Given a positive integer N, the task is to find the maximum value among all the rotations of the digits of the integer N.
Examples:
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
Explanation:
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
// Java program for the above approachimport java.util.*;class GFG{  // Function to find the maximum value// possible by rotations of digits of Nstatic 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    System.out.print(ans);}  // Driver Codepublic static void main(String[] args){    int N = 657;    findLargestRotation(N);}}  // This code is contributed by sanjoy_62. |
765
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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!
