Reversal algorithm for Array rotation

Last Updated : 06 Apr, 2023

Given an array arr[] of size N, the task is to rotate the array by d position to the left.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, d = 2
Output: 3, 4, 5, 6, 7, 1, 2
Explanation: If the array is rotated by 1 position to the left,
it becomes {2, 3, 4, 5, 6, 7, 1}.
When it is rotated further by 1 position,
it becomes: {3, 4, 5, 6, 7, 1, 2}

Input: arr[] = {1, 6, 7, 8}, d = 3
Output: 8, 1, 6, 7

Approach: We have already discussed several methods in this post. The ways discussed there are:

Using another temporary array.
Rotating one by one.
Using a juggling algorithm.

Another Approach (The Reversal Algorithm): Here we will be discussing another method which uses the concept of reversing a part of array. The intuition behind the idea is mentioned below:

Intuition:

If we observe closely, we can see that a group of array elements is changing its position. For example see the following array:
arr[] = {1, 2, 3, 4, 5, 6, 7} and d = 2. The rotated array is {3, 4, 5, 6, 7, 1, 2}

The group having the first two elements is moving to the end of the array. This is like reversing the array.

But the issue is that if we only reverse the array, it becomes {7, 6, 5, 4, 3, 2, 1}.
After rotation the elements in the chunks having the first 5 elements {7, 6, 5, 4, 3} and the last 2 elements {2, 1} should be in the actual order as of the initial array [i.e., {3, 4, 5, 6, 7} and {1, 2}]but here it gets reversed.
So if those blocks are reversed again we get the desired rotated array.

So the sequence of operations is:

Reverse the whole array
Then reverse the last 'd' elements and
Then reverse the first (N-d) elements.

As we are performing reverse operations it is also similar to the following sequence:

Reverse the first 'd' elements
Reverse last (N-d) elements
Reverse the whole array.

Algorithm: The algorithm can be described with the help of the below pseudocode:

Pseudocode:

Algorithm reverse(arr, start, end):
mid = (start + end)/2
loop from i = start to mid:
swap (arr[i], arr[end-(mid-i+1)])

Algorithm rotate(arr, d, N):
reverse(arr, 1, d) ;
reverse(arr, d + 1, N);
reverse(arr, 1, N);

Illustration:

Follow the illustration below to for better understanding of the algorithm and intuition:

For example take the array arr[] = {1, 2, 3, 4, 5, 6, 7} and d = 2.

Array

The rotated array will look like:

Rotated Array

1st Step: Consider the array as a combination of two blocks. One containing the first two elements and the other containing the remaining elements as shown above.

Considered 2 blocks

2nd Step: Now reverse the first d elements. It becomes as shown in the image

Reverse the first K elements

3rd Step: Now reverse the last (N-d) elements. It become as it is shown in the below image:

Reverse the last (N-K) elements

4th Step: Now the array is the exact reversed form of how it should be if left shifted d times. So reverse the whole array and you will get the required rotated array.

The total array is reversed

See that the array is now the same as the rotated array.

Below is the implementation of the above approach:

C++

#include  // for reverse function
#include   // for input/output operations
#include     // for vector container
using namespace std;

// Function to rotate an array by k elements to the right
void rotateArray(vector<int>& arr, int k)
{
    // Find the size of the array
    int n = arr.size();

    // Mod k with the size of the array
    // To handle the case where k is greater than the size of the array
    k %= n;

    // Reverse the entire array
    reverse(arr.begin(), arr.end());

    // Reverse the first k elements
    reverse(arr.begin(), arr.begin() + k);

    // Reverse the remaining n-k elements
    reverse(arr.begin() + k, arr.end());
}

int main()
{
    // Initialize the array
    vector<int> arr = { 1, 2, 3, 4, 5 };

    // Number of elements to rotate to the right
    int k = 2;

    // Call the rotateArray function to rotate the array
    rotateArray(arr, k);

    // Print the rotated array
    for (int i : arr) {
        cout << i << " ";
    }

    // Return 0 to indicate successful termination of the program
    return 0;
}

C++

// C++ program for reversal algorithm
// of array rotation

#include 
using namespace std;

// Function to reverse arr[] 
// from index start to end
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 left rotate arr[] of size n by d
void leftRotate(int arr[], int d, int n)
{
    if (d == 0)
        return;
    
    // In case the rotating factor is
    // greater than array length
    d = d % n;

    reverseArray(arr, 0, d - 1);
    reverseArray(arr, d, n - 1);
    reverseArray(arr, 0, n - 1);
}

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

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    // Function call
    leftRotate(arr, d, N);
    printArray(arr, N);
    return 0;
}

// C/C++ program for reversal algorithm of array rotation
#include 

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

/* Utility function to reverse arr[] from start to end */
void reverseArray(int arr[], int start, int end);

/* Function to left rotate arr[] of size n by d */
void leftRotate(int arr[], int d, int n)
{

    if (d == 0)
        return;
    // in case the rotating factor is
    // greater than array length
    d = d % n;

    reverseArray(arr, 0, d - 1);
    reverseArray(arr, d, n - 1);
    reverseArray(arr, 0, n - 1);
}

