Rotate an Array by d – Counterclockwise or Left

Try it on GfG Practice

Given an array of integers arr[] of size n, the task is to rotate the array elements to the left by d positions.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}, d = 2
Output: {3, 4, 5, 6, 1, 2}
Explanation: After first left rotation, arr[] becomes {2, 3, 4, 5, 6, 1} and after the second rotation, arr[] becomes {3, 4, 5, 6, 1, 2}

Input: arr[] = {1, 2, 3}, d = 4
Output: {2, 3, 1}
Explanation: The array is rotated as follows:

• After first left rotation, arr[] = {2, 3, 1}
• After second left rotation, arr[] = {3, 1, 2}
• After third left rotation, arr[] = {1, 2, 3}
• After fourth left rotation, arr[] = {2, 3, 1}

[Naive Approach] Rotate one by one – O(n * d) Time and O(1) Space

In each iteration, shift the elements by one position to the left in a circular fashion (the first element becomes the last). Perform this operation d times to rotate the elements to the left by d positions.

Illustration:

Let us take arr[] = {1, 2, 3, 4, 5, 6}, d = 2.

First Step:
        => Rotate to left by one position.
        => arr[] = {2, 3, 4, 5, 6, 1}
Second Step:
        => Rotate again to left by one position
        => arr[] = {3, 4, 5, 6, 1, 2}
Rotation is done 2 times.
So the array becomes arr[] = {3, 4, 5, 6, 1, 2}

// C++ Program to left rotate the array by d positions
// by rotating one element at a time

#include <bits/stdc++.h>
using namespace std;

// Function to left rotate array by d positions
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();
  
    // Repeat the rotation d times
    for (int i = 0; i < d; i++) {
      
        // Left rotate the array by one position
        int first = arr[0];
        for (int j = 0; j < n - 1; j++) {
            arr[j] = arr[j + 1];
        }
        arr[n - 1] = first;
    }
}

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;

    rotateArr(arr, d);

    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";

    return 0;
}

// C Program to left rotate the array by d positions
// by rotating one element at a time

#include <stdio.h>

// Function to left rotate array by d positions
// Repeat the rotation d times
void rotateArr(int arr[], int n, int d) {
    for (int i = 0; i < d; i++) {
      
        // Left rotate the array by one position
        int first = arr[0];
        for (int j = 0; j < n - 1; j++) {
            arr[j] = arr[j + 1];
        }
        arr[n - 1] = first;
    }
}

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

    rotateArr(arr, n, d);

    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);

    return 0;
}

// Java Program to left rotate the array by d positions
// by rotating one element at a time

import java.util.Arrays;

class GfG {
    
    // Function to left rotate array by d positions
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;
  
        // Repeat the rotation d times
        for (int i = 0; i < d; i++) {
          
            // Left rotate the array by one position
            int first = arr[0];
            for (int j = 0; j < n - 1; j++) {
                arr[j] = arr[j + 1];
            }
            arr[n - 1] = first;
        }
    }

    public static void main(String[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
    }
}

# Python Program to left rotate the array by d positions
# by rotating one element at a time

# Function to left rotate array by d positions
def rotateArr(arr, d):
    n = len(arr)
  
    # Repeat the rotation d times
    for i in range(d):
      
        # Left rotate the array by one position
        first = arr[0]
        for j in range(n - 1):
            arr[j] = arr[j + 1]
        arr[n - 1] = first

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2

    rotateArr(arr, d)

    for i in range(len(arr)):
        print(arr[i], end=" ")

// C# Program to left rotate the array by d positions
// by rotating one element at a time

using System;

class GfG {
  
    // Function to left rotate array by d positions
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;
  
        // Repeat the rotation d times
        for (int i = 0; i < d; i++) {
          
            // Left rotate the array by one position
            int first = arr[0];
            for (int j = 0; j < n - 1; j++) {
                arr[j] = arr[j + 1];
            }
            arr[n - 1] = first;
        }
    }

    static void Main() {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}

// JavaScript Program to left rotate the array by d positions
// by rotating one element at a time

// Function to left rotate array by d positions
function rotateArr(arr, d) {
    let n = arr.length;
  
    // Repeat the rotation d times
    for (let i = 0; i < d; i++) {
      
        // Left rotate the array by one position
        let first = arr[0];
        for (let j = 0; j < n - 1; j++) {
            arr[j] = arr[j + 1];
        }
        arr[n - 1] = first;
    }
}

let arr = [1, 2, 3, 4, 5, 6];
let d = 2;

rotateArr(arr, d);

console.log(arr.join(" "));

3 4 5 6 1 2 

Time Complexity: O(n*d), the outer loop runs d times, and within each iteration, the inner loop shifts all n elements of the array by one position, resulting in a total of n*d operations.
Auxiliary Space: O(1)

[Better Approach] Using Temporary Array – O(n) Time and O(n) Space

This problem can be solved using the below idea:

The idea is to use a temporary array of size n, where n is the length of the original array. If we left rotate the array by d positions, the last n – d elements will be at the front and the first d elements will be at the end.

