# Generate all rotations of a number

• Difficulty Level : Medium
• Last Updated : 10 May, 2021

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 ``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``?>`

## Javascript

 ``
Output:
`4451 4514 5144`

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