Generate all cyclic permutations of a number

03 May, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 numbervoid 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 Programint main(){    int N = 5674;    cyclic(N);    return 0;}
// Java Program to generate all// cyclic permutations of numberclass 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 Program to# generate all cyclic# permutations of numberimport 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 numberdef 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 CodeN = 5674;cyclic(N); # This code is contributed by mits
// C# Program to generate all// cyclic permutations of numberusing 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 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 $count;} // Function to generate all// cyclic permutations of a numberfunction cyclic($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?>
<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 numberfunction 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 Programvar N = 5674;cyclic(N); // This code is contributed by Amit Katiyar</script>
Output:  
 5674
 4567
 7456
 6745
Time Complexity: O(N), where N is the number of digitsAuxiliary Space: O(1)  
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nitin mittal
Mithun Kumar
amit143katiyar
singhh3010
simmytarika5
number-digits
Combinatorial
Mathematical
Mathematical
Combinatorial
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