• Copy the last (n – d) elements of original array into the first n – d positions of temporary array.
• Then copy the first d elements of the original array to the end of temporary array.
• Finally, copy all the elements of temporary array back into the original array.

Working:

1 / 4

Below is the implementation of the algorithm:

// C++ Program to left rotate the array by d positions
// using temporary array

#include <bits/stdc++.h>
using namespace std;

// Function to rotate vector
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();

    // Handle case when d > n
    d %= n;
  
    // Storing rotated version of array
    vector<int> temp(n);

    // Copy last n - d elements to the front of temp
    for (int i = 0; i < n - d; i++)
        temp[i] = arr[d + i];

    // Copy the first d elements to the back of temp
    for (int i = 0; i < d; i++)
        temp[n - d + i] = arr[i];

    // Copying the elements of temp in arr
    // to get the final rotated vector
    for (int i = 0; i < n; i++)
        arr[i] = temp[i];
}

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;

    rotateArr(arr, d);

    // Print the rotated vector
    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";

    return 0;
}

// C Program to left rotate the array by d positions
// using temporary array

#include <stdio.h>
#include <stdlib.h>

// Function to rotate array
void rotateArr(int* arr, int d, int n) {
    
    // Handle case when d > n
    d %= n;
  
    // Storing rotated version of array
    int temp[n];

    // Copy last n - d elements to the front of temp
    for (int i = 0; i < n - d; i++)
        temp[i] = arr[d + i];

    // Copy the first d elements to the back of temp
    for (int i = 0; i < d; i++)
        temp[n - d + i] = arr[i];

    // Copying the elements of temp in arr
    // to get the final rotated array
    for (int i = 0; i < n; i++)
        arr[i] = temp[i];
}

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

    rotateArr(arr, d, n);

    // Print the rotated array
    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);

    return 0;
}

// Java Program to left rotate the array by d positions
// using temporary array

import java.util.Arrays;

class GfG {
    
    // Function to rotate array
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;

        // Handle case when d > n
        d %= n;
        
        // Storing rotated version of array
        int[] temp = new int[n];

        // Copy last n - d elements to the front of temp
        for (int i = 0; i < n - d; i++)
            temp[i] = arr[d + i];

        // Copy the first d elements to the back of temp
        for (int i = 0; i < d; i++)
            temp[n - d + i] = arr[i];

        // Copying the elements of temp in arr
        // to get the final rotated array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }

    public static void main(String[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
    }
}

# Python Program to left rotate the array by d positions
# using temporary array

# Function to rotate array
def rotateArr(arr, d):
    n = len(arr)

    # Handle case when d > n
    d %= n
    
    # Storing rotated version of array
    temp = [0] * n

    # Copy last n - d elements to the front of temp
    for i in range(n - d):
        temp[i] = arr[d + i]

    # Copy the first d elements to the back of temp
    for i in range(d):
        temp[n - d + i] = arr[i]

    # Copying the elements of temp in arr
    # to get the final rotated array
    for i in range(n):
        arr[i] = temp[i]

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2

    rotateArr(arr, d)

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

// C# Program to left rotate the array by d positions
// using temporary array

using System;

class GfG {
   
    // Function to rotate array
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;

        // Handle case when d > n
        d %= n;

        // Storing rotated version of array
        int[] temp = new int[n];

        // Copy last n - d elements to the front of temp
        for (int i = 0; i < n - d; i++)
            temp[i] = arr[d + i];

        // Copy the first d elements to the back of temp
        for (int i = 0; i < d; i++)
            temp[n - d + i] = arr[i];

        // Copying the elements of temp in arr
        // to get the final rotated array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }

    static void Main(string[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        // Print the rotated array
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}

// JavaScript Program to left rotate the array by d positions
// using temporary array

// Function to rotate array
function rotateArr(arr, d) {
    let n = arr.length;

    // Handle case when d > n
    d %= n;

    // Storing rotated version of array
    let temp = new Array(n);

    // Copy last n - d elements to the front of temp
    for (let i = 0; i < n - d; i++)
        temp[i] = arr[d + i];

    // Copy the first d elements to the back of temp
    for (let i = 0; i < d; i++)
        temp[n - d + i] = arr[i];

    // Copying the elements of temp in arr
    // to get the final rotated array
    for (let i = 0; i < n; i++)
        arr[i] = temp[i];
}

const arr = [1, 2, 3, 4, 5, 6];
const d = 2;

rotateArr(arr, d);

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

3 4 5 6 1 2 

Time Complexity: O(n), as we are visiting each element only twice.
Auxiliary Space: O(n), as we are using an additional temporary array.

[Expected Approach 1] Using Juggling Algorithm – O(n) Time and O(1) Space

The idea is to use Juggling Algorithm which is based on rotating elements in cycles. Each cycle is independent and represents a group of elements that will shift among themselves during the rotation. If the starting index of a cycle is i, then next elements of the cycle can be found at indices (i + d) % n, (i + 2d) % n, (i + 3d) % n … and so on till we return to the original index i.

