**How do you rotate one array in Python?** : Step 1: Add an element to the array . Second step: Put the final component in a variable , like x. 3. Advance all elements by one position. In step 4, swap out the array’s initial element for x.

1.

In this blog post, we will be discussing how to rotate an array in Python. We will first discuss what an array is and what is meant by rotating an array. We will then see how to rotate an array in Python with some examples.

## What is an array?

An array is a data structure that stores a collection of elements. The elements in an array can be of the same data type or different data types. The size of an array is fixed.

## What is meant by rotating an array?

Rotating an array means moving the elements in the array to a different position. The rotation can be to the right or to the left.

## How do you rotate an array in Python?

There are different ways to rotate an array in Python. We will discuss two methods here.

The first method is to use the built-in function rotate(). The rotate() function takes two arguments. The first argument is the array to be rotated. The second argument is the number of places to rotate the array. If the second argument is positive, the array is rotated to the right. If the second argument is negative, the array is rotated to the left.

The second method is to use the slice operator. The slice operator can be used to select a range of elements from an array. We can use the slice operator to rotate an array. To rotate an array to the right, we can use the following code:

array[::-1]

To rotate an array to the left, we can use the following code:

array[::1]

## Examples

Let us see some examples of how to rotate an array in Python.

Example 1: Rotate an Array to the Right

In this example, we will rotate an array to the right by two places.

array = [1, 2, 3, 4, 5]

array.rotate(2)

print(array)

Output:

[4, 5, 1, 2, 3]

Example 2: Rotate an Array to the Left

In this example, we will rotate an array to the left by two places.

array = [1, 2, 3, 4, 5]

array[::-1]

print(array)

Output:

[5, 4, 3, 2, 1]

6.

In this blog post, we have seen how to rotate an array in Python. We have seen two methods to do this. The first method is to use the built-in function rotate(). The second method is to use the slice operator.

**How do you rotate an array and time in Python?**: print(arr[i]), Rotate the given array by n times toward left for i in range(0, n): Stores the first element of the array

**What is left and right rotation of array?**: Array Rotation simply means shifting the array elements to the left or right of the array by specified positions. An array can be rotated to the left(clockwise) or to the right (anti-clockwise) to the given number of positions.

## Read Detail Answer On What is left and right rotation of array?

Given an array **arr[]** of size **N** and **D** index, the task is to rotate the array by the **D **index. We have two flexibilities either to rotate them leftwards or rightwards via different ways which we are going to explore by implementing every way of rotating in both of the rotations.

**Output explanation:**

- The initial array [1, 2, 3, 4, 5]
- rotate by first index [2, 3, 4, 5, 1]
- rotate by second index [3, 4, 5, 1, 2]

Input :arr[] = {10, 34, 56, 23, 78, 12, 13, 65} D = 7Output:65 10 34 56 23 78 12 13

In this method simply create a temporary array and copy the elements of the array arr[] from 0 tothe (D-1)th index. After that move, the rest elements of the array arr[] from index D to N. Then move the temporary array elements to the original array.

Input arr[] = [1, 2, 3, 4, 5] D = 2

- Store the first d elements in a temp array:
**temp[] = [1, 2]** - Shift rest of the arr[]:
**arr[] = [3, 4, 5]** - Store back the D elements:
**arr[] = [3, 4, 5, 1, 2]**

**Example:**

## Java

`class`

`GFG {`

` `

`void`

`leftRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`int`

`temp[] = `

`new`

`int`

`[d];`

` `

`for`

`(`

`int`

`i = `

`; i < d; i++)`

` `

`temp[i] = arr[i];`

` `

`for`

`(`

`int`

`i = d; i < n; i++) {`

` `

`arr[i - d] = arr[i];`

` `

`}`

` `

`for`

`(`

`int`

`i = `

`; i < d; i++) {`

` `

`arr[i + n - d] = temp[i];`

` `

`}`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = `

`; i < n; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.leftRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Time complexity:O(N) ,Auxiliary Space:O(D)

**Way 2: **Rotate one by one

**Approach:**

Rotate the array recursively one by one element

Input arr[] = [1, 2, 3, 4, 5]D = 21

- Swap arr[0] to arr[1]
- Swap arr[1] to arr[2]
- Swap arr[N-1] to arr[N]
- Repeat 1, 2, 3 to D times

Store arr[0] in a temporary variable called temp, then move arr[1] to arr[0], arr[2] to arr[1], and temp to arr[n-1].

**Illustration:**

Let us take the same example arr[] = [1, 2, 3, 4, 5], d = 2 Rotate arr[] by one 2 times We get [2, 3, 4, 5, 1] after first rotation and [ 3, 4, 5, 1, 2] after second rotation.

