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Right rotate given Array K times using Pointers

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  • 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




// 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

Python3




# 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#




// 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.

Javascript




<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>

 
 

Output

5 6 1 2 3 4 

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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