Today's algorithm is the Rotate Array problem:

Given an array, rotate the array to the right by k steps, where k is non-negative.

Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem. Could you do it in place with O(1) extra space?.

For example, if you were given the array `[1, 2, 3, 4, 5]`

, and told to rotate it `2`

steps to the right, the output should be `[4, 5, 1, 2, 3]`

. After 1 step, the array would be `[5, 1, 2, 3, 4]`

, so that after 2 steps it would be `[4, 5, 1, 2, 3]`

.

On Leetcode, this problem is labeled "easy"--how they determine difficulty level, I'm not sure. However, I think this problem is by no means an "easy" one. There are many ways to solve this problem, which is part of why I like it, and I think each solution is complicated in its own way.

In this blog post, I'll walk through three different ways to approach and solve this problem: (1) popping and unshifting the elements in the array, (2) creating a new array where the elements start out shifted, and (3) reversing different sections of the array.

## Approach #1: Popping and Unshifting

When working with arrays, a few methods come up all the time. One of which is `.pop()`

, which "removes the last element from an array and returns that element" (you can read more about .pop() here). For example:

```
const arr = [1, 2, 3]
arr.pop() // would return 3
console.log(arr) // would return [1, 2]
```

Another common method used on arrays is `.unshift()`

. This method "adds one or more elements to the beginning of an array and returns the new length of the array" (you can read more about .unshift() here). For example:

```
const arr = [2, 3]
arr.unshift(1) // would return 3, the new length of the array
console.log(arr) // would return [1, 2, 3]
```

Rotating an array right can also be thought of as moving the elements from the back of the array to the front of the array. In this problem, we want to move elements from the back of the array to the front, doing so `k`

times. In a for loop, which will run `k`

times, we can pop the last number off the back of the array, and unshift that number to the front of the array.

For example, let's say we were given the array `nums = [1, 2, 3, 4, 5]`

, and `k = 2`

, so we should rotate the array 2 times. Using pop and unshift, we'd start by popping off the last element, `5`

, which would make `nums`

be `[1, 2, 3, 4]`

. Then, we'd unshift `5`

, putting it at the front of the array, so that `nums`

is `[5, 1, 2, 3, 4]`

.

We'd repeat this cycle once more, popping off the `4`

, making `nums = [5, 1, 2, 3]`

, and then unshifting the 4, giving us the final answer of `[4, 5, 1, 2, 3]`

.

### Coding the first approach

Before we start coding this solution, there's one more thing to note about this problem: let's say that the given array was `[1, 2]`

, and we were told to rotate it to the right 7 times. The array is less than 7 elements long, so rotating it 7 times would be a lot of unnecessary work. Therefore, before we do anything, both in this solution and in the other approaches, we should modify `k`

using modulo (%).

The modulo operator returns the remainder after dividing one number by another. For example, `10%3`

would return `1`

, because 10/3 has a remainder of 1. Similarly, in this problem, we'd want to set `k`

equal to `k % nums.length`

. Using the same example, if `k = 7`

, and `nums = [1, 2]`

, then `k = k % nums.length`

is the same as `k = 7%2`

, or `k=1`

. The first line of this solution, therefore, will be this line.

```
function rotate1(nums, k) {
k = k % nums.length;
//...
}
```

We want to do `.pop()`

and `.unshift()`

as many times as `k`

equals, so we'll make a for loop that goes on `k`

times. Inside the for loop, we'll store the result of `nums.pop()`

to a variable called `back`

. Then, we will unshift `back`

, putting in at the start of the `nums`

array.

Once the for loop stops executing, we'll return `nums`

.

```
function rotate1(nums, k) {
k = k % nums.length;
for (let i = 0; i < k; i++) {
const back = nums.pop();
nums.unshift(back);
}
return nums;
}
```

This first approach is done in linear time (O(n)) and constant space (O(1)).

## Approach #2: Creating a New Array

In the second approach, we'll be creating a new array, where the elements have moved over `k`

spaces. The idea behind this approach is that we can just iterate through the `nums`

array, and move each element `k`

spaces to the right of where it already was.

What happens if the element is supposed to move to a index that's longer than the length of the `nums`

array? In that case, you'd want to use the modulo operator, calculating the result of moving to the new distance % the length of the `nums`

array. I think this is a particularly tricky part of this approach, so I'll use an example.