**Example**

## Java

`class`

`GFG {`

` `

`void`

`leftRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = `

`; i < d; i++)`

` `

`leftRotatebyOne(arr, n);`

` `

`}`

` `

`void`

`leftRotatebyOne(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`int`

`i, temp;`

` `

`temp = arr[`

`];`

` `

`for`

`(i = `

`; i < n - `

`1`

`; i++)`

` `

`arr[i] = arr[i + `

`1`

`];`

` `

`arr[i] = temp;`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = `

`; i < n; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.leftRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Time complexity:O(N * D),Auxiliary Space:O(1)

**Way 3: **UsingJuggling Algorithm

**Approach:**

This is an extension of method 2. Instead of moving one by one, divide the array into different sets where the number of sets is equal to GCD of n and d and move the elements within sets.

If GCD is 1 as-isfor the above example array (n = 5 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.

**Example:**

## Java

`class`

`GFG {`

` `

`void`

`leftRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`d = d % n;`

` `

`int`

`i, j, k, temp;`

` `

`int`

`g_c_d = gcd(d, n);`

` `

`for`

`(i = `

`; i < g_c_d; i++) {`

` `

`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;`

` `

`}`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`size)`

` `

`{`

` `

`int`

`i;`

` `

`for`

`(i = `

`; i < size; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`int`

`gcd(`

`int`

`a, `

`int`

`b)`

` `

`{`

` `

`if`

`(b == `

`)`

` `

`return`

`a;`

` `

`else`

` `

`return`

`gcd(b, a % b);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.leftRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Time complexity:O(n) ,Auxiliary Space:O(1)

## Right Rotation of Array

**Illustration:**

Input :arr[] = {1, 2, 3, 4, 5} D = 2Output :4 5 1 2 3

**Output explanation:**

- Theinitial array [1, 2, 3, 4, 5]
- rotate first index [2, 3, 4, 5, 1]
- rotate second index [3, 4, 5, 1, 2]
- rotate third index [4, 5, 1, 2, 3]

Input: arr[] :{10, 34, 56, 23, 78, 12, 13, 65} D = 5Output :56 23 78 12 13 65 10 34

**Way 1:** Using temp array

**Approach:**

In this method simply create a temporary array and copy the elements of the array arr[] from 0 to the **N â€“ D** index. After that move, the rest elements of the array arr[] from index D to N. Then move the temporary arrayelements to the original array. It is as illustrated below illustration as follows:

Input arr[] = [1, 2, 3, 4, 5], D = 21) Store the first d elements in a temp array temp[] = [1, 2, 3]2) Shift rest of the arr[] arr[] = [4, 5]3) Store back the D elements arr[] = [4, 5, 1, 2, 3]

**Example:**

## Java

`class`

`GFG {`

` `

`void`

`rightRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`while`

`(d > n) {`

` `

`d = d - n;`

` `

`}`

` `

`int`

`temp[] = `

`new`

`int`

`[n - d];`

` `

`for`

`(`

`int`

`i = `

`; i < n - d; i++)`

` `

`temp[i] = arr[i];`

` `

`for`

`(`

`int`

`i = n - d; i < n; i++) {`

` `

`arr[i - n + d] = arr[i];`

` `

`}`

` `

`for`

`(`

`int`

`i = `

`; i < n - d; i++) {`

` `

`arr[i + d] = temp[i];`

` `

`}`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = `

`; i < n; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.rightRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Time complexity:O(N) ,Auxiliary Space:O(D)

**Way 2: **Rotate one by one** **

**Approach:** Rotate the array recursively one by one element

Input arr[] = [1, 2, 3, 4, 5], D = 2

- swap arr[N] to arr[N-1]
- swap arr[N-1] to arr[N-2]
- swap arr[2] to arr[1]
- Repeat 1, 2, 3 to D times

To rotate by one, store arr[N] in a temporary variable temp, move arr[N-1] to arr[N], arr[N-2] to arr[N-1] â€¦ and finally temp to arr[1].

Let us take the same example arr[] = [1, 2, 3, 4, 5], d = 2 Rotate arr[] by one 2 times Weget [5, 1, 2, 3, 4] after first rotation and [ 4, 5, 1, 2, 3] after second rotation.

**Example:**

## Java

`class`

`GFG {`

` `

`void`

`rightRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = n; i > d; i--)`

` `

`rightRotatebyOne(arr, n);`

` `

`}`

` `

`void`

`rightRotatebyOne(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`int`

`i, temp;`

` `

`temp = arr[`

`];`

` `

`for`

`(i = `

`; i < n - `

`1`

`; i++)`

` `

`arr[i] = arr[i + `

`1`

`];`

` `

`arr[i] = temp;`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`n)`

` `

`{`

` `

`for`

`(`

`int`

`i = `

`; i < n; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.rightRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Here we have rotated 2 elements one by one else if we would have had rotated by one element only then the array would have been **[2,3,4,5,1]**

Time complexity:O(N * D) ,Auxiliary Space:O(1)

**Way 3: **Juggling Algorithm

**Approach: **This is an extension of method 2. Instead of moving one by one, divide the array into different sets where the 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 = 5 and d =2), then elements will be moved within one set only, we just start withtemp = arr[N] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.

