Rotation in Array

Program for array rotation

Difficulty Level : Easy
Last Updated : 16 Apr, 2021

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

Rotation of the above array by 2 will make array

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 :

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// 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;

}

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

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