In my next installment of coding concepts for liberal arts programmers (*i.e. coders who do not come from a math or science background*) we will look at merging sorted arrays in Javascript. A common challenge you will encounter is the presentation of two arrays of sorted integers and the need to merge them together into one large sorted array. How might you do that? Let's take a look at one method.

In our scenario we have two lists of sorted numbers:

```
let firstList = [4, 6, 8, 9]
let secondList = [2, 3, 5, 7]
```

Before even beginning to code out our solution, if you were facing this problem in any other context what would you do first? Well, you would probably go about creating a new list taking the smallest number from each list and adding it to your new list. How would we code that? At the outset we'd want to just find the smallest number from each list and then remove it so our original lists get smaller and less complicated as we go forward:

```
function getSmallestThenRemove(firstList, secondList) {
let smallestFirstList = firstList[0];
let smallestSecondList = secondList[0];
if (smallestFirstList < smallestSecondList) {
return firstList.shift()
} else {
return secondList.shift()
}
}
```

In this function we locate the smallest number by referencing the first index place in each existing array and creating two new variables to hold that data. (*In another post we will discuss how to implement a similar function with two unsorted arrays.*) Then we check which smallest number is smaller. If the first list's smallest number is smaller than the second list's smallest number, we return the first list without that number and, if the opposite is true, we return the second list without its smallest number.

Now that we have a function that finds for us the smallest number, removes it from the list and returns the original list without that smallest number, what do we need to do next?

The next step would be to create another function that calls `getSmallestThenRemove()`

from within itself as it iterates through the two separate arrays, adding each smallest number as they are removed from their original arrays into a new merged array:

```
function mergeLists(firstList, secondList) {
let newList = [];
let iteratedNum;
while (firstList.length != 0 && secondList.length != 0) {
let iteratedNum = getSmallestThenRemove(firstList, secondList)
newList.push(iteratedNum)
}
return newList.concat(firstList).concat(secondList)
}
```

Within the `mergeLists()`

function we are doing several things:

- We create an empty array, which will be the place our new sorted array will live.
- We create a variable
`iteratedNum`

, which will hold the current number we are working with. - We work our way through the original lists until they are empty (
`!= 0`

). Each time, we define the value of`iteratedNum`

to be the return value of`getSmallestThenRemove()`

and push that value into our`newList`

. - Finally, we return the
`newList`

by concatenating the remainder of either`firstList`

or`secondList`

, because once we have worked our way through the function we will be left with one original list that is empty and the other that holds the remainder of our new sorted array.

Thus, returning back to our original two lists, once we run our new function we will return the following:

```
let firstList = [4, 6, 8, 9]
let secondList = [2, 3, 5, 7]
mergeLists(firstList, secondList)
// [2, 3, 4, 5, 6, 7, 8, 9]
```

Check back next week for another installment of coding concepts for liberal arts programmers!

## Discussion

You can also make use of ES6 to merge any number of lists in a single call:

Please mind that when using

`Array.prototype.sort`

without argument, the sorting is done via Unicode endpoints – pretty much an alphabetical sort, which means that this array:will be sorted into this one:

This is a great way to think through and implement the algorithm for solving this problem. This is my take on a simpler, more compact solution using the ES6 spread operator and the built-in

`sort`

method on JS arrays:If you can't use ES6 for whatever reason, the

`concat`

method would allow you to still keep this nice and simple:Those are very compact, love the simplicity of it! My goal in these posts is to breakdown what's happening behind the scenes a bit so I intentionally spell out the steps rather than using a built in method. The idea being that once someone has some idea what's happening "behind the scenes" then they will more fully appreciate a language's built in methods.

`sort()`

should be`sort((a, b) => a - b)`

Otherwise, surprise, surprise -

`[1, 2, 11].sort()`

->`[1, 11, 2]`

I forgot about that, good catch!

Another solution, simple and faster.

sort an array

developer.mozilla.org/en-US/docs/W...

concat 2 arrays:

developer.mozilla.org/en-US/docs/W...

Simpler yes, faster no. If you look at algorithmic complexity, the

`sort()`

function has at least`O(n*log(n))`

. Presorted merge sort (what this basicly is) has a complexity of`O(n)`

.A canonical implementation would be:

As you can see, only one loop, with

`n = |a| + |b|`

executions}

I think that as this is a series for relative beginners, it's important to emphasise the fact that often, simpler code is better than "faster" code. If you're working with arrays as small as in the example, the difference will be essentially negligible.

I usually take the rule that if there is any noticeable slowness in my code, then I go and try and optimise things like this, but in the vast majority of situations (I would suggest), this sort of complexity is overkill.

Yes of course, always write simple first, then measure performance and optimize where needed.

My point was that using properties like presorted arrays can allow you to go beyond the theoretical limit of

`O(n*log(n))`

that a normal sorting algorithmus has.Absolutely, definitely useful stuff to know!

Thanks for sharing!

Tiny little thing - you are hiding

`iteratedNum`

by defining it inside and outside while loop.Yes, you are right, thanks for noticing!

Thanks for the article. This is a good lesson on algorithm thinking and building. However, the wording in the next phrase confused me a bit:

...we return the first list without that number and, if the opposite is true, we return the second list without its smallest number...What it is returning is

the smallest number foundfrom the two lists/arrays (the iteratedNum variable would then hold this number), and at the same time modifying the original array to not have that number anymore.Just for fun

Really faster will be approach to

`concat`

and`sort`

.This works and is quite easy to understand. I just want to add, i prefer a pure function where it does always return what's expected without modifying the inputs.