Given an integerĀ 
. 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:Ā Ā
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:
- let LEN is length of integer (i.e. ceil(log10(n))).
- iterate over LEN and check whether n is even or odd.
- if n is odd return n
- else right ā rotate n once. and continue.
- if LEN is over return -1
Below is the implementation of the above approach:Ā
C++
// C++ program to find permutation of n// which is divisible by 3 but not// divisible by 6Ā
#include <bits/stdc++.h>using namespace std;Ā
// Function to find the permutationint findPermutation(int n){Ā Ā Ā Ā // length of integerĀ Ā Ā Ā int len = ceil(log10(n));Ā
Ā Ā Ā Ā for (int i = 0; i < len; i++) {Ā Ā Ā Ā Ā Ā Ā Ā // if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if (n % 2 != 0) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return n;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā n = (n / 10) + (n % 10) * pow(10, len - i - 1);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // return -1 in case no requiredĀ Ā Ā Ā // permutation existsĀ Ā Ā Ā return -1;}Ā
// Driver Codeint main(){Ā Ā Ā Ā int n = 132;Ā
Ā Ā Ā Ā cout << findPermutation(n);Ā
Ā Ā Ā Ā return 0;} |
Java
// Java program to find permutation // of n which is divisible by 3 // but not divisible by 6import java.lang.*;import java.util.*;Ā
class GFG{// Function to find the permutationstatic int findPermutation(int n){Ā Ā Ā Ā // length of integerĀ Ā Ā Ā int len = (int)Math.ceil(Math.log10(n));Ā
Ā Ā Ā Ā for (int i = 0; i < len; i++) Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if (n % 2 != 0) Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return n;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā n = (n / 10) + (n % 10) * Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (int)Math.pow(10, len - i - 1);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // return -1 in case no requiredĀ Ā Ā Ā // permutation existsĀ Ā Ā Ā return -1;}Ā
// Driver Codepublic static void main(String args[]){Ā Ā Ā Ā int n = 132;Ā
Ā Ā Ā Ā System.out.println(findPermutation(n));}}Ā
// 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 6from math import log10, ceil, powĀ
# Function to find the permutationdef findPermutation(n):Ā Ā Ā Ā Ā Ā Ā Ā Ā # length of integerĀ Ā Ā Ā len = ceil(log10(n))Ā
Ā Ā Ā Ā for i in range(0, len, 1):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if n % 2 != 0:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return nĀ Ā Ā Ā Ā Ā Ā Ā else:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā n = ((n / 10) + (n % 10) *Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā pow(10, len - i - 1))Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continueĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # return -1 in case no requiredĀ Ā Ā Ā # permutation existsĀ Ā Ā Ā return -1Ā
# Driver Codeif __name__ == '__main__':Ā Ā Ā Ā n = 132Ā
Ā Ā Ā Ā print(int(findPermutation(n)))Ā
# This code is contributed # by Surendra_Gangwar |
C#
// C# program to find permutation // of n which is divisible by 3 // but not divisible by 6using System;Ā
class GFG{// Function to find the permutationstatic int findPermutation(int n){Ā Ā Ā Ā // length of integerĀ Ā Ā Ā int len = (int)Math.Ceiling(Math.Log10(n));Ā
Ā Ā Ā Ā for (int i = 0; i < len; i++) Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if (n % 2 != 0) Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return n;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā n = (n / 10) + (n % 10) * Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (int)Math.Pow(10, len - i - 1);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // return -1 in case no requiredĀ Ā Ā Ā // permutation existsĀ Ā Ā Ā return -1;}Ā
// Driver Codepublic static void Main(){Ā Ā Ā Ā int n = 132;Ā
Ā Ā Ā Ā Console.WriteLine(findPermutation(n));}}Ā
// This code is contributed// by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP program to find permutation // of n which is divisible by 3 but// not divisible by 6Ā
// Function to find the permutationfunction findPermutation($n){Ā Ā Ā Ā // length of integerĀ Ā Ā Ā $len = ceil(log10($n));Ā
Ā Ā Ā Ā for ($i = 0; $i < $len; $i++) Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if ($n % 2 != 0) Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return (int)$n;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā $n = ($n / 10) + ($n % 10) *Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā pow(10, $len - $i - 1);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // return -1 in case no requiredĀ Ā Ā Ā // permutation existsĀ Ā Ā Ā return -1;}Ā
// Driver Code$n = 132;Ā
echo findPermutation($n);Ā
// This code is contributed by mits?> |
Javascript
<script>// java scriptĀ program to find permutation// of n which is divisible by 3 but// not divisible by 6Ā
// Function to find the permutationfunction findPermutation(n){Ā
Ā Ā Ā Ā // length of integerĀ Ā Ā Ā let len = Math.ceil(Math.log10(n));Ā
Ā Ā Ā Ā for (let i = 0; i < len; i++)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // if integer is evenĀ Ā Ā Ā Ā Ā Ā Ā if (n % 2 != 0)Ā Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // return odd integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return parseInt(n);Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // rotate integerĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā n = (n / 10) + (n % 10) *Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Math.pow(10, len - i - 1);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā continue;Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā // return -1 in case no requiredĀ Ā Ā Ā // permutation existsĀ Ā Ā Ā return -1;}Ā
// Driver Codelet n = 132;Ā
document.write( findPermutation(n));Ā
// This code is contributed by sravan kumar (vignan)</script> |
Output:Ā
213
Ā
Time complexity: O(logn) for given input number n
Auxiliary space: O(1)
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