Let's say you're starting out with the array `nums`

, which is `[1, 2, 3]`

and a blank array `arr`

, and we're told `k=2`

, so the array will move over 2 spots to the right. We can start out by moving the first element of the `nums`

array, `1`

. `1`

is at index 0 (`i = 0`

), and we'll want to move it 2 spots over. In other words, we'll want its position in the `arr`

array to be determined by `i + k`

, which is index 2.

Now, we're on index 1 of the `nums`

array, `2`

. We want to move it `k`

steps to the right, but `i + k`

is 3, and that would be longer than the length of the `nums`

array. So, to find the new spot for `2`

, we should do `(i + k) % nums.length`

, or `3 % 3`

, which is `0`

. So, we should move the element `2`

to the index `0`

in `arr`

.

Finally, we're on the index 2 of the `nums`

array, which is `3`

. We want to move it `k`

steps to the right, and `i + k`

is 4, which is longer than the length of the `nums`

array. So, to find the new spot for `3`

, we should do `(i + k) % nums.length`

, or `4 % 3`

, which is `1`

. So, we should move the element `3`

to the index `1`

in `arr`

, giving us the final result of this problem.

### Coding the second approach

To start out this solution, we'll do the same modifications to `k`

that we did in the first approach. We'll then initialize a new, empty array called `arr`

.

```
function rotate2(nums, k) {
k = k % nums.length;
let arr = [];
//...
}
```

Now, we'll use a for loop to go through every element in `nums`

. At each index, we'll place that element in the new spot in `arr`

. We can find that new spot by doing `(i + k) % nums.length`

. So, we'll set `arr[(i + k) % nums.length]`

equal to `nums[i]`

.

```
function rotate2(nums, k) {
k = k % nums.length;
let arr = [];
for (let i = 0; i < nums.length; i++) {
arr[(i + k) % nums.length] = nums[i];
}
//...
}
```

Now, `arr`

will be the rotated array that we want. However, in this problem, we should be modifying the `nums`

array, so we have to set each index in `nums`

equal to the value at that index in `arr`

. To do this, we can set up another for loop. At each index, we'll set `nums[i]`

equal to `arr[i]`

. When the for loop ends, we can return `nums`

.

```
function rotate2(nums, k) {
k = k % nums.length;
let arr = [];
for (let i = 0; i < nums.length; i++) {
arr[(i + k) % nums.length] = nums[i];
}
for (let i = 0; i < nums.length; i++) {
nums[i] = arr[i];
}
return nums;
}
```

This second approach is done in linear time (O(n)) and linear space (O(n)).

## Approach #3: Reversing the Sections

In this third approach, we're going to be reversing parts of the `nums`

array three times. The first time, we'll reverse the entire array. The second time, we'll reverse the first `k`

elements of the array. The third time, we'll reverse the final elements of the array, from `k`

to the end.

The idea behind this approach can best be seen with an example. We'll start with the array `[1, 2, 3, 4, 5]`

, and we're going to want to rotate it 2 steps. We'll start by rotating the entire array.

Now, we'll want to rotate the first `k`

elements. Since `k`

is 2, we'll rotate the elements at 0 and 1.

Finally, we'll rotate the last elements, from index `k`

to the end. This gives us the final array that we want.

### Coding the third approach

To code this solution, we'll write a function called `reverse`

within the `rotate`

function, and we'll call it three times. To start, however, we'll do the same modification to `k`

that we did in the previous two approaches.

```
function rotate3(nums, k) {
k = k % nums.length;
//...
}
```

Then, we'll call the function `reverse`

(which we'll write in a minute), and we'll call it three times. `reverse()`

will take in the array, the index to start reversing, and the index to end reversing. So, the first call to `reverse()`

will pass in `nums`

, `0`

(as the start index), and `nums.length-1`

(as the end index). The second call to `reverse()`

will pass in `nums`

, `0`

(as the start index), and `k-1`

(as the end index). The third call to `reverse()`

will pass in `nums`

, `k`

(as the start index), and `nums.length-1`

(as the end index).

```
function rotate3(nums, k) {
k = k % nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
//...
}
```

Now, we can write the function `reverse`

, whose parameters will be `nums`

, `start`

, and `end`

. In this function, we'll switch the values at the start and end index, and will move start and end toward the center. We'll keep doing this as long as start is less than end.

