Check if an integer is rotation of another given integer

Given two integers A and B, the task is to check if the integer A is rotation of the digits of the integer B or not. If found to be true, then print “Yes”. Otherwise print “No”.

Examples:

Input: A= 976, B= 679
Output: Yes

Input: A= 974, B= 2345
Output: No

Approach: Follow the steps below to solve the problem:

• If A == B, then print “Yes”.
• Calculate the count of digits of integer present in the integer A and B in variables, say dig1 and dig2.
• If dig1 != dig2 is found not be true, then print “No”.
• Initialize a variable, say temp. Assign temp = A.
• Now, iterate and perform the following operations:
  • Print “Yes” if A is equal to B and break.
  • Check if (A== temp) i.e A becomes equal to original integer temp. If found to be true, then print “No” and break.

Below is the implementation of the above approach:

C++

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

// Function to check if the integer
// A is a rotation of the integer B
int check(int A, int B)
{

    if (A == B) {
        return 1;
    }

    // Stores the count of digits in A
    int dig1 = floor(log10(A) + 1);

    // Stores the count of digits in B
    int dig2 = floor(log10(B) + 1);

    // If dig1 not equal to dig2
    if (dig1 != dig2) {
        return 0;
    }

    int temp = A;

    while (1) {

        // Stores position of first digit
        int power = pow(10, dig1 - 1);

        // Stores the first digit
        int firstdigit = A / power;

        // Rotate the digits of the integer
        A = A - firstdigit * power;
        A = A * 10 + firstdigit;

        // If A is equal to B
        if (A == B) {
            return 1;
        }
        // If A is equal to the initial
// value of integer A
        if (A == temp) {
            return 0;
        }
    }
}

// Driver Code
int main()
{
    int A = 967, B = 679;

    if (check(A, B))
        cout << "Yes";
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG {

    // Function to check if the integer
    // A is a rotation of the integer B
    static int check(int A, int B)
    {

        if (A == B) {
            return 1;
        }

        // Stores the count of digits in A
        int dig1 = (int)Math.floor(Math.log10(A) + 1);

        // Stores the count of digits in B
        int dig2 = (int)Math.floor(Math.log10(B) + 1);

        // If dig1 not equal to dig2
        if (dig1 != dig2) {
            return 0;
        }

        int temp = A;

        while (true) {

            // Stores position of first digit
            int power = (int)Math.pow(10, dig1 - 1);

            // Stores the first digit
            int firstdigit = A / power;

            // Rotate the digits of the integer
            A = A - firstdigit * power;
            A = A * 10 + firstdigit;

            // If A is equal to B
            if (A == B) {
                return 1;
            }
            // If A is equal to the initial
            // value of integer A
            if (A == temp) {
                return 0;
            }
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int A = 967, B = 679;

        if (check(A, B) == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 implementation of the approach
import math

# Function to check if the integer
# A is a rotation of the integer B
def check(A, B) :

    if (A == B) :
        return 1

    # Stores the count of digits in A
    dig1 = math.floor(math.log10(A) + 1)

    # Stores the count of digits in B
    dig2 = math.floor(math.log10(B) + 1)

    # If dig1 not equal to dig2
    if (dig1 != dig2) :
        return 0

    temp = A

    while (True) :

        # Stores position of first digit
        power = pow(10, dig1 - 1)

        # Stores the first digit
        firstdigit = A // power

        # Rotate the digits of the integer
        A = A - firstdigit * power
        A = A * 10 + firstdigit

        # If A is equal to B
        if (A == B) :
            return 1

        # If A is equal to the initial value of integer A
        if (A == temp) :
            return 0

          # Driver code
A, B = 967, 679

if (check(A, B)) :
    print("Yes")
else :
    print("No")

    # This code is contributed by divyesh072019.

C#

// C# implementation of the approach
using System;
public class GFG
{

    // Function to check if the integer
    // A is a rotation of the integer B
    static int check(int A, int B)
    {
        if (A == B)
        {
            return 1;
        }

        // Stores the count of digits in A
        int dig1 = (int)Math.Floor(Math.Log10(A) + 1);

        // Stores the count of digits in B
        int dig2 = (int)Math.Floor(Math.Log10(B) + 1);

        // If dig1 not equal to dig2
        if (dig1 != dig2)
        {
            return 0;
        }
        int temp = A;
        while (true)
        {

            // Stores position of first digit
            int power = (int)Math.Pow(10, dig1 - 1);

            // Stores the first digit
            int firstdigit = A / power;

            // Rotate the digits of the integer
            A = A - firstdigit * power;
            A = A * 10 + firstdigit;

            // If A is equal to B
            if (A == B)
            {
                return 1;
            }

            // If A is equal to the initial
            // value of integer A
            if (A == temp)
            {
                return 0;
            }
        }
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int A = 967, B = 679;
        if (check(A, B) == 1)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript implementation of the approach

    // Function to check if the integer
    // A is a rotation of the integer B
    function check(A, B)
    {
        if (A == B)
        {
            return 1;
        }

        // Stores the count of digits in A
        let dig1 = Math.floor(Math.log10(A) + 1);

        // Stores the count of digits in B
        let dig2 = Math.floor(Math.log10(B) + 1);

        // If dig1 not equal to dig2
        if (dig1 != dig2)
        {
            return 0;
        }
        let temp = A;
        while (true)
        {

            // Stores position of first digit
            let power = Math.pow(10, dig1 - 1);

            // Stores the first digit
            let firstdigit = parseInt(A / power, 10);

            // Rotate the digits of the integer
            A = A - firstdigit * power;
            A = A * 10 + firstdigit;

            // If A is equal to B
            if (A == B)
            {
                return 1;
            }

            // If A is equal to the initial
            // value of integer A
            if (A == temp)
            {
                return 0;
            }
        }
    }

    let A = 967, B = 679;
    if (check(A, B) == 1)
      document.write("Yes");
    else
      document.write("No");

  // This code is contributed by suresh07.
</script>

Output:

Yes

Time Complexity: O(digit(N))
Auxiliary Space: O(1)