Right rotation of an array

Given an array, right rotate it by k elements.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} k = 3 Output: 8 9 10 1 2 3 4 5 6 7 Input: arr[] = {121, 232, 33, 43 ,5} k = 2 Output: 43 5 121 232 33

Note : In the below solution, k is assumed to be smaller than or equal to n. We can easily modify the solutions to handle larger k values by doing k = k % n
Algorithm:

rotate(arr[], d, n) reverseArray(arr[], 0, n-1) ; reverse(arr[], 0, d-1); reverse(arr[], d, n-1);

Below is the implementation of above approach:

  • C++

// C++ program for right rotation of

// an array (Reversal Algorithm)

#include <bits/stdc++.h>

/*Function to reverse arr[]

from index start to end*/

void reverseArray( int arr[], int start,

`` int end)

{

`` while (start < end)

`` {

`` std::swap(arr[start], arr[end]);

`` start++;

`` end--;

`` }

}

/* Function to right rotate arr[]

of size n by d */

void rightRotate( int arr[], int d, int n)

{

`` reverseArray(arr, 0, n-1);

`` reverseArray(arr, 0, d-1);

`` reverseArray(arr, d, n-1);

}

/* function to print an array */

void printArray( int arr[], int size)

{

`` for ( int i = 0; i < size; i++)

`` std::cout << arr[i] << " " ;

}

// driver code

int main()

{

`` int arr[] = {1, 2, 3, 4, 5,

`` 6, 7, 8, 9, 10};

``

`` int n = sizeof (arr)/ sizeof (arr[0]);

`` int k = 3;

``

`` rightRotate(arr, k, n);

`` printArray(arr, n);

`` return 0;

}

Output:

8 9 10 1 2 3 4 5 6 7