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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

 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 ?>

Output:

4451 4514 5144

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