Java Program to Find Maximum value possible by rotating digits of a given number
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
import java.util.*;
class GFG
{
static void findLargestRotation( int num)
{
int ans = num;
int len = ( int )Math.floor((( int )Math.log10(num)) + 1 );
int x = ( int )Math.pow( 10 , len - 1 );
for ( int i = 1 ; i < len; i++) {
int lastDigit = num % 10 ;
num = num / 10 ;
num += (lastDigit * x);
if (num > ans) {
ans = num;
}
}
System.out.print(ans);
}
public static void main(String[] args)
{
int N = 657 ;
findLargestRotation(N);
}
}
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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!
Last Updated :
27 Jan, 2022
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