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:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the maximum value // possible by rotations of digits of N void findLargestRotation( int num) { // Store the required result int ans = num; // Store the number of digits int len = floor ( log10 (num) + 1); int x = 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 cout << ans; } // Driver Code int main() { int N = 657; findLargestRotation(N); return 0; } |
Java
// 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. |
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. |
C#
// C# program for the above approach using System; 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 double lg = ( double )(Math.Log10(num) + 1); int len = ( int )(Math.Floor(lg)); 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 Console.Write(ans); } // Driver Code public static void Main( string [] args) { int N = 657; findLargestRotation(N); } } // This code is contributed by souravghosh0416, |
Javascript
<script> // Javascript program for the above approach // Function to find the maximum value // possible by rotations of digits of N function findLargestRotation(num) { // Store the required result let ans = num; // Store the number of digits let len = Math.floor(Math.log10(num) + 1); let x = Math.pow(10, len - 1); // Iterate over the range[1, len-1] for (let i = 1; i < len; i++) { // Store the unit's digit let lastDigit = num % 10; // Store the remaining number num = parseInt(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 document.write(ans); } // Driver Code let N = 657; findLargestRotation(N); // This code is contributed by souravmahato348. </script> |
765
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Convert number to string, rotate the string and convert back to integer in python:
Approach:
Initialize a variable max_num with the input number n.
Convert the input number n to a string str_num.
Loop through the indices of the string str_num, starting from index 1 to the end of the string.
Inside the loop, slice the string str_num from index 1 to the end and concatenate it with the first character of the string str_num using the string concatenation operator +.
Convert the resulting rotated string to an integer rotated_num.
Check if rotated_num is greater than max_num, if yes, update max_num to rotated_num.
Return max_num.
Test the function with some inputs and measure the execution time using the time module.
C++
#include <iostream> #include <string> #include <ctime> using namespace std; // Import the standard namespace // Function to find the maximum rotation of an integer int maxRotationApproach1( int n) { int maxNum = n; string strNum = to_string(n); // Iterate through all possible rotations for ( int i = 1; i < strNum.length(); ++i) { // Rotate the string to the left strNum = strNum.substr(1) + strNum[0]; int rotatedNum = stoi(strNum); if (rotatedNum > maxNum) { maxNum = rotatedNum; } } return maxNum; } int main() { // Test the function with input N = 657 and N = 7092 clock_t start = clock (); int result1 = maxRotationApproach1(657); int result2 = maxRotationApproach1(7092); // Print the results cout << "Max Rotation for 657: " << result1 << endl; // Output: 765 cout << "Max Rotation for 7092: " << result2 << endl; // Output: 9270 return 0; } |
Java
import java.util.Arrays; public class Main { // Function to find the maximum rotation of an integer static int maxRotationApproach1( int n) { int maxNum = n; String strNum = Integer.toString(n); // Iterate through all possible rotations for ( int i = 1 ; i < strNum.length(); ++i) { // Rotate the string to the left strNum = strNum.substring( 1 ) + strNum.charAt( 0 ); int rotatedNum = Integer.parseInt(strNum); if (rotatedNum > maxNum) { maxNum = rotatedNum; } } return maxNum; } public static void main(String[] args) { // Test the function with input N = 657 and N = 7092 long startTime = System.nanoTime(); int result1 = maxRotationApproach1( 657 ); int result2 = maxRotationApproach1( 7092 ); long endTime = System.nanoTime(); // Print the results System.out.println(result1); // Output: 765 System.out.println(result2); // Output: 9270 } } |
Python3
import time def max_rotation_approach1(n): max_num = n str_num = str (n) for i in range ( 1 , len (str_num)): str_num = str_num[ 1 :] + str_num[ 0 ] rotated_num = int (str_num) if rotated_num > max_num: max_num = rotated_num return max_num # Test the function with input N = 657 and N = 7092 start = time.time() print (max_rotation_approach1( 657 )) # Output: 765 print (max_rotation_approach1( 7092 )) # Output: 9270 end = time.time() |
C#
using System; class Program { // Function to find the maximum rotation of an integer static int MaxRotationApproach1( int n) { int maxNum = n; string strNum = n.ToString(); // Iterate through all possible rotations for ( int i = 1; i < strNum.Length; ++i) { // Rotate the string to the left strNum = strNum.Substring(1) + strNum[0]; int rotatedNum = int .Parse(strNum); if (rotatedNum > maxNum) { maxNum = rotatedNum; } } return maxNum; } static void Main() { // Test the function with input N = 657 and N = 7092 DateTime start = DateTime.Now; int result1 = MaxRotationApproach1(657); int result2 = MaxRotationApproach1(7092); // Print the results Console.WriteLine( "Max Rotation for 657: " + result1); // Output: 765 Console.WriteLine( "Max Rotation for 7092: " + result2); // Output: 9270 TimeSpan elapsed = DateTime.Now - start; Console.WriteLine( "Execution time: " + elapsed.TotalMilliseconds + " ms" ); } } |
Javascript
// Function to find the maximum rotation of an integer function maxRotationApproach1(n) { let maxNum = n; let strNum = n.toString(); // Iterate through all possible rotations for (let i = 1; i < strNum.length; i++) { // Rotate the string to the left strNum = strNum.substring(1) + strNum[0]; let rotatedNum = parseInt(strNum, 10); if (rotatedNum > maxNum) { maxNum = rotatedNum; } } return maxNum; } // Main function function main() { // Test the function with input N = 657 and N = 7092 let start = new Date().getTime(); let result1 = maxRotationApproach1(657); let result2 = maxRotationApproach1(7092); // Print the results console.log( "Max Rotation for 657: " + result1); // Output: 765 console.log( "Max Rotation for 7092: " + result2); // Output: 9270 let end = new Date().getTime(); let timeTaken = end - start; console.log( "Time taken: " + timeTaken + " milliseconds" ); } main(); |
Max Rotation for 657: 765 Max Rotation for 7092: 9270
Time complexity: O(n^2), where n is the number of digits in the input number.
Auxiliary Space: O(n), where n is the number of digits in the input number.
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