Reversal algorithm for Array rotation (original) (raw)

Last Updated : 23 Jul, 2025

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& 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 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 <bits/stdc++.h> 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/C++ program for reversal algorithm of array rotation #include <stdio.h>

/*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#

// 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

d,d, d,n) { if ($d == 0) return; // in case the rotating factor is // greater than array length d=(d = (d=(d % $n); reverseArray($arr, 0, $d - 1); reverseArray($arr, d,d, d,n - 1); reverseArray($arr, 0, $n - 1); } /*Function to reverse $arr from index start to end*/ function reverseArray(&$arr, start,start, start,end) { while ($start < $end) { temp=temp = temp=arr[$start]; arr[arr[arr[start] = arr[arr[arr[end]; arr[arr[arr[end] = $temp; $start++; $end--; } } // Function to // print an array function printArray($arr, $size) { for ($i = 0; i<i < i<size; $i++) print arr[arr[arr[i]." "; } // Driver code $arr = array(1, 2, 3, 4, 5, 6, 7); n=sizeof(n = sizeof(n=sizeof(arr); $d = 2; // Function calling leftRotate($arr, d,d, d,n); printArray($arr, $n); // This code is contributed // by ChitraNayal ?>

JavaScript

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& 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 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 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(' '));

C#

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);
}

} }

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