Java Program to Rotate digits of a given number by K

Given two integers N and K, the task is to rotate the digits of N by K. If K is a positive integer, left rotate its digits. Otherwise, right rotate its digits.

Examples:

Input: N = 12345, K = 2
Output: 34512 
Explanation:
Left rotating N(= 12345) by K(= 2) modifies N to 34512. 
Therefore, the required output is 34512

Input: N = 12345, K = -3
Output: 34512 
Explanation:
Right rotating N(= 12345) by K( = -3) modifies N to 34512. 
Therefore, the required output is 34512

Approach: Follow the steps below to solve the problem:

• Initialize a variable, say X, to store the count of digits in N.
• Update K = (K + X) % X to reduce it to a case of left rotation.
• Remove the first K digits of N and append all the removed digits to the right of the digits of N.
• Finally, print the value of N.

Below is the implementation of the above approach:

Java

// Java program to implement
// the above approach

import java.io.*;

class GFG {

    // Function to find the count of
    // digits in N
    static int numberOfDigit(int N)
    {

        // Stores count of
        // digits in N
        int digit = 0;

        // Calculate the count
        // of digits in N
        while (N > 0) {

            // Update digit
            digit++;

            // Update N
            N /= 10;
        }
        return digit;
    }

    // Function to rotate the digits of N by K
    static void rotateNumberByK(int N, int K)
    {

        // Stores count of digits in N
        int X = numberOfDigit(N);

        // Update K so that only need to
        // handle left rotation
        K = ((K % X) + X) % X;

        // Stores first K digits of N
        int left_no = N / (int)(Math.pow(10,
                                         X - K));

        // Remove first K digits of N
        N = N % (int)(Math.pow(10, X - K));

        // Stores count of digits in left_no
        int left_digit = numberOfDigit(left_no);

        // Append left_no to the right of
        // digits of N
        N = (N * (int)(Math.pow(10, left_digit)))
            + left_no;

        System.out.println(N);
    }

    // Driver Code
    public static void main(String args[])
    {

        int N = 12345, K = 7;

        // Function Call
        rotateNumberByK(N, K);
    }
}

Output:

34512

Time Complexity: O(log₁₀N)
Auxiliary Space: O(1)

Please refer complete article on Rotate digits of a given number by K for more details!