Generate all cyclic permutations of a number
Last Updated :
07 Jan, 2024
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_0 , a_1 , ..., a_n , a cyclic permutation of one place to the left would yield a_1 , ..., a_n , a_0 , and a cyclic permutation of one place to the right would yield a_n , a_0 , a_1 , ....
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>
Output5674
4567
7456
6745
Time Complexity: O(N), where N is the number of digits
Auxiliary Space: O(1)
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