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Java Program to Find Maximum value possible by rotating digits of a given number

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  • Last Updated : 27 Jan, 2022

Given a positive integer N, the task is to find the maximum value among all the rotations of the digits of the integer N.


Input: N = 657
Output: 765
Explanation: All rotations of 657 are {657, 576, 765}. The maximum value among all these rotations is 765.

Input: N = 7092
Output: 9270
All rotations of 7092 are {7092, 2709, 9270, 0927}. The maximum value among all these rotations is 9270.

Approach: The idea is to find all rotations of the number N and print the maximum among all the numbers generated. Follow the steps below to solve the problem:

  • Count the number of digits present in the number N, i.e. upper bound of log10N.
  • Initialize a variable, say ans with the value of N, to store the resultant maximum number generated.
  • Iterate over the range [1, log10(N) – 1] and perform the following steps:
    • Update the value of N with its next rotation.
    • Now, if the next rotation generated exceeds ans, then update ans with the rotated value of N
  • After completing the above steps, print the value of ans as the required answer.

Below is the implementation of the above approach:


// Java program for the above approach
import java.util.*;
class GFG
// Function to find the maximum value
// possible by rotations of digits of N
static void findLargestRotation(int num)
    // Store the required result
    int ans = num;
    // Store the number of digits
    int len = (int)Math.floor(((int)Math.log10(num)) + 1);
    int x = (int)Math.pow(10, len - 1);
    // Iterate over the range[1, len-1]
    for (int i = 1; i < len; i++) {
        // Store the unit's digit
        int lastDigit = num % 10;
        // Store the remaining number
        num = num / 10;
        // Find the next rotation
        num += (lastDigit * x);
        // If the current rotation is
        // greater than the overall
        // answer, then update answer
        if (num > ans) {
            ans = num;
    // Print the result
// Driver Code
public static void main(String[] args)
    int N = 657;
// This code is contributed by sanjoy_62.




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

Please refer complete article on Maximum value possible by rotating digits of a given number for more details!

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