Dada una array arr[] de tamaño N y un valor K, la tarea es imprimir la array rotada K veces a la derecha.
Ejemplos:
Entrada: arr = {1, 3, 5, 7, 9}, K = 2
Salida: 7 9 1 3 5Entrada: arr = {1, 2, 3, 4, 5}, K = 4
Salida: 2 3 4 5 1
Algoritmo: el problema dado se puede resolver invirtiendo los subarreglos . Se pueden seguir los siguientes pasos para resolver el problema:
- Invierta todos los elementos de la array de 1 a N -1
- Invierta los elementos de la array de 1 a K – 1
- Invierta los elementos de la array de K a N -1
C++
// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to rotate the array
// to the right, K times
void RightRotate(int Array[], int N, int K)
{
// Reverse all the array elements
reverse(Array, Array + N);
// Reverse the first k elements
reverse(Array, Array + K);
// Reverse the elements from K
// till the end of the array
reverse(Array + K, Array + N);
// Print the array after rotation
for (int i = 0; i < N; i++) {
cout << Array[i] << " ";
}
cout << endl;
}
// Driver code
int main()
{
// Initialize the array
int Array[] = { 1, 2, 3, 4, 5 };
// Find the size of the array
int N = sizeof(Array) / sizeof(Array[0]);
// Initialize K
int K = 4;
// Call the function and
// print the answer
RightRotate(Array, N, K);
return 0;
}
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG
{
// Function to rotate the array
// to the right, K times
static void RightRotate(int[] Array, int N, int K)
{
// Reverse all the array elements
for (int i = 0; i < N / 2; i++) {
int temp = Array[i];
Array[i] = Array[N - i - 1];
Array[N - i - 1] = temp;
}
// Reverse the first k elements
for (int i = 0; i < K / 2; i++) {
int temp = Array[i];
Array[i] = Array[K - i - 1];
Array[K - i - 1] = temp;
}
// Reverse the elements from K
// till the end of the array
for (int i = 0; i < (K + N) / 2; i++) {
int temp = Array[(i + K) % N];
Array[(i + K) % N] = Array[(N - i + K - 1) % N];
Array[(N - i + K - 1) % N] = temp;
}
// Print the array after rotation
for (int i = 0; i < N; i++) {
System.out.print(Array[i] + " ");
}
System.out.println();
}
// Driver code
public static void main(String[] args)
{
// Initialize the array
int Array[] = { 1, 2, 3, 4, 5 };
// Find the size of the array
int N = Array.length;
// Initialize K
int K = 4;
// Call the function and
// print the answer
RightRotate(Array, N, K);
}
}
// This code is contributed by maddler.
Python3
# Python program for the above approach import math # Function to rotate the array # to the right, K times def RightRotate(Array, N, K): # Reverse all the array elements for i in range(math.ceil(N / 2)): temp = Array[i] Array[i] = Array[N - i - 1] Array[N - i - 1] = temp # Reverse the first k elements for i in range(math.ceil(K / 2)): temp = Array[i] Array[i] = Array[K - i - 1] Array[K - i - 1] = temp # Reverse the elements from K # till the end of the array for i in range(math.ceil((K + N) / 2)): temp = Array[(i + K) % N] Array[(i + K) % N] = Array[(N - i + K - 1) % N] Array[(N - i + K - 1) % N] = temp # Print the array after rotation for i in range(N): print(Array[i], end=" ") # Driver Code arr = [1, 2, 3, 4, 5] N = len(arr) K = 4 # Call the function and # print the answer RightRotate(arr, N, K) # This code is contributed by Saurabh Jaiswal
C#
// C# program for the above approach
using System;
class GFG
{
// Function to rotate the array
// to the right, K times
static void RightRotate(int []Array, int N, int K)
{
// Reverse all the array elements
for (int i = 0; i < N / 2; i++) {
int temp = Array[i];
Array[i] = Array[N - i - 1];
Array[N - i - 1] = temp;
}
// Reverse the first k elements
for (int i = 0; i < K / 2; i++) {
int temp = Array[i];
Array[i] = Array[K - i - 1];
Array[K - i - 1] = temp;
}
// Reverse the elements from K
// till the end of the array
for (int i = 0; i < (K + N) / 2; i++) {
int temp = Array[(i + K) % N];
Array[(i + K) % N] = Array[(N - i + K - 1) % N];
Array[(N - i + K - 1) % N] = temp;
}
// Print the array after rotation
for (int i = 0; i < N; i++) {
Console.Write(Array[i] + " ");
}
}
// Driver code
public static void Main()
{
// Initialize the array
int []Array = { 1, 2, 3, 4, 5 };
// Find the size of the array
int N = Array.Length;
// Initialize K
int K = 4;
// Call the function and
// print the answer
RightRotate(Array, N, K);
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// Javascript program for the above approach
// Function to rotate the array
// to the right, K times
function RightRotate(Array, N, K)
{
// Reverse all the array elements
for (let i = 0; i < N / 2; i++) {
let temp = Array[i];
Array[i] = Array[N - i - 1];
Array[N - i - 1] = temp;
}
// Reverse the first k elements
for (let i = 0; i < K / 2; i++) {
let temp = Array[i];
Array[i] = Array[K - i - 1];
Array[K - i - 1] = temp;
}
// Reverse the elements from K
// till the end of the array
for (let i = 0; i < (K + N) / 2; i++) {
let temp = Array[(i + K) % N];
Array[(i + K) % N] = Array[(N - i + K - 1) % N];
Array[(N - i + K - 1) % N] = temp;
}
// Print the array after rotation
for (let i = 0; i < N; i++) {
document.write(Array[i] + " ");
}
}
// Driver Code
let arr = [ 1, 2, 3, 4, 5 ];
let N = arr.length;
let K =4;
// Call the function and
// print the answer
RightRotate(arr, N, K);
// This code is contributed by Samim Hossain Mondal.
</script>
Producción
2 3 4 5 1
Complejidad de Tiempo : O(N)
Espacio Auxiliar : O(1)