*Binary Search* is one of the fastest searching algorithms, especially when it comes to searching large (sorted) lists.

The main goal of Binary Search is to narrow down the area you're searching in as much as possible, which leads to reducing the steps you'd take to find an item.

When implementing Binary Search you should:

1- Assume that you're working on sorted lists -- otherwise, the search won't work.

2- Specify start and end points for where you should start and end searching.

3- Pick an item from the middle of the list and compare it to the item you're searching for. Based on that comparison, you should know if the item is found, or if you need to modify your start and end points and repeat the steps.

Let's see an example.

```
function binarySearch(list, itemToFind){
// some code to return the index of itemToFind
}
let list = [10, 21, 25, 30, 32, 35, 50, 52, 55, 60];
let itemToFind = 32;
binarySearch(list, itemToFind) // should return the index of 32.
```

To implement the code in `binarySearch`

, we first need to set our start and end points. Since we need to cover the whole list, we need to specify our initial start point to be the first index of the list and end point to be the last index of the list.

```
let start = 0;
let end = list.length -1; // 9
```

Next, we need to set a middle point index and then compare its value to the item we're searching for.

```
let middle = Math.floor((start + end)/2); // 4
if (list[middle] === itemToFind) return middle;
```

Since we are searching for an item that happens to be in the middle of the list, those few lines of code will return the index of `itemToFind`

on the spot. This is called the `best-case scenario`

of an algorithm -- your first guess is the correct answer.

But of course, that rarely happens, so we need to cover the cases where we don't find our item in the middle of the list.

Let's start a new search and search for 30 this time.

Hmm, we calculated the middle point exactly as before, but unfortunately, we didn't find 30 there.

Now, we know that the middle item does not equal `itemToFind`

. But is it greater or less than `itemToFind`

?

We found 32, which is greater than 30. So what does that mean?

Since `list`

is sorted, that means that `itemToFind`

must be somewhere between `start`

and `middle`

.

**Next step**: relocate the `end`

point of the search to narrow the search window.

```
if(middle > itemToFind){
end = middle -1;
}
```

Then recalculate `middle`

and check the new middle value.

```
if (list[middle] === itemToFind) return middle;
if(middle > itemToFind) end = middle -1; // 3
middle = Math.floor((start + end)/2); // 1
```

The middle item is now `21`

. It doesn't equal 30 so we can't return its index. It's not greater than 30, so relocating `end`

to narrow the search area is not an option. However, we can relocate `start`

. Because at this point, if the item exists, it must be somewhere between `middle`

and `end`

.

```
if(list[middle] < itemToFind){
start = middle + 1;
}
```

Then recalculate `middle`

and check the new middle value.

```
if(list[middle] === itemToFind) return middle;
if(list[middle] > itemToFind) end = middle -1; // 3
if(list[middle] < itemToFind) start = middle + 1; // 2
middle = Math.floor((start + end)/2); // 2
```

We find 25. It's still less than 30. So we relocate `start`

, calculate `middle`

, and check again.

Finally, `middle`

is pointing to the item we're searching for. However, that has happened after we have exhausted all our searching options, where our search window `start`

s where it `end`

s. This means that we've found our item at the `worst-case scenario`

of the algorithm -- your last chance of guessing is the correct answer.

*Note*: The worst-case scenario also happens if `itemToFind`

doesn't exist in `list`

.

One last thing I should mention about Binary Search is that it has `O(log n)`

time complexity, which means it takes `log n`

time to find an item in the worst-case scenario.

```
// final implemtation
function binarySearch(list, itemToFind) {
let start = 0;
let end = list.length - 1;
while (start <= end) {
let middle = Math.floor((start + end) / 2);
if (list[middle] === itemToFind) return middle;
if (list[middle] > itemToFind) {
end = middle - 1;
} else {
start = middle + 1;
}
}
return -1; // not found
}
```

(thank you for reading)

## Top comments (12)

Nice explanation and congratulations on presenting an actually correct variant of binary search. 🚀

Since I believe that anything can be improved, no matter how nice it is, here is some feedback:

Issues

`list`

and`itemToFind`

should very much be parameters that are passed to the function`parseInt`

. It's really not needed and it just causes a performance overhead.`(start + end) / 2`

isn't ideal either, since with large enough arrays, this can cause some overflowing/rounding. The improved variant is`start + ((end - start ) / 2)`

Might be interesting to note`return -1`

is`return -start`

. Then, the negative return value would point at the smallest element that's >= itemToFind. And that's pretty useful when one wants to insert a new element.Thank you so much for your feedback @stefnotch . 😀

I fixed the no-parameters issue but am curious, why do you think that

`parseInt`

isn't needed here? Aren't we using it to avoid having fractions in the middle index? 🤔I see, I missed that detail!

Although,

`parseInt`

is not designed for that. It's purpose is to turn a string (text) into a number. If the intention is to chop off the fractional part, then`Math.floor`

is the standard function to use.e.g.

`start + Math.floor((end - start) / 2)`

I like your idea of returning -start.

Great explanation of this kind of search! Nice work, Hajar!

Thank you so much for reading, Andrew. I appreciate it. 🌹

thanks :)

Thank for taking the time to right a legit article instead of all the fluff that so many others produce.

Thank you, Peter. You're so kind. 😊

perfect explanation

Thank you for reading. :)

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