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Program for array rotation
  • Difficulty Level : Medium
  • Last Updated : 07 Apr, 2021
 

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

Array

Rotation of the above array by 2 will make array
 

ArrayRotation1

 



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]

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

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 : 
 

C++




// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;
 
/*Function to left Rotate arr[] of
  size n by 1*/
void leftRotatebyOne(int arr[], int n)
{
    int temp = arr[0], i;
    for (i = 0; i < n - 1; i++)
        arr[i] = arr[i + 1];
 
    arr[n-1] = temp;
}
 
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
    for (int i = 0; i < d; i++)
        leftRotatebyOne(arr, n);
}
 
/* utility function to print an array */
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; 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 left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);
 
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
    int i;
    for (i = 0; i < d; i++)
        leftRotatebyOne(arr, n);
}
 
void leftRotatebyOne(int arr[], int n)
{
    int temp = arr[0], i;
    for (i = 0; i < n - 1; i++)
        arr[i] = arr[i + 1];
    arr[n-1] = temp;
}
 
/* utility function to print an array */
void printArray(int arr[], int n)
{
    int i;
    for (i = 0; i < n; i++)
        printf("%d ", arr[i]);
}
 
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    leftRotate(arr, 2, 7);
    printArray(arr, 7);
    return 0;
}
Java



// Java program to rotate an array by
// d elements
 
class RotateArray {
    /*Function to left rotate arr[] of size n by d*/
    void leftRotate(int arr[], int d, int n)
    {
        for (int i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
 
    void leftRotatebyOne(int arr[], int n)
    {
        int i, temp;
        temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
        arr[n-1] = temp;
    }
 
    /* utility function to print an array */
    void printArray(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // 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):
    for i in range(d):
        leftRotatebyOne(arr, n)
 
# Function to left Rotate arr[] of size n by 1*/
def leftRotatebyOne(arr, n):
    temp = arr[0]
    for i in range(n-1):
        arr[i] = arr[i + 1]
    arr[n-1] = temp
         
 
# utility function to print an array */
def printArray(arr, size):
    for i in range(size):
        print ("% d"% arr[i], end =" ")
 
  
# Driver program to test above functions */
arr = [1, 2, 3, 4, 5, 6, 7]
leftRotate(arr, 2, 7)
printArray(arr, 7)
 
# 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)
    {
        for (int i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
 
    static void leftRotatebyOne(int[] arr, int n)
    {
        int i, temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
 
        arr[n-1] = temp;
    }
 
    /* utility 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 };
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}
 
// This code is contributed by Sam007
PHP



<?php
// PHP program to rotate an array
// by d elements
 
/*Function to left Rotate arr[] 
of size n by 1*/
function leftRotatebyOne(&$arr, $n)
{
    $temp = $arr[0];
    for ($i = 0; $i < $n - 1; $i++)
        $arr[$i] = $arr[$i + 1];
 
    $arr[$n-1] = $temp;
}
 
/*Function to left rotate arr[]
  of size n by d*/
function leftRotate(&$arr, $d, $n)
{
    for ($i = 0; $i < $d; $i++)
        leftRotatebyOne($arr, $n);
}
 
/* utility function to print
   an array */
function printArray(&$arr, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] . " ";
}
 
// Driver Code
$arr = array( 1, 2, 3, 4, 5, 6, 7 );
$n = sizeof($arr);
 
// Function calling
leftRotate($arr, 2, $n);
printArray($arr, $n);
 
// This code is contributed
// by ChitraNayal
?>
Javascript



<script>
 
// JavaScript program to rotate an array by
// d elements
 
 
    /* Function to left rotate arr of size n by d */
    function leftRotate(arr , d , n) {
        for (i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
 
    function leftRotatebyOne(arr , n) {
        var i, temp;
        temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
        arr[n - 1] = temp;
    }
 
    /* utility function to print an array */
    function printArray(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }
 
    // Driver program to test above functions
     
        var arr = [ 1, 2, 3, 4, 5, 6, 7 ];
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
 
// This code is contributed by todaysgaurav
 
</script>
Output :  
3 4 5 6 7 1 2 
Time complexity : O(n * d) 
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 in different sets 
where number of sets is equal to GCD of n and d and move 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;
 
/*Fuction 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);
 
/*Fuction 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]);
}
 
/*Fuction 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] + " ");
    }
 
    /*Fuction 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 =" ")
 
# Fuction 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] + " ");
    }
 
    /* Fuction 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
Output : 
3 4 5 6 7 1 2 
Time complexity : O(n) 
Auxiliary Space : O(1)
 
 
Please see 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 bug in above programs/algorithms.
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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