Reversal algorithm for right rotation of an array
Last Updated :
01 Aug, 2022
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 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 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# 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 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
?>
<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>
Output8 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.
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