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Generate all rotations of a number

Last Updated : 23 Jun, 2022
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Given an integer n, the task is to generate all the left shift numbers possible. A left shift number is a number that is generated when all the digits of the number are shifted one position to the left and the digit at the first position is shifted to the last.
Examples: 
 

Input: n = 123 
Output: 231 312
Input: n = 1445 
Output: 4451 4514 5144 
 


 


Approach: 
 

  • Assume n = 123.
  • Multiply n with 10 i.e. n = n * 10 = 1230.
  • Add the first digit to the resultant number i.e. 1230 + 1 = 1231.
  • Subtract (first digit) * 10k from the resultant number where k is the number of digits in the original number (in this case, k = 3).
  • 1231 - 1000 = 231 is the left shift number of the original number.


Below is the implementation of the above approach: 
 

C++ 
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of digits of n
int numberOfDigits(int n)
{
    int cnt = 0;
    while (n > 0) {
        cnt++;
        n /= 10;
    }
    return cnt;
}

// Function to print the left shift numbers
void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = pow(10, digits - 1);

    for (int i = 0; i < digits - 1; i++) {

        int firstDigit = num / powTen;

        // Formula to calculate left shift
        // from previous number
        int left
            = ((num * 10) + firstDigit)
              - (firstDigit * powTen * 10);
        cout << left << " ";

        // Update the original number
        num = left;
    }
}

// Driver Code
int main()
{
    int num = 1445;
    cal(num);
    return 0;
}
Java 
// Java implementation of the approach
class GFG 
{

// Function to return the count of digits of n
static int numberOfDigits(int n) 
{
    int cnt = 0;
    while (n > 0) 
    {
        cnt++;
        n /= 10;
    }
    return cnt;
}

// Function to print the left shift numbers
static void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = (int) Math.pow(10, digits - 1);

    for (int i = 0; i < digits - 1; i++)
    {
        int firstDigit = num / powTen;

        // Formula to calculate left shift
        // from previous number
        int left = ((num * 10) + firstDigit) -
                    (firstDigit * powTen * 10);
                
        System.out.print(left + " ");
                
        // Update the original number
        num = left;
    }
}

// Driver Code
public static void main(String[] args) 
{
    int num = 1445;
    cal(num);
}
}

// This code is contributed by 
// PrinciRaj1992
Python3 
# Python3 implementation of the approach

# function to return the count of digit of n
def numberofDigits(n):
    cnt = 0
    while n > 0:
        cnt += 1
        n //= 10
    return cnt
    
# function to print the left shift numbers
def cal(num):
    digit = numberofDigits(num)
    powTen = pow(10, digit - 1)
    
    for i in range(digit - 1):
        
        firstDigit = num // powTen
        
        # formula to calculate left shift 
        # from previous number
        left = (num * 10 + firstDigit - 
               (firstDigit * powTen * 10))
        print(left, end = " ")
        
        # Update the original number
        num = left
        
# Driver code
num = 1445
cal(num)

# This code is contributed
# by Mohit Kumar
C# 
// C# implementation of the approach
using System;

public class GFG{
    
// Function to return the count of digits of n
static int numberOfDigits(int n)
{
    int cnt = 0;
    while (n > 0) {
        cnt++;
        n /= 10;
    }
    return cnt;
}

// Function to print the left shift numbers
static void cal(int num)
{
    int digits = numberOfDigits(num);
    int powTen = (int)Math.Pow(10, digits - 1);

    for (int i = 0; i < digits - 1; i++) {

        int firstDigit = num / powTen;

        // Formula to calculate left shift
        // from previous number
        int left
            = ((num * 10) + firstDigit)
            - (firstDigit * powTen * 10);
        Console.Write(left +  " ");

        // Update the original number
        num = left;
    }
}

// Driver Code
    static public void Main (){
        int num = 1445;
        cal(num);
    } 
}

// This code is contributed by akt_mit....
PHP 
<?php
// PHP implementation of the approach 

// Function to return the count 
// of digits of n 
function numberOfDigits($n) 
{ 
    $cnt = 0; 
    while ($n > 0) 
    { 
        $cnt++; 
        $n = floor($n / 10); 
    } 
    return $cnt; 
} 

// Function to print the left shift numbers 
function cal($num) 
{ 
    $digits = numberOfDigits($num); 
    $powTen = pow(10, $digits - 1); 

    for ($i = 0; $i < $digits - 1; $i++)
    { 

        $firstDigit = floor($num / $powTen); 

        // Formula to calculate left shift 
        // from previous number 
        $left
            = (($num * 10) + $firstDigit) - 
               ($firstDigit * $powTen * 10); 
            
        echo $left, " "; 

        // Update the original number 
        $num = $left; 
    } 
} 

// Driver Code
$num = 1445;
cal($num); 

// This code is contributed by Ryuga
?>
JavaScript 
<script>

    // Javascript implementation of the approach
    
    // Function to return the count of digits of n
    function numberOfDigits(n)
    {
        let cnt = 0;
        while (n > 0) {
            cnt++;
            n = parseInt(n / 10, 10);
        }
        return cnt;
    }

    // Function to print the left shift numbers
    function cal(num)
    {
        let digits = numberOfDigits(num);
        let powTen = Math.pow(10, digits - 1);

        for (let i = 0; i < digits - 1; i++) {

            let firstDigit = parseInt(num / powTen, 10);

            // Formula to calculate left shift
            // from previous number
            let left = ((num * 10) + firstDigit)
                - (firstDigit * powTen * 10);
            document.write(left +  " ");

            // Update the original number
            num = left;
        }
    }
    
    let num = 1445;
    cal(num);
    
</script>

Output: 
4451 4514 5144

 

Time Complexity: O(log10n)
Auxiliary Space: O(1), since no extra space has been taken.


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