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 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
System.out.print(ans);
} // Driver Code public 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!