Dados dos números enteros N y K , la tarea es rotar los dígitos de N por K. Si K es un número entero positivo, rotar a la izquierda sus dígitos. De lo contrario, gire a la derecha sus dígitos.
Ejemplos:
Entrada: N = 12345, K = 2
Salida: 34512
Explicación
: Girar a la izquierda N(= 12345) por K(= 2) modifica N a 34512.
Por lo tanto, la salida requerida es 34512Entrada: N = 12345, K = -3
Salida: 34512
Explicación
: Girar a la derecha N(= 12345) por K( = -3) modifica N a 34512.
Por lo tanto, la salida requerida es 34512
Enfoque: siga los pasos a continuación para resolver el problema:
- Inicialice una variable, digamos X , para almacenar el recuento de dígitos en N.
- Actualice K = (K + X) % X para reducirlo a un caso de rotación a la izquierda.
- Elimine los primeros K dígitos de N y agregue todos los dígitos eliminados a la derecha de los dígitos de N .
- Finalmente, imprima el valor de N .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of
// digits in N
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
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)(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 code
int main()
{
int N = 12345, K = 7;
// Function Call
rotateNumberByK(N, K);
return 0;
}
// The code is contributed by Dharanendra L V
Java
// Java program to implement
// the above approach
import 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);
}
}
Python3
# Python3 program to implement # the above approach # Function to find the count of # digits in N def 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 K def 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 Code if __name__ == '__main__': N, K = 12345, 7 # Function Call rotateNumberByK(N, K) # This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above approach
using 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
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to find the count of
// digits in N
function numberOfDigit(N)
{
// Stores count of
// digits in N
let digit = 0;
// Calculate the count
// of digits in N
while (N > 0) {
// Update digit
digit++;
// Update N
N = Math.floor( N / 10);
}
return digit;
}
// Function to rotate the digits of N by K
function rotateNumberByK(N, K)
{
// Stores count of digits in N
let X = numberOfDigit(N);
// Update K so that only need to
// handle left rotation
K = ((K % X) + X) % X;
// Stores first K digits of N
let left_no = Math.floor (N / Math.floor(Math.pow(10,
X - K)));
// Remove first K digits of N
N = N % Math.floor(Math.pow(10, X - K));
// Stores count of digits in left_no
let left_digit = numberOfDigit(left_no);
// Append left_no to the right of
// digits of N
N = (N * Math.floor(Math.pow(10, left_digit)))
+ left_no;
document.write(N);
}
// Driver Code
let N = 12345, K = 7;
// Function Call
rotateNumberByK(N, K);
// This code is contributed by souravghosh0416.
</script>
Producción:
34512
Complejidad de tiempo: O(log 10 N)
Espacio auxiliar: O(1)