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 Nvoid 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 Codeint main(){ int N = 657; findLargestRotation(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the maximum value// possible by rotations of digits of Nstatic 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 Codepublic 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 Ndef 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 CodeN = 657findLargestRotation(N)# This code is contributed by rohitsingh07052. |
C#
// C# program for the above approachusing System;class GFG{// Function to find the maximum value// possible by rotations of digits of Nstatic 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 Codepublic 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 Nfunction 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 Codelet 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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