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 codeint 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!
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