Given an integer
Examples:
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 – rotate n once. and continue.
- if LEN is over return -1
Below is the implementation of the above approach:
// C++ program to find permutation of n // which is divisible by 3 but not // divisible by 6 #include <bits/stdc++.h> 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 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 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# 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 // PHP program to find permutation // of n which is divisible by 3 but // not divisible by 6 // Function to find the permutation function findPermutation( $n )
{ // length of integer
$len = ceil (log10( $n ));
for ( $i = 0; $i < $len ; $i ++)
{
// if integer is even
if ( $n % 2 != 0)
{
// return odd integer
return (int) $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 $n = 132;
echo findPermutation( $n );
// This code is contributed by mits ?> |
<script> // java script program to find permutation // of n which is divisible by 3 but // not divisible by 6 // Function to find the permutation function findPermutation(n)
{ // length of integer
let len = Math.ceil(Math.log10(n));
for (let i = 0; i < len; i++)
{
// if integer is even
if (n % 2 != 0)
{
// return odd integer
return parseInt(n);
}
else
{
// rotate integer
n = (n / 10) + (n % 10) *
Math.pow(10, len - i - 1);
continue ;
}
}
// return -1 in case no required
// permutation exists
return -1;
} // Driver Code let n = 132; document.write( findPermutation(n)); // This code is contributed by sravan kumar (vignan) </script> |
Output:
213
Time complexity: O(logn) for given input number n
Auxiliary space: O(1)