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Jared Nielsen

Posted on • Originally published at jarednielsen.com

# How to Code the Merge Sort Algorithm in JavaScript and Python

If you want to learn how to code, you need to learn algorithms. Learning algorithms improves your problem solving skills by revealing design patterns in programming. In this tutorial, you will learn how to code the merge sort algorithm in JavaScript and Python.

## How to Code the Merge Sort Algorithm

Programming is problem solving. There are four steps we need to take to solve any programming problem:

1. Understand the problem

2. Make a plan

3. Execute the plan

4. Evaluate the plan

### Understand the Problem

Why do we need another sorting algorithm? What's the problem with the Bubble, Insertion, and Selection sorting algorithms? Why is their performance so slow? They each use nested iteration. If our goal is to improve the performance of our sorting algorithm, it follows that we need to do less iteration. But how do we do that?

To understand our problem, we first need to define it. Let’s reframe the problem as acceptance criteria, but this time let's up the ante:

GIVEN an array of unsorted numbers
WHEN my algorithm sorts the numbers
THEN the performance is faster than quadratic time complexity

That’s our general outline. We know our input conditions (an unsorted array) and our output requirements (a sorted array), and our goal is to organize the elements in the array in ascending, or non-descending, order with an order better than O(n^2).

Let’s make a plan!

### Make a Plan

Let’s revisit our computational thinking heuristics as they will aid and guide is in making a plan. They are:

• Decomposition

• Pattern recognition

• Abstraction

• Algorithm

If we are writing a sorting algorithm, we need to start with something to sort. Let’s declare an array of ‘unsorted’ integers:

``````[10, 1, 9, 2, 8, 3, 7, 4, 6, 5]
``````

When we are decomposing a problem, we want to break it into the types of problems that need to be solved. We also want to break it into the smallest problems that can be solved.

What’s the smallest problem we can solve?

Two numbers. We can shift the first two off our array…

[10, 1]

… and swap them:

[1, 10]

What is actually happening when we swap values in an array?

We need to slice the values into their own arrays, then concatenate the two arrays in sequential order.

Using our smallest problem as an example, `[10, 1]`, we would break it into the following:

``````[10], [1]
``````

Then do some conditional checkeroos and concatenate the two arrays back into one:

``````[1] +  [10] = [1, 10]
``````

Where have we seen this or something like it before?

Divide and conquer!

As we recalled above, in divide and conquer algorithms, we choose a pivot, and continuously divide the problem in two until we find a solution.

If you recall, our `pivot` is the length of the array divided by two. So, for our smallest problem where the length of our array is 2, our `pivot` will be 1.

Let's start pseudocoding this:

``````IF THE LENGTH OF arr IS LESS THAN OR EQUAL TO 1
RETURN arr

SET pivot TO THE LENGTH OF arr DIVIDED BY 2
``````

Once we set our `pivot`, we can divide the array in two.

``````IF THE LENGTH OF arr IS LESS THAN OR EQUAL TO 1
RETURN arr

SET pivot TO THE LENGTH OF arr DIVIDED BY 2

SET left TO THE ELEMENTS PRECEDING pivot
SET right TO THE ELEMENTS SUCCEEDING pivot
``````

Now what?

Now we need to run those conditional checkerinos and merge the two arrays.

Where have we see this or something like it before?

Merging two arrays!

Because we are pragmatic programmers, we are simply going to "import" that algorithm into this one. We'll do that by wrapping the pseudocode in a "FUNCTION":

``````FUNCTION merge(left, right)

SET result TO AN EMPTY ARRAY

WHILE THERE ARE ELEMENTS IN left OR right

IF THERE ARE ELEMENTS IN BOTH ARRAYS
IF THE FIRST VALUE IN left IS LESS THAN THE FIRST VALUE IN right
SHIFT THE FIRST VALUE OUT OF left AND PUSH IT INTO result
ELSE
SHIFT THE FIRST VALUE OUT OF right AND PUSH IT INTO result
ELSE IF THERE ARE ONLY ELEMENTS IN left
SHIFT THE FIRST VALUE OUT OF left AND PUSH IT INTO result
ELSE
SHIFT THE FIRST VALUE OUT OF right AND PUSH IT INTO result

RETURN result
``````

