# Implementing String repeat() function in JS

As per MDN,

The

`repeat()`

method constructs and returns a new string which contains the specified number of copies of the string on which it was called, concatenated together.

Now one might think that there is a really straightforward to implement this. Yes there is but if asked in an interview and you go with the straightforward way, they will be like :-

How do I know this ?

Because I got **mehhhhd......**

So that's why we are going to see few approaches to solve it. The real optimized approach was not intuitive to me and is still something I am trying to wrap my head around. But I came up with a middle-ground approach that works better than the **meh!!** one.

And once, again we will take a synonym for `repeat`

.
**Google time** !!!!

`replicate`

sounds cool.

Alright let's go implement `String.prototype.replicate`

now :-

### The Meh Approach

```
String.prototype.replicate = function(count) {
let input = this;
let result = "";
for (let index = 0; index < count; index++) {
result += input;
}
return result;
}
```

**Meh explanation :-**
We initialize `result`

to `""`

and start a `for`

loop in which we iterate till `count`

and simply keep appending the `input`

to the `result`

variable. Very straightforward but **meh!!**.

**Meh benchmark :-**

**100 % slower with 108 operations per second compared to 9202566.4 operations per second **. Let me cry in the corner.

### The Little Less Meh Approach

```
String.prototype.replicate = function(count) {
let input = this;
let result = this.valueOf();
for (var index = 2; index < count; index*=2) {
result += result;
}
let remainingCount = count - index/2;
return remainingCount > 0 ? result + input.replicate(remainingCount) : result;
}
```

**Little Less Meh explanation :-**

- Let's consider the case of
`'hey'.replicate(10)`

:-- We have
`input`

initialized to`this`

and`result`

initialized to`this.valueOf()`

. The`valueOf()`

bit helps in decreasing the implicit conversion time that's happening whenever later`result`

will be concatenated to itself. - Now the
`for`

loop stuff -`index`

is intialized to`2`

.`index`

should be less than`count`

`index`

should be multiplied each time by`2`

`result`

will be appended to itself each time in the iteration:-`result`

for`index = 2`

will become`heyhey`

`result`

for`index = 4`

will become`heyheyheyhey`

`result`

for`index = 8`

will become`heyheyheyheyheyheyheyhey`

`index`

will become`16`

which is greater than`10`

and we exit the loop.

`remainingCount`

will be`10`

-`16/2`

=`2`

;- When
`remainingCount`

will be greater than`0`

, we will recurse by calling`input.replicate(remainingCount)`

and add its result to current`result`

or simply return`result`

.

- We have

**Little Less Meh benchmark :-**

**76.79% slower with 2109699.5 operations per second compared to 9091332.85 operations per second **. That's still relatively slower than the native one but way way way faster than what we had initially.

Earlier performing the repetitions itself was **O(count)** but now the same is somewhere down the line of **O(log(x)+log(y) +....+log(k))** but not completely **O(log(count))**.

In `'hey'.replicate(10)`

scenario :-

- First time,
**O(log(8))**work is done and then in next recursive step**O(log(2))**i.e`O(log(8) + log(2))`

. And if I am doing maths correct,

`log(a) + log(b) = log(ab)`

That means `O(log(8) + log(2))`

is `O(log(16))`

which is greater than `O(log(10))`

(the optimal solution).

### The legendary optimal solution I would have never landed upon without the internet

```
String.prototype.replicate = function(count) {
let result = ''
let pattern = this.valueOf();
while (count > 0) {
if (count & 1)
result += pattern;
count >>= 1
if (count) pattern += pattern;
}
return result;
};
```

** Noob explanation :-**
I am still trying to understand the intuition behind this solution but I think it's to do with the fact that every number can be represented in a binary form. So let's say `count`

is **5** then it can be represented as `101`

in binary. So it's possible for us to repeat the string `count`

times by just resorting to **binary calculations**. If we try to differentiate between **4** and **5**, we know there is an extra **1** in latter case. Now instead of seeing the above code as some **binary work of art**, replace **count&1** by **count%2!==0** and **count>>=1** by **count=Math.floor(count/2)**. What this means is that, whenever `count`

is odd, we would want to save the `pattern`

till now in `result`

variable. What is `pattern`

? `pattern`

is repeated concatenation of itself similar to our earlier algorithm so it will always repeat in powers of 2. It's necessary to take care of the situation when `count`

is **not divisible by 2** and store the current `pattern`

in `result`

as we go until `count`

becomes 0.

Did you expect a better explanation ? I can't give it right now since I am a **noob** in binary land. But maybe somewhere in a parallel universe I invented this Algo and helped Brendan Eich get rid of `typeof null`

-> `object`

🤷♂️.

** Best benchmark yet :- **

** Still 29% slower ? WTH. But hey, I ain't competing with JavaScript engines here **.

### The Bonus MDN polyfill

```
String.prototype.replicate = function(count) {
var str = '' + this;
count = +count;
count = Math.floor(count);
if (str.length == 0 || count == 0)
return '';
var maxCount = str.length * count;
count = Math.floor(Math.log(count) / Math.log(2));
while (count) {
str += str;
count--;
}
str += str.substring(0, maxCount - str.length);
return str;
}
```

** Expected an explanation ? I don't care and you will see why 👇 **

** The mandatory benchmark :- **

** 99.94 % slower with 5211.6 operations per second compared to 8344361.29 operations per second **. And there is definite reason why it is even slower than what I came up with. What I think is happening is that upto a **power of 2** which is less than `count`

, we use the same ideology as in the optimal solution for concatenating and doubling length of `str`

every time. But after that for the remaining length, it uses `substring`

and appends that to `str`

again. It's the second step of `substring`

which makes it a costly operation. Though it does better than the initial **Meh** solution of **108** ops/s, it's still no where near around the best optimal solution I found online or even mine 😎.

**MDN : 0**
**Lakshya : 1**

JK. The site is and hopefully remains a gold mine ❤️.

Here are the overall benchmarks :-

Have something to add on ? Feel free to

### Thank you for your time :D

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