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

/*Function to reverse arr[] from index start to end*/
void reverseArray(int arr[], int start, int end)
{
    int temp;
    while (start < end) {
        temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

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

    leftRotate(arr, d, n);
    printArray(arr, n);
    return 0;
}

Java

// Java program for reversal algorithm of array rotation
import java.io.*;

class LeftRotate {
    /* Function to left rotate arr[] of size n by d */
    static void leftRotate(int arr[], int d)
    {

        if (d == 0)
            return;

        int n = arr.length;
        // in case the rotating factor is
        // greater than array length
        d = d % n;
        reverseArray(arr, 0, d - 1);
        reverseArray(arr, d, n - 1);
        reverseArray(arr, 0, n - 1);
    }

    /*Function to reverse arr[] from index start to end*/
    static void reverseArray(int arr[], int start, int end)
    {
        int temp;
        while (start < end) {
            temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }

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

    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
        int n = arr.length;
        int d = 2;

        leftRotate(arr, d); // Rotate array by d
        printArray(arr);
    }
}
/*This code is contributed by Devesh Agrawal*/

Python3

# Python program for reversal algorithm of array rotation

# Function to reverse arr[] from index start to end


def reverseArray(arr, start, end):
    while (start < end):
        temp = arr[start]
        arr[start] = arr[end]
        arr[end] = temp
        start += 1
        end = end-1

# Function to left rotate arr[] of size n by d


def leftRotate(arr, d):

    if d == 0:
        return
    n = len(arr)
    # in case the rotating factor is
    # greater than array length
    d = d % n
    reverseArray(arr, 0, d-1)
    reverseArray(arr, d, n-1)
    reverseArray(arr, 0, n-1)

# Function to print an array


def printArray(arr):
    for i in range(0, len(arr)):
        print (arr[i],end=' ')


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

leftRotate(arr, d)  # Rotate array by 2
printArray(arr)

# This code is contributed by Devesh Agrawal

// C# program for reversal algorithm
// of array rotation
using System;

class GFG {
    /* Function to left rotate arr[]
    of size n by d */
    static void leftRotate(int[] arr, int d)
    {

        if (d == 0)
            return;
        int n = arr.Length;
          // in case the rotating factor is
        // greater than array length
        d = d % n;
        reverseArray(arr, 0, d - 1);
        reverseArray(arr, d, n - 1);
        reverseArray(arr, 0, n - 1);
    }

    /* Function to reverse arr[] from
    index start to end*/
    static void reverseArray(int[] arr, int start,
                             int end)
    {
        int temp;
        while (start < end) {
            temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }

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

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

        leftRotate(arr, d); // Rotate array by 2
        printArray(arr);
    }
}

// This code is contributed by Sam007

PHP


// PHP program for reversal 
// algorithm of array rotation

/* Function to left rotate
   arr of size n by d */
function leftRotate(&$arr, $d, $n)
{

    if ($d == 0)
        return;
      // in case the rotating factor is
      // greater than array length
      $d = ($d % $n);
    reverseArray($arr, 0, $d - 1);
    reverseArray($arr, $d, $n - 1);
    reverseArray($arr, 0, $n - 1);
}

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

// Driver code
$arr = array(1, 2, 3, 
             4, 5, 6, 7);
$n = sizeof($arr);
$d = 2;

// Function calling
leftRotate($arr, $d, $n);
printArray($arr, $n);
    
// This code is contributed
// by ChitraNayal
?>

JavaScript

<script>
      // JavaScript program for reversal algorithm
      // of array rotation

      /*Function to reverse arr[] from index start to end*/
      function reverseArray(arr, start, end) {
        while (start < end) {
          var temp = arr[start];
          arr[start] = arr[end];
          arr[end] = temp;
          start++;
          end--;
        }
      }

      /* Function to left rotate arr[] of size n by d */
      function leftRotate(arr, d, n) {
        if (d == 0) return;
        // in case the rotating factor is
        // greater than array length
        d = d % n;

        reverseArray(arr, 0, d - 1);
        reverseArray(arr, d, n - 1);
        reverseArray(arr, 0, n - 1);
      }

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

      /* Driver program to test above functions */

      var arr = [1, 2, 3, 4, 5, 6, 7];
      var n = arr.length;
      var d = 2;

      // Function calling
      leftRotate(arr, d, n);
      printArray(arr, n);
      
      // This code is contributed by rdtank.
    </script>

Output

3 4 5 6 7 1 2

Time Complexity: O(N)
Auxiliary Space: O(1)

Another Method :- Using C++ STL reverse

C++

#include  // for reverse function
#include   // for input/output operations
#include     // for vector container
using namespace std;

// Function to rotate an array by k elements to the right
void rotateArray(vector<int>& arr, int k)
{
    // Find the size of the array
    int n = arr.size();

    // Mod k with the size of the array
    // To handle the case where k is greater than the size of the array
    k %= n;

