Generate all cyclic permutations of a number
Given a number N, our task is to generate all the possible cyclic permutations of the number.
A cyclic permutation shifts all the elements of a set by a fixed offset. For a set with elements ,
, …,
, a cyclic permutation of one place to the left would yield
, …,
,
, and a cyclic permutation of one place to the right would yield
,
,
, ….
Examples:
Input : 123 Output : 123 312 231 Input : 5674 Output : 5674 4567 7456 6745
The idea is to generate next permutation of a number using below formula.
rem = num % 10; div = num / 10; num = (pow(10, n - 1)) * rem + div;
While repeating above steps, if we come back to original number, we stop and return.
C++
// Program to generate all cyclic permutations // of number #include <bits/stdc++.h> using namespace std; // Function to count the total number of digits // in a number. int countdigits( int N) { int count = 0; while (N) { count++; N = N / 10; } return count; } // Function to generate all cyclic permutations // of a number void cyclic( int N) { int num = N; int n = countdigits(N); while (1) { cout << num << endl; // Following three lines generates a // circular permutation of a number. int rem = num % 10; int div = num / 10; num = ( pow (10, n - 1)) * rem + div ; // If all the permutations are checked // and we obtain original number exit // from loop. if (num == N) break ; } } // Driver Program int main() { int N = 5674; cyclic(N); return 0; } |
Java
// Java Program to generate all // cyclic permutations of number public class GFG { // Function to count the total number // of digits in a number. static int countdigits( int N) { int count = 0 ; while (N> 0 ) { count++; N = N / 10 ; } return count; } // Function to generate all cyclic // permutations of a number static void cyclic( int N) { int num = N; int n = countdigits(N); while ( true ) { System.out.println(num); // Following three lines generates a // circular permutation of a number. int rem = num % 10 ; int dev = num / 10 ; num = ( int )((Math.pow( 10 , n - 1 )) * rem + dev); // If all the permutations are // checked and we obtain original // number exit from loop. if (num == N) break ; } } // Driver Program public static void main (String[] args) { int N = 5674 ; cyclic(N); } } /* This code is contributed by Mr. Somesh Awasthi */ |
Python3
# Python3 Program to # generate all cyclic # permutations of number import math # Function to count the # total number of digits # in a number. def countdigits(N): count = 0 ; while (N): count = count + 1 ; N = int (math.floor(N / 10 )); return count; # Function to generate # all cyclic permutations # of a number def cyclic(N): num = N; n = countdigits(N); while ( 1 ): print ( int (num)); # Following three lines # generates a circular # permutation of a number. rem = num % 10 ; div = math.floor(num / 10 ); num = ((math. pow ( 10 , n - 1 )) * rem + div); # If all the permutations # are checked and we obtain # original number exit from loop. if (num = = N): break ; # Driver Code N = 5674 ; cyclic(N); # This code is contributed by mits |
C#
// C# Program to generate all // cyclic permutations of number using System; class GFG { // Function to count the total number // of digits in a number. static int countdigits( int N) { int count = 0; while (N > 0) { count++; N = N / 10; } return count; } // Function to generate all cyclic // permutations of a number static void cyclic( int N) { int num = N; int n = countdigits(N); while ( true ) { Console.WriteLine(num); // Following three lines generates a // circular permutation of a number. int rem = num % 10; int dev = num / 10; num = ( int )((Math.Pow(10, n - 1)) * rem + dev); // If all the permutations are // checked and we obtain original // number exit from loop. if (num == N) break ; } } // Driver Program public static void Main () { int N = 5674; cyclic(N); } } // This code is contributed by nitin mittal |
PHP
<?php // PHP Program to generate all // cyclic permutations of number // Function to count the total // number of digits in a number. function countdigits( $N ) { $count = 0; while ( $N ) { $count ++; $N = floor ( $N / 10); } return $count ; } // Function to generate all // cyclic permutations of a number function cyclic( $N ) { $num = $N ; $n = countdigits( $N ); while (1) { echo ( $num ); echo "\n" ; // Following three lines generates a // circular permutation of a number. $rem = $num % 10; $div = floor ( $num / 10); $num = (pow(10, $n - 1)) * $rem + $div ; // If all the permutations are checked // and we obtain original number exit // from loop. if ( $num == $N ) break ; } } // Driver Code $N = 5674; cyclic( $N ); // This code is contributed by nitin mittal ?> |
Javascript
<script> // javascript Program to generate all // cyclic permutations of number // Function to count the total number // of digits in a number. function countdigits(N) { var count = 0; while (N>0) { count++; N = parseInt(N / 10); } return count; } // Function to generate all cyclic // permutations of a number function cyclic(N) { var num = N; var n = countdigits(N); while ( true ) { document.write(num+ "<br>" ); // Following three lines generates a // circular permutation of a number. var rem = num % 10; var dev = parseInt(num / 10); num = parseInt(((Math.pow(10, n - 1)) * rem + dev)); // If all the permutations are // checked and we obtain original // number exit from loop. if (num == N) break ; } } // Driver Program var N = 5674; cyclic(N); // This code is contributed by Amit Katiyar </script> |
5674 4567 7456 6745
Time Complexity: O(N), where N is the number of digits
Auxiliary Space: O(1)
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