Program for array rotation

Write a function rotate(arr[], d, n) that rotates arr[] of size n by d elements. 

Rotation of the above array by 2 will make an array

Method 1 (Using temp array) :

Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store the first d elements in a temp array
   temp[] = [1, 2]
2) Shift rest of the arr[]
   arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
   arr[] = [3, 4, 5, 6, 7, 1, 2]

Below is the implementation of the above approach : 

Python3

def rotate(L,d,n):
    k=L.index(d)
    new_lis=[]
    new_lis=L[k+1:]+L[0:k+1]
    return new_lis

d = 2
n = 7
print(rotate([1,2,3,4,5,6,7],d,n))

Output

[3, 4, 5, 6, 7, 1, 2]

Time complexity : O(n) 
Auxiliary Space : O(n)

Method 2 (Rotate one by one):

leftRotate(arr[], d, n)
start
  For i = 0 to i < d
    Left rotate all elements of arr[] by one
end

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2 
Rotate arr[] by one 2 times 
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.

Below is the implementation of the above approach:

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int a[] = { 1, 2, 3, 4, 5, 6, 7 };
        int n = a.length;
        // Rotate 2 times
        int d = 2;

        int p = 1;
        while (p <= d) {
            int last = a[0];
            for (int i = 0; i < n - 1; i++) {
                a[i] = a[i + 1];
            }
            a[n - 1] = last;
            p++;
        }

        System.out.println("Array after Rotation :");
        for (int i = 0; i < n; i++) {
            System.out.print(a[i] + " ");
        }
    }
}
// contributed by keerthikarathan123

Output

Array after Rotation :
3 4 5 6 7 1 2 

Time Complexity: O(N*d), Where N is the length of the given array and d is the rotation number.
Auxiliary Space: O(1)

Method 3 (A Juggling Algorithm) :
This is an extension of method 2. Instead of moving one by one, divide the array into different sets 
where the number of sets is equal to the GCD of n and d and moves the elements within sets. 
If GCD is 1 as-is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.
Here is an example for n =12 and d = 3. GCD is 3 and 

Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

a) Elements are first moved in first set – (See below 
   diagram for this movement)

      arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}

b)    Then in second set.
          arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}

c)    Finally in third set.
          arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}

Below is the implementation of the above approach :

C++

// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
    if (b == 0)
        return a;

    else
        return gcd(b, a % b);
}

/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (int i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        int temp = arr[i];
        int j = i;

        while (1) {
            int k = j + d;
            if (k >= n)
                k = k - n;

            if (k == i)
                break;

            arr[j] = arr[k];
            j = k;
        }
        arr[j] = temp;
    }
}

// Function to print an array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function calling
    leftRotate(arr, 2, n);
    printArray(arr, n);

    return 0;
}

C

// C program to rotate an array by
// d elements
#include <stdio.h>

/* function to print an array */
void printArray(int arr[], int size);

/*Function to get gcd of a and b*/
int gcd(int a, int b);

/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
    int i, j, k, temp;
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        temp = arr[i];
        j = i;
        while (1) {
            k = j + d;
            if (k >= n)
                k = k - n;
            if (k == i)
                break;
            arr[j] = arr[k];
            j = k;
        }
        arr[j] = temp;
    }
}

/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int n)
{
    int i;
    for (i = 0; i < n; i++)
        printf("%d ", arr[i]);
}

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    leftRotate(arr, 2, 7);
    printArray(arr, 7);
    getchar();
    return 0;
}

Java

// Java program to rotate an array by
// d elements
class RotateArray {
    /*Function to left rotate arr[] of siz n by d*/
    void leftRotate(int arr[], int d, int n)
    {
        /* To handle if d >= n */
        d = d % n;
        int i, j, k, temp;
        int g_c_d = gcd(d, n);
        for (i = 0; i < g_c_d; i++) {
            /* move i-th values of blocks */
            temp = arr[i];
            j = i;
            while (true) {
                k = j + d;
                if (k >= n)
                    k = k - n;
                if (k == i)
                    break;
                arr[j] = arr[k];
                j = k;
            }
            arr[j] = temp;
        }
    }

    /*UTILITY FUNCTIONS*/

    /* function to print an array */
    void printArray(int arr[], int size)
    {
        int i;
        for (i = 0; i < size; i++)
            System.out.print(arr[i] + " ");
    }

    /*Function to get gcd of a and b*/
    int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
        RotateArray rotate = new RotateArray();
        int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
        rotate.leftRotate(arr, 2, 7);
        rotate.printArray(arr, 7);
    }
}

// This code has been contributed by Mayank Jaiswal

Python3

# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
    d = d % n
    g_c_d = gcd(d, n)
    for i in range(g_c_d):

        # move i-th values of blocks
        temp = arr[i]
        j = i
        while 1:
            k = j + d
            if k >= n:
                k = k - n
            if k == i:
                break
            arr[j] = arr[k]
            j = k
        arr[j] = temp

# UTILITY FUNCTIONS
# function to print an array
def printArray(arr, size):
    for i in range(size):
        print ("% d" % arr[i], end =" ")

# Function to get gcd of a and b
def gcd(a, b):
    if b == 0:
        return a;
    else:
        return gcd(b, a % b)

# Driver program to test above functions
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)

# This code is contributed by Shreyanshi Arun

C#

// C# program for array rotation
using System;

class GFG {
    /* Function to left rotate arr[]
    of size n by d*/
    static void leftRotate(int[] arr, int d,
                           int n)
    {
        int i, j, k, temp;
        /* To handle if d >= n */
        d = d % n;
        int g_c_d = gcd(d, n);
        for (i = 0; i < g_c_d; i++) {
            /* move i-th values of blocks */
            temp = arr[i];
            j = i;
            while (true) {
                k = j + d;
                if (k >= n)
                    k = k - n;
                if (k == i)
                    break;
                arr[j] = arr[k];
                j = k;
            }
            arr[j] = temp;
        }
    }

    /*UTILITY FUNCTIONS*/
    /* Function to print an array */
    static void printArray(int[] arr, int size)
    {
        for (int i = 0; i < size; i++)
            Console.Write(arr[i] + " ");
    }

    /* Function to get gcd of a and b*/
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }

    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}

// This code is contributed by Sam007

Javascript

<script>

// JavaScript program to rotate an array by
// d elements

/*Function to get gcd of a and b*/
function gcd( a, b){
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

/*Function to left rotate arr[] of siz n by d*/
function leftRotate(arr, d, n){
    /* To handle if d >= n */
    d = d % n;
    let g_c_d = gcd(d, n);
    for (let i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        let temp = arr[i];
        let j = i;

        while (1) {
            let k = j + d;
            if (k >= n)
                k = k - n;

            if (k == i)
                break;

            arr[j] = arr[k];
            j = k;
        }
        arr[j] = temp;
    }
}

// Function to print an array
function printArray(arr, size){
    for (let i = 0; i < size; i++)
        document.write(arr[i] +" ");
}

/* Driver program to test above functions */
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);

</script>

Output

3 4 5 6 7 1 2 

Time complexity : O(n) 
Auxiliary Space : O(1)

Please see the following posts for other methods of array rotation: 
Block swap algorithm for array rotation
Reversal algorithm for array rotation
Please write comments if you find any bugs in the above programs/algorithms.