Now we can simply call our `merge` function in our merge sort algorithm.

``````IF THE LENGTH OF arr IS LESS THAN OR EQUAL TO 1
RETURN arr

SET pivot TO THE LENGTH OF arr DIVIDED BY 2

SET left TO THE ELEMENTS PRECEDING pivot
SET right TO THE ELEMENTS SUCCEEDING pivot

RETURN merge(left, right)
``````

Will this work?

Yes, but only if our array contains two elements.

What happens if we double our problem?

``````[10, 1, 9, 2]
``````

If we follow the logic or our algorithm, we'll divide the array in two:

``````left = [10, 1]
right = [9, 2]
``````

We'll then call our `merge` function like so:

``````merge(left, right)
``````

Within our `merge` function, we will see that the first value in `right` is less than the first value in `left`, so we'll push that to `result`:

``````[9]
``````

Uh oh! We're already in trouble!

In the next iteration we'll see that the first (and only) value in `right` is 2, which is less than the first value in `left`, 10, so we'll push 2 to `result`:

``````[9, 2]
``````

If we follow the logic, we'll end up with a "sorted" array that looks like this:

``````[9, 2, 10, 1]
``````

Not sorted, but definitely swapped!

What's the solution?

Our `merge sort` algorithm was beautiful when we were only working with two values.

How do we continually break a problem down into smaller, yet simliar, problems?

Recursion!

We can make recursive calls to `merge sort` to continually divide our problem in half until we reach our base case, at which point we'll return. And conquer!

Let's update our pseudocode:

``````FUNCTION merge sort
IF THE LENGTH OF arr IS LESS THAN OR EQUAL TO 1
RETURN arr

SET pivot TO THE LENGTH OF arr DIVIDED BY 2

SET left TO THE ELEMENTS PRECEDING pivot
SET right TO THE ELEMENTS SUCCEEDING pivot

RETURN merge(merge sort(left), merge sort(right))
``````

### Execute the Plan

Now it’s simply a matter of translating our pseudocode into the syntax of our programming language.

#### How to Code the Merge Sort Algorithm in JavaScript

``````const merge = (left, right) => {

let result = [];

while(left.length || right.length) {

if(left.length && right.length) {
if(left[0] < right[0]) {
result.push(left.shift())
} else {
result.push(right.shift())
}
} else if(left.length) {
result.push(left.shift())
} else {
result.push(right.shift())
}
}
return result;
};

const mergeSort = (arr) =>{
if(arr.length <= 1) {
return arr;
}

const pivot = arr.length / 2 ;
const left = arr.slice(0, pivot);
const right = arr.slice(pivot, arr.length);

return merge(mergeSort(left), mergeSort(right));
};
``````

#### How to Code the Merge Sort Algorithm in Python

Now let’s see it in Python…

``````def merge(left, right):
result = []

while (len(left) or len(right)):
if (len(left) and len(right)):
if (left[0] < right[0]):
result.append(left.pop(0))
else:
result.append(right.pop(0))
elif (len(left)):
result.append(left.pop(0))
else:
result.append(right.pop(0))
return result

def merge_sort(list):
if (len(list) <= 1):
return list

pivot = len(list) // 2
left = list[0:pivot]
right = list[pivot:]

return merge(merge_sort(left),merge_sort(right))
``````

### Evaluate the Plan

Can we do better?

It depends!

This is a standard implementation of merge sort, but there are still more sort algorithms to learn.

We'll look at quick sort in the next tutorial.

#### What is the Big O Of Merge Sort Sequence?

The order of our `mergeSort` algorithm is O(n log n), or log linear. You can learn more in my article on the topic: Big O Log Linear Time Complexity.

## A is for Algorithms

Give yourself an A. Grab your copy of A is for Algorithms