Rotate digits of a given number by K
Given two integers N and K, the task is to rotate the digits of N by K. If K is a positive integer, left rotate its digits. Otherwise, right rotate its digits.
Examples:
Input: N = 12345, K = 2
Output: 34512
Explanation:
Left rotating N(= 12345) by K(= 2) modifies N to 34512.
Therefore, the required output is 34512Input: N = 12345, K = -3
Output: 34512
Explanation:
Right rotating N(= 12345) by K( = -3) modifies N to 34512.
Therefore, the required output is 34512
Approach: Follow the steps below to solve the problem:
- Initialize a variable, say X, to store the count of digits in N.
- Update K = (K + X) % X to reduce it to a case of left rotation.
- Remove the first K digits of N and append all the removed digits to the right of the digits of N.
- Finally, print the value of N.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the count of// digits in Nint numberOfDigit(int N){ // Stores count of // digits in N int digit = 0; // Calculate the count // of digits in N while (N > 0) { // Update digit digit++; // Update N N /= 10; } return digit;}// Function to rotate the digits of N by Kvoid rotateNumberByK(int N, int K){ // Stores count of digits in N int X = numberOfDigit(N); // Update K so that only need to // handle left rotation K = ((K % X) + X) % X; // Stores first K digits of N int left_no = N / (int)(pow(10, X - K)); // Remove first K digits of N N = N % (int)(pow(10, X - K)); // Stores count of digits in left_no int left_digit = numberOfDigit(left_no); // Append left_no to the right of // digits of N N = (N * (int)(pow(10, left_digit))) + left_no; cout << N;}// Driver codeint main(){ int N = 12345, K = 7; // Function Call rotateNumberByK(N, K); return 0;}// The code is contributed by Dharanendra L V |
chevron_right
filter_none
Java
// Java program to implement// the above approachimport java.io.*;class GFG { // Function to find the count of // digits in N static int numberOfDigit(int N) { // Stores count of // digits in N int digit = 0; // Calculate the count // of digits in N while (N > 0) { // Update digit digit++; // Update N N /= 10; } return digit; } // Function to rotate the digits of N by K static void rotateNumberByK(int N, int K) { // Stores count of digits in N int X = numberOfDigit(N); // Update K so that only need to // handle left rotation K = ((K % X) + X) % X; // Stores first K digits of N int left_no = N / (int)(Math.pow(10, X - K)); // Remove first K digits of N N = N % (int)(Math.pow(10, X - K)); // Stores count of digits in left_no int left_digit = numberOfDigit(left_no); // Append left_no to the right of // digits of N N = (N * (int)(Math.pow(10, left_digit))) + left_no; System.out.println(N); } // Driver Code public static void main(String args[]) { int N = 12345, K = 7; // Function Call rotateNumberByK(N, K); }} |
chevron_right
filter_none
Python3
# Python3 program to implement# the above approach# Function to find the count of# digits in Ndef numberOfDigit(N): # Stores count of # digits in N digit = 0 # Calculate the count # of digits in N while (N > 0): # Update digit digit += 1 # Update N N //= 10 return digit# Function to rotate the digits of N by Kdef rotateNumberByK(N, K): # Stores count of digits in N X = numberOfDigit(N) # Update K so that only need to # handle left rotation K = ((K % X) + X) % X # Stores first K digits of N left_no = N // pow(10, X - K) # Remove first K digits of N N = N % pow(10, X - K) # Stores count of digits in left_no left_digit = numberOfDigit(left_no) # Append left_no to the right of # digits of N N = N * pow(10, left_digit) + left_no print(N)# Driver Codeif __name__ == '__main__': N, K = 12345, 7 # Function Call rotateNumberByK(N, K) # This code is contributed by mohit kumar 29 |
chevron_right
filter_none
C#
// C# program to implement// the above approachusing System;class GFG { // Function to find the count of // digits in N static int numberOfDigit(int N) { // Stores count of // digits in N int digit = 0; // Calculate the count // of digits in N while (N > 0) { // Update digit digit++; // Update N N /= 10; } return digit; } // Function to rotate the digits of N by K static void rotateNumberByK(int N, int K) { // Stores count of digits in N int X = numberOfDigit(N); // Update K so that only need to // handle left rotation K = ((K % X) + X) % X; // Stores first K digits of N int left_no = N / (int)(Math.Pow(10, X - K)); // Remove first K digits of N N = N % (int)(Math.Pow(10, X - K)); // Stores count of digits in left_no int left_digit = numberOfDigit(left_no); // Append left_no to the right of // digits of N N = (N * (int)(Math.Pow(10, left_digit))) + left_no; Console.WriteLine(N); } // Driver Code public static void Main(string []args) { int N = 12345, K = 7; // Function Call rotateNumberByK(N, K); }}// This code is contributed by AnkThon |
chevron_right
filter_none
Output:
34512
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
