# Java Program to Rotate digits of a given number by K

• Last Updated : 27 Jan, 2022

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!

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