So, we'll write a while loop, that will keep going as long as start is less than end. Inside the loop, we'll keep a temporary variable that will store the value of the nums array at the start index. Then, we'll set the value at the start index equal to the value at the end index, and the value at the end index equal to the temporary variable. We'll move start toward the middle by incrementing it, and we'll move the end toward the middle by decrementing it. Finally, when the while loop is done executing, we'll return `nums`

to the `rotate`

function.

```
function rotate3(nums, k) {
k = k % nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
//...
function reverse(nums, start, end) {
while (start < end) {
let temporary = nums[start];
nums[start] = nums[end];
nums[end] = temporary;
start++;
end--;
}
return nums;
}
}
```

Once each `reverse()`

function is done executing, the final thing to do will be to return `nums`

.

```
function rotate3(nums, k) {
k = k % nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
return nums;
function reverse(nums, start, end) {
while (start < end) {
let temporary = nums[start];
nums[start] = nums[end];
nums[end] = temporary;
start++;
end--;
}
return nums;
}
}
```

This solution is done in linear time (O(n)) and constant space (O(1)).

--

Let me know in the comments if you have any questions or ideas for other ways to solve this!

## Discussion (18)

After reading the problem, just on top of my head, this can be easily solved with a ring buffer. Best of all, it can be implemented in any language.

Quick and really dirty and not exactly efficient :D

But it works (mostly)

A good point, one of those if you knew the actual problem you'd find a way of solving it better than actually manipulating an array which feels "expensive".

Yeah immer.js allows you to mutate arrays and then it works out an immutable version from it. That's very cool, but still has a cost as I replied in the comments to that article with a sandbox example showing (on my machine at least) significant frame drops in some use cases.

It so happened the example I had given and then implemented in immer is a kind of kryptonite to immer.js (because it happens over multiple cycles) but yeah, I see things on DEV all the time that use spread in loops - I just think, hope no one calls that example with 1 million entries not the example 10!

Hi,

I followed this article and thank you for doing it.

Can I ask you how did you think about them?

I mean, you have used some particular pseudocode or resource?

You know what a pleasure to solve these problems , but make a huge effort yet.

P.S Sorry for my bad English

I would have though the shortest version of (1) would be:

Though I guess the splice allocates an array of size n % length

Glad you like it :) Honestly only do that when it matters though - I mean I write code that uses all of the slow things when I know it will not be handling lots of data because it's quicker to write and that is the important thing for the task at hand!

I guess my point was that the array is a specific data structure, I expect it to be optimised for being an array. If my problem is something that includes a "rotation" then a rolling buffer may well be a better algorithm - requires no shifting, just a very slightly more complicated retrieve operation and then move a head and tail pointer.

For example: if you are constantly inserting all over a list and only rarely retrieving all the items then no contiguous array, no matter how optimised, will be anywhere near the performance of a linked list.

If I want to implement a thing that has the last 100 values in it then shifting an array for every insertion is massively more costly than increment and modulus. Getting a new array with slice is allocating memory every time. It's easy to write the code, but the consequences may be significant under load.

It's probably the game programmer in me coming out. Don't allocate things because something has to garbage collect them later. That takes time, causes glitches etc.

I wrote the below a while ago about this. It really doesn't matter if you are sure the code isn't going to work on giant data or be called frequently. If that is the case then my "dont allocate/copy/garbage collect" OCD kicks in :)

## Tight Code 1: When immutability goes bad

## Mike Talbot ãƒ» May 27 ãƒ» 3 min read

There's also the reference implementation for C++'s rotate() method

It has the ability to rotate a smaller sub-section of an array as well. The array that started as

`[array[0], ... array[first], array[first+1], ..., array[middle], array[middle + 1], .... array[last -1] (array[last], which might be the magical just off the edge n+1th item), ...]`

will become

`[array[0], ... array[middle], array[middle+1], ..., array[last -1], array[first], array[first + 1], .... (array[last], which is the first index that does not move), ...]`

C++ has some non-javascriptisms, so let's a js transliteration:

Nice post! I have two solutions, neither of which is better than yours, but still might be interesting.

Approach 1: Functional style, easy to read but slow

Approach 2: using recursion and modifying the array in place

Nice approach, just wanted to point out the typo, in the 3rd approach it says coding the second approach and in the following paragraph instead of num.0 it is num,0.

I never thought of using modulo. Nice post!