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Maximum value possible by rotating digits of a given number

  • Difficulty Level : Expert
  • Last Updated : 07 Apr, 2021

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>
 
// Javasript 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>
Output: 
765

 

Time Complexity: O(log10N)
Auxiliary Space: O(1)

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