Program for array rotation

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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[i] = 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[i] = 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[i] = 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[i] = 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]; 
  
    $arr[$i] = $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 
?> 

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.

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My Personal Notes

Nov	DEC	Jan
	20
2019	2020	2021

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

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C

Java

Python3

C#

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