Generate all rotations of a number
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