## Java

`class`

`GFG {`

` `

`void`

`rightRotate(`

`int`

`arr[], `

`int`

`d, `

`int`

`n)`

` `

`{`

` `

`d = n - d;`

` `

`d = d % n;`

` `

`int`

`i, j, k, temp;`

` `

`int`

`g_c_d = gcd(d, n);`

` `

`for`

`(i = `

`; i < g_c_d; i++) {`

` `

`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;`

` `

`}`

` `

`}`

` `

`void`

`printArray(`

`int`

`arr[], `

`int`

`size)`

` `

`{`

` `

`int`

`i;`

` `

`for`

`(i = `

`; i < size; i++)`

` `

`System.out.print(arr[i] + `

`" "`

`);`

` `

`}`

` `

`int`

`gcd(`

`int`

`a, `

`int`

`b)`

` `

`{`

` `

`if`

`(b == `

`)`

` `

`return`

`a;`

` `

`else`

` `

`return`

`gcd(b, a % b);`

` `

`}`

` `

`public`

`static`

`void`

`main(String[] args)`

` `

`{`

` `

`GFG rotate = `

`new`

`GFG();`

` `

`int`

`arr[] = { `

`1`

`, `

`2`

`, `

`3`

`, `

`4`

`, `

`5`

`};`

` `

`rotate.rightRotate(arr, `

`2`

`, arr.length);`

` `

`rotate.printArray(arr, arr.length);`

` `

`}`

`}`

Time complexity:O(n) ,Auxiliary Space:O(1)

**How do you rotate an array?**: By moving the array’s elements to a position before them, which can be done by looping through the array and performing the operation arr[j] = arr[j 1], the array can be left rotated. The final element of the rotated array will be added after the array’s first element.

## Read Detail Answer On How do you rotate an array?

The elements of an array in this program must be rotated by the predetermined number of times to the left. Each element of the array will be rotated to the left by one position, and the first element of the array will be added to the end of the list. This procedure will be carried out a certain number of times.

Consider above array, if n is 1 then, all elements of the array will be moved to its left by one position such that second element of the array will take the first position, the third element will be moved to the second position and so on. The first element of the array will be added to the last of the array.

### Algorithm

### Solution

### Python

**Output:**

Original array: 1 2 3 4 5 Array after left rotation: 4 5 1 2 3

### PHP

**Output:**

Original array: 1 2 3 4 5 Array after left rotation: 4 5 1 2 3

Next Topic#

## Additional Question — How do you rotate one array in Python?

### How do you rotate an array 90 degrees?

Rotate Matrix 90 Degree Clockwise or Right Rotation The rotation of a matrix involves two steps: First, find the transpose of the given matrix. Swap the elements of the first column with the last column (if the matrix is of 3*3). The second column remains the same.

### How do I rotate an array to the right?

The array can be right rotated by shifting its elements to a position next to them which can be accomplished by looping through the array in reverse order (loop will start from the length of the array -1 to 0) and perform the operation arr[j] = arr[j-1].

### How we can rotate the array in Java?

To rotate by one, move arr[N] to arr[N-1], arr[N-2] to arr[N-1], and then temp to arr[1]. Temporary variable temp is used to store arr[N]. Following the first rotation, we obtain [5, 1, 2, 3, 4] and following the second rotation, [4, 5, 1, 2, 3].

### How do you rotate an array list in java?

There is no rotate method in Java’s Arrays class. Collections can be used. An array can be quickly rotated using the rotate() function.

### What are the steps to rotate an array by K?

bring up Java. util. arrays; class Main. An array can be rotated one position to the right using this function. rightRotateByOne(int[] A) public static void int last = A[A. length – 1]; where (int i = A. A[i 1] = A[i]; A[0] = last; length – 2; i >= 0; i–)

### How do you rotate an array in C#?

The problem asks us to rotate an array k times given an array and a number, k. Call the reverse method from 0 to the array length in the first step of the C rotation of an array k times. Call the reverse method from 0 to k-1 in step two. Call the k-1 to array length reverse method in the third step.

### How do you reverse an element in an array?

To reverse an array in Java, there are three different techniques. using a for loop to iterate through the array and copy the elements into a different array in the opposite direction. using a technique known as “in-place reversal” in which the elements are switched to put them in the opposite order. utilizing the interface’s reverse method for lists, the Collections.

### How do you rotate a right array in C++?

when (int i = 0; i.

### How do you move values in an array in Java?

Java Shift an ArrayUse a temporary variable and the for loop to perform a Java array shift. In Java 8, you can shift an array by using the skip() method. Use the Collections. Java’s rotate(List list, int distance) method rotates an array.

### How do you shift a list in Python?

Using Python’s list pop(), insert(), and append() functions is the simplest way to shift values in a list. To shift a list, you can also use the deque() data structure from the Python collections module. In Python, a list can also be advanced or reverted using list slicing.

### How do you shift an array position?

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