### Linear and Binary Search in JavaScript

#### Steph on May 16, 2018

This week I started reading Grokking's Algorithms, an illustrated guide for programmers and other curious people. So far it's a fantastic read --... [Read Full]

Hi, Stephanie...

I think, you may interested with this one. One year ago, when I want to understand C, I decided to do a very naive experiment to compare some classical sort algorithms. If you interested, you can read it at here.

It is very naive by the way, but if you have better idea how do similar experiments, that would be great.

Steph Author

Hey Hafidz,

Thank you...

It just bonus assignment since a lecturer who teach algorithm course found that few of her student bored with the course (I think).

A mouth ago, I just realize that this empirical approach very useful in parallel programming. That's Gustavson Law that suggest us to use latency as benchmark.

love this steph!

great post !

at our company(startup) we deal with large data sets that are only getting larger and implementing this on our backend (rails api) will really speed up our methods!

thanks so much

FYI ruby has already a built in binary search.

Nice implementation, and thanks for the book recommendation, I just have one question, how would you implement that with objects?

Steph Author

Do you mean how would you pass in an array of objects? I modified the function a bit here to allow you to input a sorted array of objects, the key you are targeting, and the value you are searching for. (This assumes that the array is already sorted by the specific key you're looking at).

``````function binarySearch(sortedArrayofObjects, keyToTarget, valueToFind) {
var lowIndex = 0;
var highIndex = sortedArrayofObjects.length - 1;
while (lowIndex <= highIndex) {
var midIndex = Math.floor((lowIndex + highIndex) / 2);
if (sortedArrayofObjects[midIndex][keyToTarget] == valueToFind) {
return midIndex;
} else if (sortedArrayofObjects[midIndex][keyToTarget] < valueToFind) {
lowIndex = midIndex + 1;
} else {
highIndex = midIndex - 1;
}
} return null;
}

var sortedORainbowObj = [
{
id: 5,
color: "blue"
},
{
id: 4,
color: "green"
},
{
id: 6,
color: "indigo"
},
{
id: 2,
color: "orange"
},
{
id: 1,
color: "red"
},
{
id: 7,
color: "violet"
},
{
id: 3,
color: "yellow"
}
];

console.log(binarySearch(sortedORainbowObj, "color", "yellow")); // returns 6
console.log(binarySearch(sortedORainbowObj, "color", "maroon")); // returns null

``````

Nice function, I'm looking forward to your next post.

The clean way to extend function to generic types is to override a compare function I'd say

I fully agree, nonetheless that would require using typescript, I had thought about making use of the `valueOf` function, however I like the way @stepho did it since it is easier to reuse, even more for js newcomers.

Linear searches are slower than binary searches but they aren't inherently bad and binary searches aren't always better.

If you have a use case where you often need to add elements to your array, keeping your array sorted can be very expensive. Sorting algorithms are usually O(n*log n) complexity and that may be slower than performing a simple linear search on your array.

So depending on your use case linear search can be perfectly fine.

Steph Author

That's true, it depends on your data set. Assuming it's sorted and you're only worried about the run time of the search, a linear search is O(n) while a binary search is O(log n).

Nice write up Stephanie, clear, concise, and a great step by step guide.

Steph Author

Thanks, Kyle!

This is amazing and easy to understand! Thank you so much for this!

Excuse me for a dumb question, I am very very new to JS but does it mean that comparison is performed not against the number of letters in a word but for the word content, so word “aaa” would be “smaller” than “aab”?

I’m referring to this:

——-
if the current element is less than (alphabetically before) the element you're searching for, increase the lowIndex to one more than the midIndex

Steph Author

Hey Serge, no worries at all! Yes, "aaa" would be before "aab" in an alphabetical sort. The length of the word doesn't necessarily matter unless you are comparing something like "aaa" vs "aa", where "aa" would come before "aaa". A real life example of an alphabetical sort would be words in a dictionary or names in a phonebook. Feel free to copy and modify the code below and play with it in something like jsbin.com (use just the javascript and console tools).

``````var arr = ["aaa", "aza", "aab", "aa", "ab"];
arr.sort(); //returns ["aa", "aaa", "aab", "ab", "aza"]
``````

Neat and lucid explanation. Thanks for taking the time to do that. I read grokking algorithms and loved it as well. But it's always nice to get a refresher.

Nice post, thank you.

In case anyone is interested here is the code of this book made by its author in several languages

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