So for any index i, we know that after rotation, the element that will occupy this position is arr[(i + d) % n]. Consequently, for every index in the cycle, we will place the element that should be in that position after the rotation is completed.

Please refer Juggling Algorithm for Array Rotation to know more about the implementation.

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

[Expected Approach 2] Using Reversal Algorithm – O(n) Time and O(1) Space

The idea is based on the observation that if we left rotate the array by d positions, the last (n – d) elements will be at the front and the first d elements will be at the end.

• Reverse the subarray containing the first d elements of the array.
• Reverse the subarray containing the last (n – d) elements of the array.
• Finally, reverse all the elements of the array.

Working:

1 / 4

Below is the implementation of the algorithm:

// C++ Code to left rotate an array using Reversal Algorithm

#include <bits/stdc++.h>

using namespace std;

// Function to rotate an array by d elements to the left
void rotateArr(vector<int>& arr, int d) {
    int n = arr.size();
  
    // Handle the case where d > size of array
    d %= n;
  
    // Reverse the first d elements
    reverse(arr.begin(), arr.begin() + d);

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

int main() {
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int d = 2;
    
      rotateArr(arr, d);
  
    for (int i = 0; i < arr.size(); i++) 
        cout << arr[i] << " ";
    return 0;
}

// C Code to left rotate an array using Reversal Algorithm

#include <stdio.h>

// Function to reverse a portion of the array
void reverse(int* arr, int start, int end) {
    while (start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

// Function to rotate an array by d elements to the left
void rotateArr(int* arr, int n, int d) {
    
    // Handle the case where d > size of array
    d %= n;

    // Reverse the first d elements
    reverse(arr, 0, d - 1);

    // Reverse the remaining n-d elements
    reverse(arr, d, n - 1);

    // Reverse the entire array
    reverse(arr, 0, n - 1);
}

int main() {
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;
    
    rotateArr(arr, n, d);
  
    for (int i = 0; i < n; i++) 
        printf("%d ", arr[i]);
    return 0;
}

// Java Code to left rotate an array using Reversal Algorithm

import java.util.Arrays;

class GfG {

    // Function to rotate an array by d elements to the left
    static void rotateArr(int[] arr, int d) {
        int n = arr.length;

        // Handle the case where d > size of array
        d %= n;

        // Reverse the first d elements
        reverse(arr, 0, d - 1);

        // Reverse the remaining n-d elements
        reverse(arr, d, n - 1);

        // Reverse the entire array
        reverse(arr, 0, n - 1);
    }

    // Function to reverse a portion of the array
    static void reverse(int[] arr, int start, int end) {
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }

    public static void main(String[] args) {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
    }
}

# Python Code to left rotate an array using Reversal Algorithm

# Function to rotate an array by d elements to the left
def rotateArr(arr, d):
    n = len(arr)

    # Handle the case where d > size of array
    d %= n

    # Reverse the first d elements
    reverse(arr, 0, d - 1)

    # Reverse the remaining n-d elements
    reverse(arr, d, n - 1)

    # Reverse the entire array
    reverse(arr, 0, n - 1)

# Function to reverse a portion of the array
def reverse(arr, start, end):
    while start < end:
        arr[start], arr[end] = arr[end], arr[start]
        start += 1
        end -= 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6]
    d = 2
    
    rotateArr(arr, d)
  
    for i in range(len(arr)):
        print(arr[i], end=" ")

// C# Code to left rotate an array using Reversal Algorithm

using System;

class GfG {
  
    // Function to rotate an array by d elements to the left
    static void rotateArr(int[] arr, int d) {
        int n = arr.Length;

        // Handle the case where d > size of array
        d %= n;

        // Reverse the first d elements
        reverse(arr, 0, d - 1);

        // Reverse the remaining n-d elements
        reverse(arr, d, n - 1);

        // Reverse the entire array
        reverse(arr, 0, n - 1);
    }

    // Function to reverse a portion of the array
    static void reverse(int[] arr, int start, int end) {
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }

    static void Main() {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int d = 2;

        rotateArr(arr, d);

        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}

// JavaScript Code to left rotate an array using Reversal Algorithm

// Function to rotate an array by d elements to the left
function rotateArr(arr, d) {
    let n = arr.length;

    // Handle the case where d > size of array
    d %= n;

    // Reverse the first d elements
    reverse(arr, 0, d - 1);

    // Reverse the remaining n-d elements
    reverse(arr, d, n - 1);

    // Reverse the entire array
    reverse(arr, 0, n - 1);
}

// Function to reverse a portion of the array
function reverse(arr, start, end) {
    while (start < end) {
        let temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

const arr = [1, 2, 3, 4, 5, 6];
const d = 2;

rotateArr(arr, d);

console.log(arr.join(" "));

3 4 5 6 1 2 

Time complexity: O(n), as we are visiting each element exactly twice.
Auxiliary Space: O(1)

kartik

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