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>

Output:

765

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