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:
Python3
# Python program for the above approach # Function to find the maximum value # possible by rotations of digits of N def findLargestRotation(num): # Store the required result ans = num # Store the number of digits length = len ( str (num)) x = 10 * * (length - 1 ) # Iterate over the range[1, len-1] for i in range ( 1 , length): # Store the unit's digit 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 print (ans) # Driver Code N = 657 findLargestRotation(N) # This code is contributed by rohitsingh07052. |
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!
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