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) reverse(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) { // if in case d>n,this will give segmentation fault. d=d%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; } |
Java
// Java program for right rotation of // an array (Reversal Algorithm) import java.io.*; class GFG { // Function to reverse arr[] // from index start to end static void reverseArray( int arr[], int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } } // Function to right rotate // arr[] of size n by d static 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 static void printArray( int arr[], int size) { for ( int i = 0 ; i < size; i++) System.out.print(arr[i] + " " ); } public static void main (String[] args) { int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 }; int n = arr.length; int k = 3 ; rightRotate(arr, k, n); printArray(arr, n); } } // This code is contributed by Gitanjali. |
Python3
# Python3 program for right rotation of # an array (Reversal Algorithm) # Function to reverse arr # from index start to end def reverseArray( arr, start, end): while (start < end): arr[start], arr[end] = arr[end], arr[start] start = start + 1 end = end - 1 # Function to right rotate arr # of size n by d def rightRotate( arr, d, n): reverseArray(arr, 0 , n - 1 ); reverseArray(arr, 0 , d - 1 ); reverseArray(arr, d, n - 1 ); # function to print an array def printArray( arr, size): for i in range ( 0 , size): print (arr[i], end = ' ' ) # Driver code arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 ] n = len (arr) k = 3 # Function call rightRotate(arr, k, n) printArray(arr, n) # This article is contributed # by saloni1297 |
C#
// C# program for right rotation of // an array (Reversal Algorithm) using System; class GFG { // Function to reverse arr[] // from index start to end static void reverseArray( int []arr, int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } } // Function to right rotate // arr[] of size n by d static 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 static void printArray( int []arr, int size) { for ( int i = 0; i < size; i++) Console.Write(arr[i] + " " ); } // Driver code public static void Main () { int []arr = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; int n = arr.Length; int k = 3; rightRotate(arr, k, n); printArray(arr, n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program for right rotation of // an array (Reversal Algorithm) /*Function to reverse arr[] from index start to end*/ function reverseArray(& $arr , $start , $end ) { while ( $start < $end ) { $temp = $arr [ $start ]; $arr [ $start ] = $arr [ $end ]; $arr [ $end ] = $temp ; $start ++; $end --; } } /* Function to right rotate arr[] of size n by d */ function rightRotate(& $arr , $d , $n ) { reverseArray( $arr , 0, $n - 1); reverseArray( $arr , 0, $d - 1); reverseArray( $arr , $d , $n - 1); } /* function to print an array */ function printArray(& $arr , $size ) { for ( $i = 0; $i < $size ; $i ++) echo $arr [ $i ] . " " ; } // Driver code $arr = array (1, 2, 3, 4, 5, 6, 7, 8, 9, 10); $n = sizeof( $arr ); $k = 3; rightRotate( $arr , $k , $n ); printArray( $arr , $n ); // This code is contributed by ita_c ?> |
Javascript
<script> // JavaScript program for right rotation of // an array (Reversal Algorithm) /*Function to reverse arr[] from index start to end*/ function reverseArray(arr, start, end){ while (start < end){ let temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } return arr; } /* Function to right rotate arr[] of size n by d */ function rightRotate(arr, d, n){ arr = reverseArray(arr, 0, n-1); arr = reverseArray(arr, 0, d-1); arr = reverseArray(arr, d, n-1); return arr; } /* function to print an array */ function printArray( arr, size){ for (let i = 0; i < size; i++) document.write( arr[i] + " " ); } // driver code let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; let n = arr.length; let k = 3; arr = rightRotate(arr, k, n); printArray(arr, n); </script> |
Output
8 9 10 1 2 3 4 5 6 7
Time Complexity: O(n), as we are using a while loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.