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++
#include <bits/stdc++.h>
void reverseArray( int arr[], int start,
int end)
{
while (start < end)
{
std::swap(arr[start], arr[end]);
start++;
end--;
}
}
void rightRotate( int arr[], int d, int n)
{
d=d%n;
reverseArray(arr, 0, n-1);
reverseArray(arr, 0, d-1);
reverseArray(arr, d, n-1);
}
void printArray( int arr[], int size)
{
for ( int i = 0; i < size; i++)
std::cout << arr[i] << " " ;
}
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
import java.io.*;
class GFG
{
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--;
}
}
static void rightRotate( int arr[], int d, int n)
{
reverseArray(arr, 0 , n - 1 );
reverseArray(arr, 0 , d - 1 );
reverseArray(arr, d, n - 1 );
}
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);
}
}
|
Python3
def reverseArray( arr, start, end):
while (start < end):
arr[start], arr[end] = arr[end], arr[start]
start = start + 1
end = end - 1
def rightRotate( arr, d, n):
reverseArray(arr, 0 , n - 1 );
reverseArray(arr, 0 , d - 1 );
reverseArray(arr, d, n - 1 );
def printArray( arr, size):
for i in range ( 0 , size):
print (arr[i], end = ' ' )
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 ]
n = len (arr)
k = 3
rightRotate(arr, k, n)
printArray(arr, n)
|
C#
using System;
class GFG {
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--;
}
}
static void rightRotate( int []arr, int d, int n)
{
reverseArray(arr, 0, n - 1);
reverseArray(arr, 0, d - 1);
reverseArray(arr, d, n - 1);
}
static void printArray( int []arr, int size)
{
for ( int i = 0; i < size; i++)
Console.Write(arr[i] + " " );
}
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);
}
}
|
PHP
<?php
function reverseArray(& $arr , $start , $end )
{
while ( $start < $end )
{
$temp = $arr [ $start ];
$arr [ $start ] = $arr [ $end ];
$arr [ $end ] = $temp ;
$start ++;
$end --;
}
}
function rightRotate(& $arr , $d , $n )
{
reverseArray( $arr , 0, $n - 1);
reverseArray( $arr , 0, $d - 1);
reverseArray( $arr , $d , $n - 1);
}
function printArray(& $arr , $size )
{
for ( $i = 0; $i < $size ; $i ++)
echo $arr [ $i ] . " " ;
}
$arr = array (1, 2, 3, 4, 5,
6, 7, 8, 9, 10);
$n = sizeof( $arr );
$k = 3;
rightRotate( $arr , $k , $n );
printArray( $arr , $n );
?>
|
Javascript
<script>
function reverseArray(arr, start, end){
while (start < end){
let temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
return arr;
}
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 printArray( arr, size){
for (let i = 0; i < size; i++)
document.write( arr[i] + " " );
}
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.
Last Updated :
01 Aug, 2022
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