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.