js-sorted-set

Sorted Set

A sorted set is a data structure with these guarantees:

• It's a set: it can only contain any given item once.
• It's sorted: you can iterate over all its items in order.

As an illustration, let's build a simple sorted set out of an Array:

Operation	Syntax (simple JavaScript Array)
Create	const set = []
Insert	set.push(value)
Remove	set.splice(set.indexOf(value), 1)
Iterate	set.sort(); set.forEach(doSomething)
Find	set.sort(); const index = set.indexOf(value)
Previous	const previousIndex = index - 1
Next	const nextIndex = index + 1
Test	const isInSet = set.indexOf(value) != -1

... this works, but it's a bit cryptic and some operations--notably iterate-- will be very slow with large sets.

Usage

You can npm install js-sorted-set. Alternatively, just download sorted-set.js from this directory.

To use it on a Website built with Webpack or Rollup:

import SortedSet from 'js-sorted-set'

To use it in Node:

const SortedSet = require('js-sorted-set')

Or, to pollute your global scope, insert this in your HTML:

<script src="path/to/sorted-set.js"></script>

Now that you have the SortedSet class, here's how to use it:

const set = new SortedSet({ comparator: function(a, b) { return b - a }})
set.insert(5)
set.insert(3)
set.insert(2)
set.remove(3)
const yes = set.contains(2)
console.log(set.map(function(x) { return x * 2 })) // returns [ 20, 4 ]

Operations

The SortedSet API:

Operation	Syntax (js-sorted-set)	Notes
Create	const set = new SortedSet()
Insert	set.insert(value)
Remove	set.remove(value)
Clear	set.clear()
Length	set.length
Test	set.contains(value)	Returns true or false
Iterate	set.forEach(doSomething)	Plus set.map() and other iterative methods, returning Arrays and scalars

Find, Previous and Next work with an Iterator pattern. An iterator is an immutible pointer into the space "between" two items in the set.

const iterator = set.beginIterator() // points to the left of the leftmost item
const iterator2 = iterator.next() // points to the left of the second item
const value = iterator.value(), value2 = iterator2.value()
const end = set.endIterator() // points to the right of the final item
const value2 = end.value() // null, because there is no item

Here's the full SortedSet iterator API:

Operation	Syntax (js-sorted-set)	Notes
Length	const len = set.length
Find	const iterator = set.findIterator(value)	iterator points to the left of value. If value is not in set, iterator points to the left of the first item greater than value. If value is greater than the final item in set, iterator points to the right of the final item.
Begin	const iterator = set.beginIterator()	If set is empty, this is equivalent to const iterator = set.endIterator()
End	const iterator = set.endIterator()	Points past the end of set there is never a value here
Value	const value = iterator.value()	For an end iterator, returns null
Forward	const iterator2 = iterator.next()	If iterator is an end iterator, returns null
Backward	const iterator2 = iterator.previous()	If iterator is a begin iterator, returns null
Can go forward	const isBegin = !iterator.hasPrevious()
Can go backward	const isEnd = !iterator.hasNext()	Remember, if iterator is pointing to the left of the final item in set, then hasNext() will return true -- even though iterator.next().value() === null

All iterators on set become invalid as soon as something calls set.insert() or set.remove().

Options

How exactly will these elements be ordered? Let's add a comparator option. This is the argument we would pass to Array.prototype.sort:

const compareNumbers = (a, b) => a - b
const set = new SortedSet({ comparator: compareNumbers })

How to handle insert conflicts? We'll also add a onInsertConflict option that provides users with a way to specify what to do in case an item is inserted that matches another item already present within the set. Such behavior must be specified as a function taking in the conflicting items and returning a replacement item for the previously inserted one. The SortedSet class ships with three implementations such behavior:

SortedSet.OnInsertConflictThrow     // throws an error
SortedSet.OnInsertConflictReplace   // keeps the new item
SortedSet.OnInsertConflictIgnore    // keeps the previous item

Unless differently specified through the onInsertConflict option, the SortedSet class will default to SortedSet.OnInsertConflictThrow:

const set = new SortedSet({ 
    onInsertConflict: SortedSet.OnInsertConflictThrow
})
set.insert("foo")
set.insert("foo") // throw an error

Finally, some algorithms ask for really fast replacement mechanisms. So let's add a setValue() method to the iterator, which puts the onus on the user to keep things ordered.

Because this is a particularly dangerous API to use, you must set the option allowSetValue: true when creating the SortedSet.

const set = new SortedSet({ allowSetValue: true })
set.insert("foo")
set.insert("bar")
set.insert("baz")

// Shortcut API
const iterator = set.findIterator("bar")
iterator.setValue("baq") // It must stay ordered! Do not set "bbq" here!
// The shortcut executes very quickly, but if the user makes a mistake,
// future operations will likely fail

// iterator.setValue("baq") here is equivalent to:
// set.remove("bar")
// set.insert("baq")

Strategies

We can be somewhat efficient in an Array approach by avoiding sort() calls. This strategy keeps the array ordered at all times by inserting and removing elements into and out from the correct array indices. The downside: large swaths of the array must be rewritten during each insert and remove.

We can also create a simple binary tree. insert() and remove() won't overwrite the entire array each time, so this can be faster. But it's far slower to seek through a binary tree, because it can spread out very far across memory so the processor won't cache it well. Also, depending on the order in which elements were input, inserting a single item into the tree can actually be slower than rewriting an entire Array.

Finally, we can improve upon the binary tree by balancing it. This guarantees a certain maximum number of reads and writes per operation. Think of it this way: if you're lucky, a simple binary tree's operations can be extremely fast; if you're unlucky, they can be extremely slow; you'll usually be unlucky. A balanced tree makes all operations somewhat fast.

The balanced tree (which, incidentally, is a Left-Leaning Red-Black tree) is the default, because its speed is the most predictable.

Create the sets like this:

const set = new SortedSet({ strategy: SortedSet.ArrayStrategy }) // Array
const set = new SortedSet({ strategy: SortedSet.BinaryTreeStrategy }) // simple binary tree
const set = new SortedSet({ strategy: SortedSet.RedBlackTreeStrategy }) // default

Use the ArrayStrategy if your set will only have a few values at a time. Use the BinaryTreeStrategy if you've run lots of tests and can prove it's faster than the others. If neither of these conditions applies, use the default, RedBlackTreeStrategy.

You'll see running times like this:

Operation	Array	Binary tree	Red-black tree
Create	O(1)	O(1)	O(1)
Length	O(1)	O(1)	O(1)
Clear	O(1)	O(n) (in garbage collector)	O(n) (in garbage collector)
Insert	O(n) (often slow)	O(n) (often slow)	O(lg n) (fast)
Remove	O(n) (often slow)	O(n) (often slow)	O(lg n) (fast)
Iterate	O(n) (fast)	O(n) (slowest)	O(n) (slower than Array)
Find, Test	O(lg n) (fastest)	O(n) (slowest)	O(lg n) (slower than Array)

According to some simple jsPerf tests, you should use ArrayStrategy if you plan on maintaining about 100 to 1,000 items in your set. At that size, ArrayStrategy's insert() and remove() are fastest in today's browsers; and ArrayStrategy's iteration is faster at all sizes.

Contributing

1. Fork this repository
2. Run npm install
3. Write the behavior you expect in test/
4. Edit files in src/ until npm test says you're done
5. Run npm run build to update build products
6. Submit a pull request

License

I, Adam Hooper, the sole author of this project, waive all my rights to it and release it under the Public Domain. Do with it what you will.

My hope is that a JavaScript implementation of red-black trees somehow makes the world a better place.