Today we’re going to take a look at a very common challenge used in job interviews. This challenge is inspired by CodeSignal. Let’s see what it says:

A bracket sequence is called regular if it is possible to insert some numbers and signs into the sequence in such a way that the new sequence will represent a correct arithmetic expression.

For a string consisting of only

`(`

’s and`)`

’s, determine if it is a regular bracket sequence or not.

Example:Given

`()()`

we could insert`(1 + 2) * (2 + 4)`

which is a valid arithmetic expression, but`(()))`

is not as it has an extra closing bracket:`)`

.

As you might notice, the problem doesn’t seem so hard, but it might take a while to fully understand the way to approach it correctly.

First, let’s write a function which will take as an input a `string`

consisting of only round brackets:

```
function regularBrackets(brackets) {}
```

Now… let’s use a loop to iterate through all these brackets:

```
function regularBrackets(brackets) {
for (var i = 0; i < brackets.length; i++) {}
}
```

Let’s also add a variable `equal`

initialized with `0`

which will be incremented every time we find an *opening* bracket and decremented every time we find a *closing* bracket. At the end, if equal will be `0`

we’ll know that we have a *regular brackets sequence*, otherwise we’ll know that we don’t have one. Pretty easy, right? 😇

```
function regularBrackets(brackets) {
var equal = 0;
for (var i = 0; i < brackets.length; i++) {
brackets[i] === '(' ? (equal += 1) : (equal -= 1);
}
return equal === 0;
}
```

As you can see we used a ternary operator, as a shortcut for the if-else statements in the for loop. We check to see if the current bracket from the `brackets`

string is an *opening* bracket… if it is, we increment `equal`

by one, otherwise we decrement `equal`

by one as it means that we have a *closing* bracket.

In the return statement, we check to see if `equal`

is `0`

. If it is, that means that we have a valid *regular brackets sequence* and it returns `true`

, otherwise it returns `false`

.

Now let’s test it:

```
1. regularBrackets('(((())))'); // returns true
2. regularBrackets('()(()'); // returns false
3. regularBrackets(')('); // returns true
```

Well… Everything is good for `1.`

and `2.`

but `3.`

is wrong, isn’t it? Hmm… Why is that? 🤔 As you can see we have first a closing bracket, which is wrong and it should return `false`

.

Well, that’s because we haven’t checked if our `equal`

variable ever goes under `0`

. Because if it does, that means we have a closing bracket before an opening one.

**Be aware** as this is the tricky part of this challenge. This is why the interviewers like to use this challenge often.

To fix this we need a simple condition after our ternary operator that lies in the for loop, which will check if the `equal`

variable ever becomes `-1`

, and if it does, we’re simply going to return `false`

:

```
function regularBrackets(brackets) {
var equal = 0;
for (var i = 0; i < brackets.length; i++) {
brackets[i] === '(' ? (equal += 1) : (equal -= 1);
if (equal === -1) return false;
}
return equal === 0 ? true : false;
}
```

**Perfect**! Now this will work correctly for all the cases! Yey! 😄

## Conclusion

We managed to get around with this problem by simply using a variable, but what about when we have multiple types of brackets like: `(`

, `[`

and `{`

? Can we solve this just by using a variable? Hmm… 🤔

Well… for this we’re going to need to use a `stack`

. 🤓 More about in the next article.

I hope you enjoyed this little challenge, and if you did, I’d appreciate if you share it with the world! 🌐

Happy coding! 😇