    // Reverse the entire array
    reverse(arr.begin(), arr.end());

    // Reverse the first k elements
    reverse(arr.begin(), arr.begin() + k);

    // Reverse the remaining n-k elements
    reverse(arr.begin() + k, arr.end());
}

int main()
{
    // Initialize the array
    vector<int> arr = { 1, 2, 3, 4, 5 };

    // Number of elements to rotate to the right
    int k = 2;

    // Call the rotateArray function to rotate the array
    rotateArray(arr, k);

    // Print the rotated array
    for (int i : arr) {
        cout << i << " ";
    }

    // Return 0 to indicate successful termination of the program
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Collections;

public class Main {
    public static void main(String[] args)
    {
        // Initialize the array
        ArrayList<Integer> arr = new ArrayList<>();
        arr.add(1);
        arr.add(2);
        arr.add(3);
        arr.add(4);
        arr.add(5);
        // Number of elements to rotate to the right
        int k = 2;

        // Call the rotateArray function to rotate the array
        rotateArray(arr, k);

        // Print the rotated array
        for (int i : arr) {
            System.out.print(i + " ");
        }
    }

    // Function to rotate an array by k elements to the
    // right
    public static void rotateArray(ArrayList<Integer> arr,
                                   int k)
    {
        // Find the size of the array
        int n = arr.size();

        // Mod k with the size of the array
        // To handle the case where k is greater than the
        // size of the array
        k %= n;

        // Reverse the entire array
        Collections.reverse(arr);

        // Reverse the first k elements
        for (int i = 0; i < k / 2; i++) {
            int temp = arr.get(i);
            arr.set(i, arr.get(k - i - 1));
            arr.set(k - i - 1, temp);
        }

        // Reverse the remaining n-k elements
        for (int i = k; i < (n + k) / 2; i++) {
            int temp = arr.get(i);
            arr.set(i, arr.get(n + k - i - 1));
            arr.set(n + k - i - 1, temp);
        }
    }
}

Python3

# Function to rotate an array by k elements to the right
def rotateArray(arr, k):
    # Find the size of the array
    n = len(arr);

    # Mod k with the size of the array
    # To handle the case where k is greater than the size of the array
    k %= n;

    # Reverse the entire array
    arr[0:n] = arr[0:n][::-1]

    # Reverse the first k elements
    arr[0:k] = arr[0:k][::-1]

    # Reverse the remaining n-k elements
    arr[k:n] = arr[k:n][::-1]

# Initialize the array
arr = [ 1, 2, 3, 4, 5 ];

# Number of elements to rotate to the right
k = 2;

# Call the rotateArray function to rotate the array
rotateArray(arr, k);

# Print the rotated array
for i in range(0,len(arr)):
    print(arr[i], end= " ");

JavaScript

// Define the function to rotate an array
function rotateArray(arr, k) {
// Find the length of the array
const n = arr.length;

// Mod k with the length of the array
// To handle the case where k is greater than the
// length of the array
k %= n;

// Reverse the entire array
arr.reverse();

// Reverse the first k elements
for (let i = 0; i < k / 2; i++) {
const temp = arr[i];
arr[i] = arr[k - i - 1];
arr[k - i - 1] = temp;
}

// Reverse the remaining n-k elements
for (let i = k; i < (n + k) / 2; i++) {
const temp = arr[i];
arr[i] = arr[n + k - i - 1];
arr[n + k - i - 1] = temp;
}
}

// Initialize the array
const arr = [1, 2, 3, 4, 5];

// Number of elements to rotate to the right
const k = 2;

// Call the rotateArray function to rotate the array
rotateArray(arr, k);

// Print the rotated array
console.log(arr.join(' '));

using System;
using System.Linq;

namespace ArrayRotation {
class Program {
    static void Main(string[] args)
    {
        // Initialize the array
        int[] arr = { 1, 2, 3, 4, 5 };

        // Number of elements to rotate to the right
        int k = 2;

        // Call the RotateArray function to rotate the array
        RotateArray(ref arr, k);

        // Print the rotated array
        Console.WriteLine(string.Join(" ", arr));

        // Wait for the user to end the program
        Console.ReadKey();
    }

    // Function to rotate an array by k elements to the
    // right
    static void RotateArray(ref int[] arr, int k)
    {
        // Find the size of the array
        int n = arr.Length;

        // Mod k with the size of the array
        // To handle the case where k is greater than the
        // size of the array
        k %= n;

        // Reverse the first n-k elements
        Array.Reverse(arr, 0, n - k);

        // Reverse the remaining k elements
        Array.Reverse(arr, n - k, k);

        // Reverse the entire array
        Array.Reverse(arr);
    }
}
}

Output

4 5 1 2 3

Time Complexity: O(N)
Auxiliary Space: O(1)

Print left rotation of array in O(n) time and O(1) space

kartik

Improve

Article Tags :

Practice Tags :

Reversal algorithm for Array rotation

Similar Reads

Basic operations in Array

Easy problems on Array

Intermediate problems on Array

Hard problems on Array