C++ Program for Reversal algorithm for right rotation of an array
Given an array, right rotate it by k elements.
After K=3 rotation
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
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Reversal algorithm for right rotation of an array for more details!