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:
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++
// 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
// 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
<?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 ?> |
Javascript
<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> |
213
Time complexity: O(logn) for given input number n
Auxiliary space: O(1)
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