Hello everyone! Welcome to the first in a series! I'm going to try to explain a common software engineering interview question to better understand it and hopefully remember it when the time comes!

These problems will primarily be solved in JavaScript, as that is my language of choice when testing (and I just want to become a good front-end dev. 🥺)

##
**Q: Find the second largest number in a given array.**

**Params**: We are given an array of whole, positive integers (no negative numbers or floats). We are to write a function and return the second largest integer.

##
**Let's start!**

We'll write the skeleton of our function, setting the input/argument as **"arr"** for *array*:

```
function secondLargest(arr){};
```

Then, we'll need to set two empty variables: *largest* and *second*.

**Why?** We will need placeholders for both our prospective **largest** and **second** largest numbers as we loop through our array.

We want to keep track of each integer that is in the array and measure the value against the others

```
function secondLargest(arr){
let largest = '';
let second = '';
}
```

...Which brings us to our next step: **create a for-loop!**

As we iterate through the array, we will measure each value against each other, comparing the variable "largest" to the current iteration value (arr[i]).

```
function secondLargest(arr){
let largest = '';
let second = '';
//
for(let i=0; i < arr.length; i++){};
//
};
```

To compare, we will create an **if-statement** comparing **largest** to **arr[i]**.

If

arr[i]is greater thanlargest, thenreplace itwith the current value ofarr[i]by redeclaringlargestand setting it equal toarr[i]

```
function secondLargest(arr){
let largest = '';
let second = '';
for(let i=0; i < arr.length; i++){
//
if(arr[i] > largest){
largest = arr[i]
};
//
};
};
```

We found the largest number! But how do we get the **second** largest?

We did find it already (kind of): we'll just set the former "largest" number to the "second" variable.

HOWEVER, we must declare the **second** variable BEFORE we declare the new **largest** number, simply because *order matters*-- JavaScript executes code from the top-down.

```
function secondLargest(arr){
let largest = '';
let second = '';
for(let i=0; i < arr.length; i++){
if(arr[i] > largest){
//
second = largest;
//
largest = arr[i];
};
};
};
```

Speaking of order and specificity, it's time we find the "true" second-largest number in the array.

Let's create another if-statement with more specific parameters:

If the

arr[i]is GREATER THANsecondANDLESS THANlargest, then setarr[i]assecond

```
function secondLargest(arr){
let largest = '';
let second = '';
for(let i=0; i < arr.length; i++){
if(arr[i] > largest){
second = largest;
largest = arr[i];
};
//
if(arr[i] > second && arr[i]< largest){
second = arr[i];
};
//
};
};
```

Finally, we'll return our *second* variable to complete the requirement.

```
function secondLargest(arr){
let largest = '';
let second = '';
for(let i=0; i < arr.length; i++){
if(arr[i] > largest){
second = largest;
largest = arr[i];
};
if(arr[i] > second && arr[i]< largest){
second = arr[i];
};
};
//
return second;
//
};
```

And there you have it! It's a fairly simple solution, a bit long since we're using a traditional for-loop, but it works!

Feel free to post your own solutions in your programming language of choice in the comments!

### Thanks for reading!

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## Top comments (16)

Sure good point:

Just for the sake of brevity, but not improved performance:

Daaamn that’s slick. 👌🏾👌🏾👌🏾

Yeah but I bet yours is faster :)

The initializers should probably be undefined since the function will return the wrong result if the array consists of fewer than 2 items or if all items in the array consist of the same value.

Gotta watch out for those edge cases!

My solution using using the Ramda library

Hi Cat!

Amazing solution!

There are several testcases that would not not pass.

in case [0,1] -> 0 is the second largest, but function returns ' '.

in case [1, 0] -> 0 again is the second largest, but the function will return 1.

in case [1,1,1,0] -> same as in the previous testcase.

This happens because of comparison of 0 with an empty string.

' ' === 0 // false

' ' > 0 // false

' ' < 0 // false

The small change that I would do:

You know that arr[0] element exists for sure if array is not empty.

Sorting will allow you to start with the smallest element whatever it is and find the second largest.

Small note for other solutions:

You cannot just sort an array and take a second element from the end ;)

in case of [0,1,2,2,2,2] code will return 2. You also need to make numbers unique.

This challenge also tests how creative you can be in your testcases and the ability to think what can potentially break your code.

How about checking array is valid (as an numeric array which length > 0) via:

if (!Array.isArray(array) || !array.length || array.some(isNaN)) {

return ("Not a valid array")

}

Regarding the 0 comparison, how about:

let max = -Infinity, second = -Infinity

Ooooh got it. Darn it, HackerRank, for giving me a false-positive. ;____;

I'll refactor the code above and credit you! Thanks Lia! You da best.

As a learning lesson, this example could be improved:

First, there's a bug in that it doesn't handle cases like

`[2, 5, 10, 9, 10, 7]`

where the numbers aren't unique. (I double-checked, uniqueness was not part of your Params description above.) In that case, the code above will return 9, even though it's clearly not the second-largest.Second, if you fix that by using

`>=`

in the first test, the second one isn't even needed. The storing of`second`

in the first test takes care of the second largest, even if there's only one element in the array.So the

`for`

loop could be simplified to just:Why not return arr.sortDescending()[1] ?

Subscribed!

if(arr.length > 1)

return arr.sort((a,b)=> {return b-a})[1]

Good point! Let me try that out and I'll fix it up. :) Thanks Sebastian!

Well the problem was toooo simple.

You may want to make it recursive for multi-dimensional arrays