``// divisible by 6` `#include ``using` `namespace` `std;` `// Function to find the permutation``int` `findPermutation(``int` `n)``{``    ``// length of integer``    ``int` `len = ``ceil``(``log10``(n));` `    ``for` `(``int` `i = 0; i < len; i++) {``        ``// if integer is even``        ``if` `(n % 2 != 0) {``            ``// return odd integer``            ``return` `n;``        ``}``        ``else` `{``            ``// rotate integer``            ``n = (n / 10) + (n % 10) * ``pow``(10, len - i - 1);``            ``continue``;``        ``}``    ``}` `    ``// return -1 in case no required``    ``// permutation exists``    ``return` `-1;``}` `// Driver Code``int` `main()``{``    ``int` `n = 132;` `    ``cout << findPermutation(n);` `    ``return` `0;``}`

## Java

 `// Java program to find permutation``// of n which is divisible by 3``// but not divisible by 6``import` `java.lang.*;``import` `java.util.*;` `class` `GFG``{``// Function to find the permutation``static` `int` `findPermutation(``int` `n)``{``    ``// length of integer``    ``int` `len = (``int``)Math.ceil(Math.log10(n));` `    ``for` `(``int` `i = ``0``; i < len; i++)``    ``{``        ``// if integer is even``        ``if` `(n % ``2` `!= ``0``)``        ``{``            ``// return odd integer``            ``return` `n;``        ``}``        ``else``        ``{``            ``// rotate integer``            ``n = (n / ``10``) + (n % ``10``) *``                ``(``int``)Math.pow(``10``, len - i - ``1``);``            ``continue``;``        ``}``    ``}` `    ``// return -1 in case no required``    ``// permutation exists``    ``return` `-``1``;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``132``;` `    ``System.out.println(findPermutation(n));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## Python3

 `# Python3 program to find permutation``# of n which is divisible by 3 but``# not divisible by 6``from` `math ``import` `log10, ceil, ``pow` `# Function to find the permutation``def` `findPermutation(n):``    ` `    ``# length of integer``    ``len` `=` `ceil(log10(n))` `    ``for` `i ``in` `range``(``0``, ``len``, ``1``):``        ` `        ``# if integer is even``        ``if` `n ``%` `2` `!``=` `0``:``            ` `            ``# return odd integer``            ``return` `n``        ``else``:``            ` `            ``# rotate integer``            ``n ``=` `((n ``/` `10``) ``+` `(n ``%` `10``) ``*``                  ``pow``(``10``, ``len` `-` `i ``-` `1``))``            ``continue``        ` `    ``# return -1 in case no required``    ``# permutation exists``    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `132` `    ``print``(``int``(findPermutation(n)))` `# This code is contributed``# by Surendra_Gangwar`

## C#

 `// C# program to find permutation``// of n which is divisible by 3``// but not divisible by 6``using` `System;` `class` `GFG``{``// Function to find the permutation``static` `int` `findPermutation(``int` `n)``{``    ``// length of integer``    ``int` `len = (``int``)Math.Ceiling(Math.Log10(n));` `    ``for` `(``int` `i = 0; i < len; i++)``    ``{``        ``// if integer is even``        ``if` `(n % 2 != 0)``        ``{``            ``// return odd integer``            ``return` `n;``        ``}``        ``else``        ``{``            ``// rotate integer``            ``n = (n / 10) + (n % 10) *``                ``(``int``)Math.Pow(10, len - i - 1);``            ``continue``;``        ``}``    ``}` `    ``// return -1 in case no required``    ``// permutation exists``    ``return` `-1;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 132;` `    ``Console.WriteLine(findPermutation(n));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## PHP

 ``

## Javascript

 ``
Output:
`213`

My Personal Notes arrow_drop_up