Not an "Easy" Algorithm: Rotating an Array, Three Ways
Tags: javascript, codenewbie, leetcode, algorithms
Published: 2020-06-12 17:27:35

Today's algorithm is the [Rotate Array problem](https://leetcode.com/problems/rotate-array/):
>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](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/pop)). 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](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/unshift)). 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]`. 

![First line: `nums` on the left, and to the right is a series of blocks for `[1, 2, 3, 4, 5]`. Second line: `nums.pop()` on the left, and `5` in a blue block on its own. Third line: `nums` on the left, and to the right is a series of blocks for `[1, 2, 3, 4]`. Fourth line: `nums.unshift(5)` on the left, and on the right is a series of blocks for `[5, 1, 2, 3, 4]`, and the `5` is blue.](https://dev-to-uploads.s3.amazonaws.com/i/i19zj0wy55573rtnvjoq.png)

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]`.

![First line: `nums.pop()` on the left, and to the right is `4` in a blue block on its own. Second line: `nums` on the left, and to the right is a series of blocks for `[5, 1, 2, 3]`. Third line: `nums.unshift(4)` on the left, and on the right is a series of blocks for `[4, 5, 1, 2, 3]`, and the `4` is blue.](https://dev-to-uploads.s3.amazonaws.com/i/i19zj0wy55573rtnvjoq.png)

### 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.


```javascript
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`.

```javascript
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.

![First row: `nums` on the left, and a series of blocks for `[1, 2, 3]` on the right. Second row: `arr` on the left, and an empty block representing an empty array on the right. Third row: `i = 0; i + k = 2` on the left, and a blue block for `1` on the right. Fourth row: `arr` on the left, and a block with two blank spaces, followed by a blue `1` on the right (representing an array `[<empty>, <empty>, 1]`).](https://dev-to-uploads.s3.amazonaws.com/i/8j0vol0adcm0nr4dh5on.png)

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`.

![First row: `i = 1; i + k = 3; nums.length = 3; 3%3 = 0` on the left, and a blue block for `2` on the right. Second row: `arr` on the left, and a block with the first space as a blue box with `2`, second box is blank, and third box is `1`, on the right (representing an array `[2, <empty>, 1]`).](https://dev-to-uploads.s3.amazonaws.com/i/wyyhg0ij9yuhvqmmd74c.png)

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.

![First row: `i = 2; i + k = 4; nums.length = 3; 4%3 = 1` on the left, and a blue block for `3` on the right. Second row: `arr` on the left, and a block with the first space as `2`, second space is a blue box `3`, and third box is `1`, on the right (representing an array `[2, 3, 1]`).](https://dev-to-uploads.s3.amazonaws.com/i/uw8s3y4hwqcbomto2wfr.png)

### 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`.

```javascript
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]`.

```javascript
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`.


```javascript
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.

![Two arrays represented as a two blocks, with a red array going from the left to right. The entire array on the left has a blue border and is filled with white, and is `[1, 2, 3, 4, 5]`. The array on the right has a black border and is filled with blue, and is `[5, 4, 3, 2, 1]`.](https://dev-to-uploads.s3.amazonaws.com/i/9lzo75p64l2kmdepaice.png)

Now, we'll want to rotate the first `k` elements. Since `k` is 2, we'll rotate the elements at 0 and 1.

![Two arrays represented as a two blocks, with a red array going from the left to right. The first two elements in the array on the left have a blue border and are filled with white. The last three elements have a black border and are filled with white. The array on the left is `[5, 4, 3, 2, 1]`. The array on the right has the first two elements filled with blue with a black border. The last three elements are filled with white with a black border. The array is `[4, 5, 3, 2, 1]`.](https://dev-to-uploads.s3.amazonaws.com/i/o0er5hzw7ga30dx5uw40.png)

Finally, we'll rotate the last elements, from index `k` to the end. This gives us the final array that we want.

![Two arrays represented as a two blocks, with a red array going from the left to right. The last three elements in the array on the left have a blue border and are filled with white. The first two elements have a black border and are filled with white. The array on the left is `[4, 5, 3, 2, 1]`. The array on the right has the first two elements filled with white and have a black border. The last three elements have a black border and are filled with blue. The array is `[4, 5, 1, 2, 3]`.](https://dev-to-uploads.s3.amazonaws.com/i/firkny4nzbs8wk7sbbws.png)

### 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.

```javascript
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).

```javascript
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.

```javascript
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`.


```javascript
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!