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
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!
