Hello Everyone,
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