Check whether all the rotations of a given number is greater than or equal to the given number or not

Given an integer x, the task is to find if every k-cycle shift on the element produces a number greater than or equal to the same element.
A k-cyclic shift of an integer x is a function that removes the last k digits of x and inserts them in its beginning.
For example, the k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2. Print Yes if the given condition is satisfied else print No.

Examples:

Input: x = 123
Output : Yes
The k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2.
Both 312 and 231 are greater than 123.

Input: 2214
Output: No
The k-cyclic shift of 2214 when k=2 is 1422 which is smaller than 2214

Approach: Simply find all the possible k cyclic shifts of the number and check if all are greater than the given number or not.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
void CheckKCycles(int n, string s) 
{
    bool ff = true;
    int x = 0;
    for (int i = 1; i < n; i++) 
    {
  
        // Splitting the number at index i 
        // and adding to the front 
        x = (s.substr(i) + s.substr(0, i)).length();
  
        // Checking if the value is greater than 
        // or equal to the given value 
        if (x >= s.length()) 
        {
            continue;
        }
        ff = false;
        break;
    }
    if (ff) 
    {
        cout << ("Yes");
    }
    else
    {
        cout << ("No");
    }
}
  
// Driver code
int main() 
{
    int n = 3;
    string s = "123";
    CheckKCycles(n, s);
    return 0;
}
  
/* This code contributed by Rajput-Ji */

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation of the approach 


class GFG  


{ 


  


    static void CheckKCycles(int n, String s)  


    { 


        boolean ff = true; 


        int x = 0; 


        for (int i = 1; i < n; i++)  


        { 


  


            // Splitting the number at index i  


            // and adding to the front  


            x = (s.substring(i) + s.substring(0, i)).length(); 


  


            // Checking if the value is greater than  


            // or equal to the given value  


            if (x >= s.length())  


            { 


                continue; 


            } 


            ff = false; 


            break; 


        } 


        if (ff)  


        { 


            System.out.println("Yes"); 


        } 


        else 


        { 


            System.out.println("No"); 


        } 


  


    } 


  


    // Driver code 


    public static void main(String[] args)  


    { 


        int n = 3; 


        String s = "123"; 


        CheckKCycles(n, s); 


    } 


} 


  


/* This code contributed by PrinciRaj1992 */



















                        chevron_right







                        filter_none









Python














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 implementation of the approach 


def CheckKCycles(n, s): 


    ff = True


    for i in range(1, n): 


  


        # Splitting the number at index i  


        # and adding to the front 


        x = int(s[i:] + s[0:i]) 


  


        # Checking if the value is greater than  


        # or equal to the given value 


        if (x >= int(s)): 


            continue


        ff = False


        break


    if (ff): 


        print("Yes") 


    else: 


        print("No") 


  


n = 3


s = "123"


CheckKCycles(n, s) 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# implementation of the approach 


using System; 


  


class GFG  


{ 


  


    static void CheckKCycles(int n, String s)  


    { 


        bool ff = true; 


        int x = 0; 


        for (int i = 1; i < n; i++)  


        { 


  


            // Splitting the number at index i  


            // and adding to the front  


            x = (s.Substring(i) + s.Substring(0, i)).Length; 


  


            // Checking if the value is greater than  


            // or equal to the given value  


            if (x >= s.Length)  


            { 


                continue; 


            } 


            ff = false; 


            break; 


        } 


        if (ff)  


        { 


            Console.WriteLine("Yes"); 


        } 


        else


        { 


            Console.WriteLine("No"); 


        } 


  


    } 


  


    // Driver code 


    public static void Main(String[] args)  


    { 


        int n = 3; 


        String s = "123"; 


        CheckKCycles(n, s); 


    } 


} 


  


// This code has been contributed by 29AjayKumar 



















                        chevron_right







                        filter_none









PHP





= strlen($s))

		{

			continue;

		}

		$ff = false;

		break;

	}

	if ($ff)

	{

		print(“Yes”);

	}

	else

	{

		print(“No”);

	}

}


// Driver code

$n = 3;

$s = “123”;

CheckKCycles($n, $s);


// This code contributed by mits

?>





Output:


Yes






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
indrajit1
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.