Right rotate given Array K times using Pointers
Last Updated :
19 Jan, 2022
Given an array arr[] of size N and an integer K, the task is to right rotate the array K times.
Examples:
Input: arr[] = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5
Explanation: After 1st rotation - {9, 1, 3, 5, 7}
After 2nd rotation - {7, 9, 1, 3, 5}
Input: {1, 2, 3, 4, 5, 6}, K = 2
Output: 5 6 1 2 3 4
Approach: The naive approach and approach based on reversing parts of the array is discussed here.
Pointer based approach: The base of this concept is the reversal algorithm for array rotation. The array is divided into two parts where the first part is of size (N-K) and the end part is of size K. These two parts are individually reversed. Then the whole array is reversed.
Below is the implementation of the above approach:
C++
// C++ code to implement above approach
#include <iostream>
using namespace std;
// Function to print the array
void print(int arr[], int N)
{
for (int i = 0; i < N; i++)
cout << *(arr + i) << " ";
}
// Function to reverse the array
// from start to end index
void reverse(int arr[], int start, int end)
{
int temp;
int size = end - start;
// Reversal based on pointer approach
for (int i = 0; i < (size / 2); i++) {
temp = *(arr + i + start);
*(arr + i + start) = *(arr + start
+ size - i - 1);
*(arr + start + size - i - 1) = temp;
}
}
// Function to right rotate the array K times
void right(int arr[], int K, int N)
{
reverse(arr, 0, N - K);
reverse(arr, N - K, N);
reverse(arr, 0, N);
print(arr, N);
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
right(arr, K, N);
return 0;
}
// Java code to implement above approach
import java.util.*;
class GFG{
// Function to print the array
static void print(int arr[], int N)
{
for (int i = 0; i < N; i++)
System.out.print(arr[i]+ " ");
}
// Function to reverse the array
// from start to end index
static int[] reverse(int arr[], int start, int end)
{
int temp;
int size = end - start;
// Reversal based on pointer approach
for (int i = 0; i < (size / 2); i++) {
temp = arr[ i + start];
arr[i + start] = arr[start
+ size - i - 1];
arr[start + size - i - 1] = temp;
}
return arr;
}
// Function to right rotate the array K times
static void right(int arr[], int K, int N)
{
arr = reverse(arr, 0, N - K);
arr = reverse(arr, N - K, N);
arr = reverse(arr, 0, N);
print(arr, N);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = arr.length;
int K = 2;
right(arr, K, N);
}
}
// This code is contributed by 29AjayKumar
# Python code to implement above approach
# Function to print array
def print1(arr, N):
for i in range(N):
print(arr[i], end = " ");
# Function to reverse the array
# from start to end index
def reverse(arr, start, end):
temp = 0;
size = end - start;
# Reversal based on pointer approach
for i in range(size//2):
temp = arr[i + start];
arr[i + start] = arr[start + size - i - 1];
arr[start + size - i - 1] = temp;
return arr;
# Function to right rotate the array K times
def right(arr, K, N):
arr = reverse(arr, 0, N - K);
arr = reverse(arr, N - K, N);
arr = reverse(arr, 0, N);
print1(arr, N);
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5, 6];
N = len(arr);
K = 2;
right(arr, K, N);
# This code is contributed by 29AjayKumar
// C# code to implement above approach
using System;
class GFG {
// Function to print the array
static void print(int[] arr, int N)
{
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Function to reverse the array
// from start to end index
static void reverse(int[] arr, int start, int end)
{
int temp;
int size = end - start;
// Reversal based on pointer approach
for (int i = 0; i < (size / 2); i++) {
temp = arr[i + start];
arr[i + start] = arr[start + size - i - 1];
arr[start + size - i - 1] = temp;
}
}
// Function to right rotate the array K times
static void right(int[] arr, int K, int N)
{
reverse(arr, 0, N - K);
reverse(arr, N - K, N);
reverse(arr, 0, N);
print(arr, N);
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6 };
int N = arr.Length;
int K = 2;
right(arr, K, N);
}
}
// This code is contributed by ukasp.
<script>
// JavaScript code to implement above approach
// Function to print the array
const print = (arr, N) => {
for (let i = 0; i < N; i++)
document.write(`${arr[i]} `);
}
// Function to reverse the array
// from start to end index
const reverse = (arr, start, end) => {
let temp;
let size = end - start;
// Reversal based on pointer approach
for (let i = 0; i < parseInt(size / 2); i++) {
temp = arr[i + start];
arr[i + start] = arr[start + size - i - 1];
arr[start + size - i - 1] = temp;
}
}
// Function to right rotate the array K times
const right = (arr, K, N) => {
reverse(arr, 0, N - K);
reverse(arr, N - K, N);
reverse(arr, 0, N);
print(arr, N);
}
// Driver code
let arr = [1, 2, 3, 4, 5, 6];
let N = arr.length;
let K = 2;
right(arr, K, N);
// This code is contributed by rakeshsahni
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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