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 | var set = []; |
Insert | set.push(value); |
Remove | set.splice(set.indexOf(value), 1); |
Iterate | set.sort(); set.forEach(doSomething); |
Find | set.sort(); var index = set.indexOf(value); |
Previous | var previousIndex = index - 1; |
Next | var nextIndex = index + 1; |
Test | var 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.
Download sorted-set.js
. Alternatively, install through Bower:
bower install js-sorted-set
Include it through RequireJS.
Then write code like this:
require([ 'vendor/sorted-set' ], function(SortedSet) {
var set = new SortedSet({ comparator: function(a, b) { return b - a; });
set.insert(5);
set.insert(3);
set.insert(2);
set.remove(3);
var yes = set.contains(2);
console.log(set.map(function(x) { return x * 2; })); // returns [ 20, 4 ]
});
If you don't like RequireJS, you can download the standalone version,
sorted-set.no-require.js
, and write:
var set = new SortedSet({ comparator: function(a, b) { return b - a; });
set.insert(5);
set.insert(3);
set.insert(2);
set.remove(3);
var yes = set.contains(2);
console.log(set.map(function(x) { return x * 2; })); // returns [ 20, 4 ]
The SortedSet API:
Operation | Syntax (js-sorted-set) | Notes |
---|---|---|
Create | var set = new SortedSet(); |
|
Insert | set.insert(value); |
|
Remove | set.remove(value); |
|
Test | set.contains(value); |
Returns true or false |
Iterate | set.forEach(doSomething); |
Plus set.map() and other iterative methods, returning Array s 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.
var iterator = set.beginIterator(); // points to the left of the leftmost item
var iterator2 = iterator.next(); // points to the left of the second item
var value = iterator.value(), value2 = iterator2.value();
var end = set.endIterator(); // points to the right of the final item
var value2 = end.value(); // null, because there is no item
Here's the full SortedSet iterator API:
Operation | Syntax (js-sorted-set) | Notes |
---|---|---|
Find | var 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 | var iterator = set.beginIterator(); |
If set is empty, this is equivalent to var iterator = set.endIterator(); |
End | var iterator = set.endIterator(); |
Points past the end of set ; there is never a value here |
Value | var value = iterator.value(); |
For an end iterator, returns null |
Forward | var iterator2 = iterator.next(); |
If iterator is an end iterator, returns null |
Backward | var iterator2 = iterator.previous(); |
If iterator is a begin iterator, returns null |
Can go forward | var isBegin = !iterator.hasPrevious(); |
|
Can go backward | var 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()
.
How exactly will these elements be ordered? Let's add a comparator
option.
This is the argument we would pass to
Array.prototype.sort:
var compareNumbers = function(a, b) { return a - b; };
var set = new SortedSet({ comparator: compareNumbers });
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.
var set = new SortedSet({ allowSetValue: true });
set.insert("foo");
set.insert("bar");
set.insert("baz");
// Shortcut API
var 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");
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:
var set = new SortedSet({ strategy: SortedSet.ArrayStrategy }); // Array
var set = new SortedSet({ strategy: SortedSet.BinaryTreeStrategy }); // simple binary tree
var 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) |
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.
- Fork this repository
- Run
npm install
- Write the behavior you expect in
spec-coffee/
- Edit files in
coffee/
untilgrunt test
says you're done - Run
grunt
to updatesorted-set.js
andsorted-set.min.js
- Submit a pull request
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.